CBSE Class 10 Mathematics Arithmetic Progression Worksheet Set 04

ARITHMETIC PROGRESSIONS

Very Short Answer type Questions

Question. Find the sum of the first 22 terms of the AP : 8, 3, –2, . . .
Answer : – 979

Question. If the 3^rd and the 9^th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?
Answer : 5^th term

Question. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Answer : n²

Question. How many terms of the AP : 24, 21, 18, . . . must be taken so that their sum is 78?
Answer : 4 or 13

Question. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
Answer : 234

Question. For what value of n, are the n^th terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?
Answer : 13

Short Answer type Questions

Question. lf 2x, x + 10, 3x + 2 are in A.P., find the value of x.
Answer : lf 2x, x + 10, 3x + 2 are in A.P.,we have to find the value of x.
Since, 2x, x + 10, 3x + 2 are in A.P.therefore 2 (x + 10) = 2x + 3x + 2
2x + 20 = 5x + 2
3x = 18 x = 6

Question. Find 7th term from the end of the AP : 7, 10 , 13,....,184. 
Answer : Given, AP is 7, 10 , 13,....,184.
we have to find 7th term from the end reversing the AP , 184,.....,13,10,7.
now, = common difference = 7-10 = -3
7th term from the beginning of AP =a+(7-1)d=a+6d
=184+(6 ×(-3))
= 184 - 18 = 166

Question. Some children planned for a donation to an orphanage. The total amount to be donated was Rs. 2400. Because four children failed to pay donation due to some reason, the amount of donation to be paid by each child get increased by Rs. 50. How many children donated the fund? Which values do these children possess?
Answer: Let the total number of children originally planned be \( x \).
The share of each child originally was \( \frac{2400}{x} \) rupees.
Since four children failed to pay, the number of children who donated becomes \( x - 4 \).
The new share of each child is \( \frac{2400}{x-4} \) rupees.
According to the given condition:
\( \frac{2400}{x-4} - \frac{2400}{x} = 50 \)
Dividing both sides by 50:
\( \frac{48}{x-4} - \frac{48}{x} = 1 \)
\( 48 \left( \frac{x - (x - 4)}{x(x-4)} \right) = 1 \)
\( 48 \left( \frac{4}{x^2 - 4x} \right) = 1 \)
\( 192 = x^2 - 4x \)
\( x^2 - 4x - 192 = 0 \)
\( (x - 16)(x + 12) = 0 \)
Since the number of children cannot be negative, we take \( x = 16 \).
Therefore, the number of children who actually donated is \( 16 - 4 = 12 \) children.
The values shown by these children are charity, empathy, and social responsibility.
In simple words: Originally 16 children planned to donate, but only 12 children actually paid. They showed kindness and a willingness to help people in need.

Exam Tip: Always state the quadratic variable's real-world constraints clearly - for instance, the number of children must be a positive integer, so discard any negative solutions.

Question 1. S17 − S16 = ?
(a) S1
(b) T1
(c) T17
(d) 2d
Answer: (c) T17
In simple words: If you subtract the sum of the first 16 terms from the sum of the first 17 terms of a sequence, you are left with only the 17th term.

Exam Tip: Remember the standard identity \( S_n - S_{n-1} = T_n \), which relates the sum of terms to individual terms in any sequence.

Question 2. T10 of an AP = 31 & T20 = 71 then T30 = ?
(a) 171
(b) 222
(c) 112
(d) 111
Answer: (d) 111
In simple words: By finding the starting number and how much it increases each time, we can calculate that the 30th number in this pattern is 111.

Exam Tip: Set up simultaneous linear equations for \( a \) and \( d \) using \( T_n = a + (n-1)d \), then solve them to find any other term.

Question 3. No. of terms of the sequence −12, −9, −6, −3, …….. must be added to make the sum 54 are _____
(a) 33
(b) −46
(c) 10
(d) 12
Answer: (d) 12
In simple words: We need to add up the first 12 numbers in this sequence to get a total sum of 54.

Exam Tip: When solving the quadratic equation for the number of terms \( n \), discard any fractional or negative values since \( n \) must be a positive integer.

Question 4. If ‘n’ times the ‘n’th term an AP is equal to m times its ‘m’th term, then (m + n)th term is _____
(a) 0
(b) 1
(c) −1
(d) 2
Answer: (a) 0
In simple words: For any arithmetic progression, if this specific relationship holds, the sum-positioned term is always zero.

Exam Tip: Expand the given relation \( n[a + (n-1)d] = m[a + (m-1)d] \) and group terms to factor out \( (n-m) \). Since \( m \neq n \), you can safely divide by it.

Question 5. The sum of three numbers in an AP is (−3), and their product is 8. Then numbers are
(a) −4, −1, 2
(b) −4, −1, −2
(c) 4, 1, 2
(d) 2, 1, −4
Answer: (a) −4, −1, 2
In simple words: These three numbers increase by 3 each time. Adding them gives -3, and multiplying them together gives 8.

Exam Tip: When dealing with three numbers in AP, represent them as \( a-d \), \( a \), and \( a+d \) to simplify the summation calculation.

Question 6. The sum of four numbers is 20 and sum of whose squares is 120. This nos. are _____
(a) 2, 4, −6, −8
(b) 2, 4, 6, 8
(c) 8, 6, 2, −1
(d) 8, 6, −2, 1
Answer: (b) 2, 4, 6, 8
In simple words: The numbers 2, 4, 6, and 8 add up to 20, and their squared values add up to 120.

Exam Tip: For multiple choice questions, testing the options directly is often much faster than solving the algebraic equations.

Question 7. The nth term of a pattern of numbers is 2n + 1. The common difference of this AP is ________.
(a) 2
(b) −2
(c) 3
(d) −3
Answer: (a) 2
In simple words: The numbers in this pattern increase by 2 each time.

Exam Tip: The coefficient of \( n \) in any linear nth-term expression \( T_n = An + B \) is always the common difference \( d \).

Question 8. Which term of the AP 3, 8, 13, 18………… is 78 ?
(a) t15
(b) t16
(c) t17
(d) t18
Answer: (b) t16
In simple words: If you keep adding 5 starting from 3, the 16th number you write down will be 78.

Exam Tip: Use the formula \( T_n = a + (n-1)d \), substitute the known values, and solve directly for \( n \).

Question 9. Which term of the sequence 114, 109, 104, ………….. is the first negative term ?
(a) t23
(b) t24
(c) t25
(d) None of these
Answer: (b) t24
In simple words: This pattern goes down by 5 each time. The 24th term is the first one that drops below zero.

Exam Tip: Set up an inequality \( a + (n-1)d < 0 \) and solve for the smallest integer value of \( n \).

Question 10. The sum of first 18 terms of an AP whose nth term is 3 − 2n is
(a) −288
(b) 250
(c) −278
(d) −260
Answer: (a) −288
In simple words: When we add up the first 18 numbers of this decreasing pattern, the total sum is -288.

Exam Tip: Find the first term \( T_1 \) and the 18th term \( T_{18} \) first, then use the sum formula \( S_n = \frac{n}{2}(a + l) \) for a faster calculation.

Question 11. If 2x, x + 10, 3x + 2 are in AP, then the value of x is
(a) 6
(b) 5
(c) 4
(d) 3
Answer: (a) 6
In simple words: For these three expressions to form an arithmetic sequence, the variable \( x \) must equal 6.

Exam Tip: If three terms \( A, B, C \) are in AP, then \( 2B = A + C \). This relation is extremely useful for finding unknown values.

Question 12. The 9th term of an AP is 499 and 499th term is 9. The term which is equal to zero is
(a) t508
(b) t805
(c) t504
(d) t501
Answer: (a) t508
In simple words: This pattern decreases gradually, and the 508th number in the sequence will be exactly 0.

Exam Tip: If \( T_p = q \) and \( T_q = p \), then the \( (p+q) \)-th term is always equal to 0.

Question 13. If in an AP a = 1, tn = 20 and Sn = 399 then n is
(a) 19
(b) 21
(c) 38
(d) 42
Answer: (c) 38
In simple words: There are 38 terms in this progression that add up to a total sum of 399.

Exam Tip: Use the formula \( S_n = \frac{n}{2}(a + t_n) \) directly when the first and last terms are known.

Question 14. Two APs have the same common difference. The first term of one of these −1 and that of the other is −8. Then the difference between their 4th term is
(a) −1
(b) −8
(c) 7
(d) −9
Answer: (c) 7
In simple words: Since both patterns grow at the exact same rate, the difference between any corresponding terms will always be the same as the difference between their first terms, which is 7.

Exam Tip: If two APs have the same common difference, the difference between any of their corresponding terms is constant and equals the difference between their first terms.

Question 15. Sum of n terms of the series \( \sqrt{2} + \sqrt{8} + \sqrt{18} + \sqrt{32} + \dots \) is
(a) \( \frac{n(n+1)}{2} \)
(b) \( \frac{2n(n+1)}{3} \)
(c) \( \frac{n(n+1)}{\sqrt{2}} \)
(d) None of these
Answer: (c) \( \frac{n(n+1)}{\sqrt{2}} \)
In simple words: We can simplify the terms to \( \sqrt{2}, 2\sqrt{2}, 3\sqrt{2} \), and so on. Factoring out \( \sqrt{2} \) gives the sum of the first \( n \) numbers.

Exam Tip: Always simplify radical expressions first to identify the common term or progression properties.

Question 16. The sum of first 50 odd natural numbers is
(a) 2500
(b) 2400
(c) 2600
(d) 2300
Answer: (a) 2500
In simple words: Adding up the first 50 odd numbers (like 1, 3, 5, etc.) gives a total sum of 2500.

Exam Tip: The sum of the first \( n \) odd natural numbers is simply \( n^2 \). Memorizing this formula saves precious time during exams.

Question 17. If \( \frac{a^{n+1} + b^{n+1}}{a^n + b^n} \) is the AM between a and b, then the value of n is
(a) 1
(b) −1
(c) 0
(d) None of these
Answer: (c) 0
In simple words: When \( n = 0 \), this fraction simplifies directly to the average of \( a \) and \( b \), which is their arithmetic mean.

Exam Tip: Instead of cross-multiplying and solving, substitute the given options into the expression to find which value of \( n \) yields \( \frac{a+b}{2} \).

Question 18. If an AP a = −2.5, d = 0.5, an = 7.5 then n is
(a) 20
(b) 21
(c) 3.6
(d) −3.6
Answer: (b) 21
In simple words: If you start at -2.5 and add 0.5 repeatedly, you will reach 7.5 on the 21st term.

Exam Tip: Be careful with sign operations when working with negative first terms in the general term formula.

Question 19. The 16th term of the AP : \( 15, \frac{25}{2}, 10, \frac{15}{2}, 5, \dots \) is
(a) \( \frac{45}{2} \)
(b) \( -\frac{45}{2} \)
(c) \( \frac{105}{2} \)
(d) \( -\frac{105}{2} \)
Answer: (b) \( -\frac{45}{2} \)
In simple words: This pattern decreases by 2.5 each time. The 16th number in this sequence is \( -22.5 \), which is the fraction \( -\frac{45}{2} \).

Exam Tip: Keep the common difference in fractional form (e.g., \( -\frac{5}{2} \)) to make the final algebraic calculations cleaner and easier to simplify.

Question 20. The common difference of an AP in which a25 − a12 = −52 is
(a) 4
(b) −4
(c) −3
(d) 3
Answer: (b) −4
In simple words: Since there are 13 steps between the 12th term and the 25th term, dividing the total change of -52 by 13 tells us that the numbers decrease by 4 at each step.

Exam Tip: Remember that \( a_p - a_q = (p-q)d \). Using this direct relationship lets you bypass finding the first term \( a \) entirely.

Question 21. Two AP’s have the same common difference. The first term of one of these is −1 and that of the other is −8. Then the difference between their 4th term is
(a) −1
(b) −8
(c) 7
(d) −9
Answer: (c) 7
In simple words: Since both patterns change at the same rate, the distance between corresponding terms always stays equal to the distance between their first terms, which is 7.

Exam Tip: The difference between any corresponding terms of two APs with equal common difference will always be \( a_1 - a_2 \).

Question 22. If \( \frac{6}{5}, a, 4 \) are in AP, the value of a is
(a) 1
(b) 13
(c) \( \frac{13}{5} \)
(d) \( \frac{26}{5} \)
Answer: (c) \( \frac{13}{5} \)
In simple words: The middle number \( a \) must be the exact average of the two outer numbers, which is \( \frac{13}{5} \).

Exam Tip: Calculate the arithmetic mean of the outer terms to find the value of the middle term in any three-term AP sequence.

SA-1

Question 1. If 18, a, b, −3 are in AP then find a + b.
Answer: Since \( 18, a, b, -3 \) are in an Arithmetic Progression, the common difference \( d \) remains constant throughout.
The third term is \( b \) and the first term is \( 18 \). The fourth term is \( -3 \).
We can find the common difference using the first and last terms:
\( T_4 = a_1 + 3d \)
\( -3 = 18 + 3d \)
\( 3d = -21 \implies d = -7 \)
Now, we can find \( a \) and \( b \):
\( a = 18 + d = 18 - 7 = 11 \)
\( b = a + d = 11 - 7 = 4 \)
Therefore, the sum is:
\( a + b = 11 + 4 = 15 \)
In simple words: The sequence decreases by 7 each time, making the terms 18, 11, 4, and -3. Adding the two middle terms gives 15.

Exam Tip: In any finite AP, the sum of terms equidistant from the beginning and end is constant, so \( a_1 + a_n = a_2 + a_{n-1} \), which gives \( 18 + (-3) = a + b = 15 \) instantly.

Question 2. How many integers which are multiples of 4 lie between 10 and 250 ?
Answer: The first multiple of 4 greater than 10 is 12.
The last multiple of 4 less than 250 is 248.
This forms an AP: \( 12, 16, 20, \dots, 248 \)
Here, first term \( a = 12 \), common difference \( d = 4 \), and the \( n \)-th term \( a_n = 248 \).
Using the general term formula:
\( a_n = a + (n-1)d \)
\( 248 = 12 + (n-1)4 \)
\( 236 = (n-1)4 \)
\( n-1 = 59 \)
\( n = 60 \)
There are 60 integers.
In simple words: The first multiple is 12 and the last is 248. By counting them step-by-step by 4s, we find there are 60 such numbers.

Exam Tip: Be careful with the word "between", which means the boundary numbers are not included (though in this case 10 and 250 are not multiples of 4 anyway).

Question 3. The sum of three numbers is 27 and their product is 405. Find the numbers.
Answer: Let the three numbers in AP be \( a-d \), \( a \), and \( a+d \).
According to the first condition, their sum is:
\( (a-d) + a + (a+d) = 27 \)
\( 3a = 27 \implies a = 9 \)
According to the second condition, their product is:
\( (a-d) \cdot a \cdot (a+d) = 405 \)
\( a(a^2 - d^2) = 405 \)
Substitute \( a = 9 \) into the equation:
\( 9(81 - d^2) = 405 \)
\( 81 - d^2 = 45 \)
\( d^2 = 36 \implies d = \pm 6 \)
If \( d = 6 \), the numbers are \( 9-6, 9, 9+6 \), which are \( 3, 9, 15 \).
If \( d = -6 \), the numbers are \( 9 - (-6), 9, 9 + (-6) \), which are \( 15, 9, 3 \).
The required numbers are 3, 9, 15.
In simple words: The three numbers are 3, 9, and 15. They add up to 27 and multiply to 405.

Exam Tip: Representing three numbers in AP as \( a-d \), \( a \), and \( a+d \) makes it incredibly easy to find the middle term immediately when the sum is given.

Question 4. Justify that 1 + n + n2 is the nth term of an AP or not.
Answer: Let the \( n \)-th term of the sequence be \( T_n = 1 + n + n^2 \).
Let's find the first few terms of the sequence by substituting values of \( n \):
For \( n = 1 \): \( T_1 = 1 + 1 + 1^2 = 3 \)
For \( n = 2 \): \( T_2 = 1 + 2 + 2^2 = 7 \)
For \( n = 3 \): \( T_3 = 1 + 3 + 3^2 = 13 \)
Let us calculate the differences between consecutive terms:
\( T_2 - T_1 = 7 - 3 = 4 \)
\( T_3 - T_2 = 13 - 7 = 6 \)
Since the difference between consecutive terms is not constant (\( 4 \neq 6 \)), the given expression is not the \( n \)-th term of an AP.
In simple words: The first three numbers in this pattern are 3, 7, and 13. Since the difference increases from 4 to 6, this is not an arithmetic sequence.

Exam Tip: Remember that the \( n \)-th term of any Arithmetic Progression is always a linear polynomial in \( n \) (i.e., of the form \( An + B \)). If there is an \( n^2 \) term, it cannot be an AP.

Question 5. a100 is the 100th term of the AP 3, 3.11, 3.22, 3.33, ………. and b100 is the 100th term of the AP P, P +.11, p+.22, p+.33, ……….. a100 − b100 = .22345. So find P.
Answer: For the first AP: \( 3, 3.11, 3.22, \dots \)
First term \( a_1 = 3 \), common difference \( d = 0.11 \).
The 100th term is:
\( a_{100} = a_1 + 99d = 3 + 99(0.11) \)
For the second AP: \( P, P + 0.11, P + 0.22, \dots \)
First term \( b_1 = P \), common difference \( d = 0.11 \).
The 100th term is:
\( b_{100} = b_1 + 99d = P + 99(0.11) \)
According to the question:
\( a_{100} - b_{100} = 0.22345 \)
\( [3 + 99(0.11)] - [P + 99(0.11)] = 0.22345 \)
\( 3 - P = 0.22345 \)
\( P = 3 - 0.22345 = 2.77655 \)
Therefore, the value of \( P \) is 2.77655.
In simple words: Because both sequences increase at the exact same speed, the gap between their 100th terms is the same as the gap between their first terms. Solving this gives \( P = 2.77655 \).

Exam Tip: When two progressions share the exact same common difference, the difference between any of their matching terms is always equal to the difference between their very first terms.

Question 6. Split 207 into 3 parts such that these are in AP and the product of the two smaller parts is 4623.
Answer: Let the three parts of \( 207 \) in AP be \( a-d \), \( a \), and \( a+d \).
The sum of these three parts is \( 207 \):
\( (a-d) + a + (a+d) = 207 \)
\( 3a = 207 \implies a = 69 \)
The two smaller parts are \( a-d \) and \( a \) (assuming \( d > 0 \)).
Their product is given as \( 4623 \):
\( a(a-d) = 4623 \)
Substitute \( a = 69 \):
\( 69(69 - d) = 4623 \)
\( 69 - d = \frac{4623}{69} \)
\( 69 - d = 67 \)
\( d = 2 \)
Therefore, the three parts are:
\( a-d = 69 - 2 = 67 \)
\( a = 69 \)
\( a+d = 69 + 2 = 71 \)
The three parts are 67, 69, and 71.
In simple words: We split 207 into 67, 69, and 71. These numbers are in AP, and multiplying the two smaller ones (67 and 69) gives 4623.

Exam Tip: Always verify your final answers by checking if they satisfy all the original conditions of the problem.

Question 7. Raj deposited Rs.1000 at CI at the rate of 10% per annum. The amount at the end of first year, second year, third year …………. form an AP or not. Justify it.
Answer: The principal amount is Rs. 1000 and the compound interest rate is \( 10\% \) per annum.
The amount at the end of year \( n \) is calculated using the formula:
\( A_n = P \left(1 + \frac{R}{100}\right)^n \)
At the end of the 1st year:
\( A_1 = 1000(1.1)^1 = \text{Rs. } 1100 \)
At the end of the 2nd year:
\( A_2 = 1000(1.1)^2 = \text{Rs. } 1210 \)
At the end of the 3rd year:
\( A_3 = 1000(1.1)^3 = \text{Rs. } 1331 \)
Now, let's find the differences between the consecutive years' amounts:
\( A_2 - A_1 = 1210 - 1100 = 110 \)
\( A_3 - A_2 = 1331 - 1210 = 121 \)
Since the common difference is not constant (\( 110 \neq 121 \)), the amounts at the end of consecutive years do not form an AP.
In simple words: In compound interest, the money grows faster each year because you earn interest on top of interest. Since the increase is not a fixed amount, it does not form an AP.

Exam Tip: Remember that simple interest values form an AP, whereas compound interest values grow exponentially and therefore do not form an AP.

Question 8. Find the 11th term from the last term of the AP 10, 7, 4 ……., −62.
Answer: The given AP is \( 10, 7, 4, \dots, -62 \).
First term \( a = 10 \), common difference \( d = 7 - 10 = -3 \), and last term \( l = -62 \).
To find the term from the end, we can reverse the AP.
In the reversed AP, the first term \( a' = -62 \), and the common difference \( d' = -d = 3 \).
Using the general term formula for the 11th term:
\( T_{11} = a' + (11-1)d' \)
\( T_{11} = -62 + 10(3) \)
\( T_{11} = -62 + 30 = -32 \)
Therefore, the 11th term from the last term is -32.
In simple words: Working backwards from -62 by adding 3 each time, the 11th number we reach is -32.

Exam Tip: Use the direct formula for finding the \( m \)-th term from the end: \( T_{m\text{ from end}} = l - (m-1)d \), where \( l \) is the last term and \( d \) is the original common difference.

Question 9. Is 301 a term of 5, 11, 17, 23, ……… ? Justify your answer.
Answer: The given sequence is \( 5, 11, 17, 23, \dots \).
Here, the first term \( a = 5 \) and the common difference \( d = 11 - 5 = 6 \).
Let us assume that \( 301 \) is the \( n \)-th term of this AP:
\( T_n = a + (n-1)d = 301 \)
\( 5 + (n-1)6 = 301 \)
\( (n-1)6 = 296 \)
\( n-1 = \frac{296}{6} = 49.33 \)
\( n = 50.33 \)
Since the number of terms \( n \) must be a positive integer, \( 301 \) cannot be a term of this sequence.
In simple words: If we keep adding 6 starting from 5, we skip right over 301. Since the step number isn't a whole number, 301 is not in the list.

Exam Tip: Any valid term in an AP must correspond to a whole number index \( n \). If your calculation gives a decimal, the number is definitely not a term.

Question 10. The 6th term from the end of the AP 17, 14, 11, ………., −40, is −15. Is it true or not ? Give reason.
Answer: The given AP is \( 17, 14, 11, \dots, -40 \).
Here, the first term \( a = 17 \), the common difference \( d = 14 - 17 = -3 \), and the last term \( l = -40 \).
The \( n \)-th term from the end is calculated using:
\( T_{n\text{ from end}} = l - (n-1)d \)
For the 6th term from the end:
\( T_{6\text{ from end}} = -40 - (6-1)(-3) \)
\( T_{6\text{ from end}} = -40 - 5(-3) \)
\( T_{6\text{ from end}} = -40 + 15 = -25 \)
Since the calculated term is \( -25 \), and the statement claims it is \( -15 \), the statement is false.
In simple words: Counting back 6 steps from -40, we get -25, not -15. So the given statement is incorrect.

Exam Tip: Be extra careful with double-negative signs during calculation (such as subtracting a negative common difference).

Question 11. If the 9th term of an AP is zero, prove that its 29th term is twice its 19th term.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the AP.
Given that the 9th term is zero:
\( T_9 = a + 8d = 0 \implies a = -8d \)
Now, let us find the 19th term:
\( T_{19} = a + 18d \)
Substitute \( a = -8d \):
\( T_{19} = -8d + 18d = 10d \)
Next, let us find the 29th term:
\( T_{29} = a + 28d \)
Substitute \( a = -8d \):
\( T_{29} = -8d + 28d = 20d \)
Comparing the two terms:
\( T_{29} = 20d = 2(10d) = 2 \cdot T_{19} \)
Hence proved.
In simple words: Since the 9th term is 0, we can write everything in terms of the jump distance. The 29th term turns out to be twice as large as the 19th term.

Exam Tip: Expressing terms of the AP in terms of the common difference \( d \) is a neat algebraic way to easily handle proofs.

Question 12. The angles of a triangle are in AP the greatest angle is twice the least. Find the angles.
Answer: Let the three angles of the triangle in AP be \( a-d \), \( a \), and \( a+d \) (in degrees).
Since the sum of angles in a triangle is always \( 180^\circ \):
\( (a-d) + a + (a+d) = 180^\circ \)
\( 3a = 180^\circ \implies a = 60^\circ \)
So the least angle is \( 60^\circ - d \) and the greatest angle is \( 60^\circ + d \).
According to the given condition:
\( 60^\circ + d = 2(60^\circ - d) \)
\( 60^\circ + d = 120^\circ - 2d \)
\( 3d = 60^\circ \implies d = 20^\circ \)
Now, we can find the three angles:
\( a-d = 60^\circ - 20^\circ = 40^\circ \)
\( a = 60^\circ \)
\( a+d = 60^\circ + 20^\circ = 80^\circ \)
The angles of the triangle are \( 40^\circ \), \( 60^\circ \), and \( 80^\circ \).
In simple words: The three angles inside the triangle are 40 degrees, 60 degrees, and 80 degrees. They form a steady pattern and the biggest one is twice the size of the smallest.

Exam Tip: Don't forget to include the degree symbol (\( ^\circ \)) on your final answers when dealing with angles to avoid losing minor presentation marks.

Question 13. Determine an AP where 3rd term is 16 and when 5th term is subtracted from 7th term, we get 12.
Answer: Let the first term be \( a \) and the common difference be \( d \).
The 3rd term of the AP is \( 16 \):
\( T_3 = a + 2d = 16 \quad \text{--- (Equation 1)} \)
Subtracting the 5th term from the 7th term yields \( 12 \):
\( T_7 - T_5 = 12 \)
\( (a + 6d) - (a + 4d) = 12 \)
\( 2d = 12 \implies d = 6 \)
Now substitute \( d = 6 \) back into Equation 1:
\( a + 2(6) = 16 \)
\( a + 12 = 16 \implies a = 4 \)
The AP is: \( 4, 10, 16, 22, \dots \)
In simple words: The sequence starts at 4 and increases by 6 each time, giving us the pattern: 4, 10, 16, 22, and so on.

Exam Tip: Remember that the difference between any two terms of an AP is always \( T_x - T_y = (x-y)d \), which lets you find \( d \) instantly.

Question 14. If the sum of the first q terms of an AP is 2q + 3q2. What is its common difference ?
Answer: The sum of the first \( q \) terms is \( S_q = 3q^2 + 2q \).
For \( q = 1 \):
\( S_1 = 3(1)^2 + 2(1) = 5 \)
Since the sum of the first term is just the first term itself, \( a_1 = 5 \).
For \( q = 2 \):
\( S_2 = 3(2)^2 + 2(2) = 12 + 4 = 16 \)
The sum of the first two terms is \( a_1 + a_2 = 16 \).
Now, let's find the second term \( a_2 \):
\( a_2 = S_2 - S_1 = 16 - 5 = 11 \)
The common difference \( d \) is:
\( d = a_2 - a_1 = 11 - 5 = 6 \)
Therefore, the common difference is 6.
In simple words: We find the first number is 5, and the sum of the first two is 16, meaning the second number is 11. The jump from 5 to 11 is 6.

Exam Tip: In any sum-of-n-terms expression \( S_n = An^2 + Bn \), the common difference is always equal to twice the coefficient of \( n^2 \) (i.e., \( 2A \)). Here, \( 2 \times 3 = 6 \).

Question 15. How many terms are there in the AP : −1, −5/6, −2/3, −1/2, ...., 10/3 ?
Answer: The given AP is \( -1, -\frac{5}{6}, -\frac{2}{3}, -\frac{1}{2}, \dots, \frac{10}{3} \).
First term \( a = -1 \).
The common difference \( d \) is:
\( d = -\frac{5}{6} - (-1) = -\frac{5}{6} + 1 = \frac{1}{6} \)
Let there be \( n \) terms in the AP, with the last term \( a_n = \frac{10}{3} \).
Using the general term formula:
\( a_n = a + (n-1)d \)
\( \frac{10}{3} = -1 + (n-1)\frac{1}{6} \)
\( \frac{10}{3} + 1 = (n-1)\frac{1}{6} \)
\( \frac{13}{3} = (n-1)\frac{1}{6} \)
\( n-1 = \frac{13}{3} \times 6 \)
\( n-1 = 26 \)
\( n = 27 \)
There are 27 terms in the AP.
In simple words: The sequence starts at -1 and increases by \( \frac{1}{6} \) at each step. It takes 27 terms to reach \( \frac{10}{3} \).

Exam Tip: Convert all terms to a common denominator to make visual checking and fractional operations much simpler.

SA-II

Question 1. Divide 32 into four parts which are in AP such that the product of extremes is to the product of means is 7 : 15.
Answer: Let the four parts of \( 32 \) in AP be represented as \( a-3d \), \( a-d \), \( a+d \), and \( a+3d \).
Their total sum is \( 32 \):
\( (a-3d) + (a-d) + (a+d) + (a+3d) = 32 \)
\( 4a = 32 \implies a = 8 \)
The extremes are \( a-3d \) and \( a+3d \). Their product is:
\( (a-3d)(a+3d) = a^2 - 9d^2 = 64 - 9d^2 \)
The means are \( a-d \) and \( a+d \). Their product is:
\( (a-d)(a+d) = a^2 - d^2 = 64 - d^2 \)
According to the given ratio:
\( \frac{64 - 9d^2}{64 - d^2} = \frac{7}{15} \)
\( 15(64 - 9d^2) = 7(64 - d^2) \)
\( 960 - 135d^2 = 448 - 7d^2 \)
\( 512 = 128d^2 \)
\( d^2 = 4 \implies d = \pm 2 \)
If we take \( d = 2 \), the four parts are:
\( a-3d = 8 - 3(2) = 2 \)
\( a-d = 8 - 2 = 6 \)
\( a+d = 8 + 2 = 10 \)
\( a+3d = 8 + 3(2) = 14 \)
The four parts are 2, 6, 10, and 14.
In simple words: We split 32 into the numbers 2, 6, 10, and 14. These are in AP, and the ratio of the multiplied outer numbers to the multiplied inner numbers is 7 to 15.

Exam Tip: When choosing four terms in an AP, use the terms \( a-3d, a-d, a+d, a+3d \) with a common difference of \( 2d \) to eliminate the \( d \) variable when summing.

Question 2. Four numbers are in AP. Whose sum is 20 and the sum of whose square is 120. Find the numbers
Answer: Let the four numbers in AP be represented as \( a-3d \), \( a-d \), \( a+d \), and \( a+3d \).
Their sum is \( 20 \):
\( (a-3d) + (a-d) + (a+d) + (a+3d) = 20 \)
\( 4a = 20 \implies a = 5 \)
The sum of their squares is \( 120 \):
\( (a-3d)^2 + (a-d)^2 + (a+d)^2 + (a+3d)^2 = 120 \)
Expanding these terms yields:
\( 4a^2 + 20d^2 = 120 \)
Substitute \( a = 5 \):
\( 4(25) + 20d^2 = 120 \)
\( 100 + 20d^2 = 120 \)
\( 20d^2 = 20 \)
\( d^2 = 1 \implies d = \pm 1 \)
If we take \( d = 1 \), the numbers are:
\( a-3d = 5 - 3 = 2 \)
\( a-d = 5 - 1 = 4 \)
\( a+d = 5 + 1 = 6 \)
\( a+3d = 5 + 3 = 8 \)
The four numbers are 2, 4, 6, and 8.
In simple words: The numbers are 2, 4, 6, and 8. They add up to 20, and squaring each of them and adding those up gives 120.

Exam Tip: Expand terms carefully. Cross-product terms like \( -6ad \) and \( +6ad \) will cancel out nicely when summing symmetric terms.

Question 3. Find the sum of first 20 terms of an AP in which 3rd term is 7 and 7th term is two more than thrice of its 3rd term.
Answer: Let the first term of the AP be \( a \) and the common difference be \( d \).
The 3rd term is 7:
\( T_3 = a + 2d = 7 \quad \text{--- (Equation 1)} \)
The 7th term is two more than thrice of its 3rd term:
\( T_7 = 3(T_3) + 2 \)
\( T_7 = 3(7) + 2 = 23 \)
Since \( T_7 = a + 6d \):
\( a + 6d = 23 \quad \text{--- (Equation 2)} \)
Subtract Equation 1 from Equation 2:
\( (a + 6d) - (a + 2d) = 23 - 7 \)
\( 4d = 16 \implies d = 4 \)
Substitute \( d = 4 \) back into Equation 1:
\( a + 2(4) = 7 \)
\( a + 8 = 7 \implies a = -1 \)
Now, find the sum of the first 20 terms (\( S_{20} \)):
\( S_n = \frac{n}{2}[2a + (n-1)d] \)
\( S_{20} = \frac{20}{2}[2(-1) + (20-1)4] \)
\( S_{20} = 10[-2 + 19(4)] \)
\( S_{20} = 10[-2 + 76] = 10(74) = 740 \)
Therefore, the sum of the first 20 terms is 740.
In simple words: We find that the starting number of this sequence is -1 and it increases by 4 each time. Adding up the first 20 numbers gives a total of 740.

Exam Tip: Translate the word problem carefully into clear equations before attempting to perform any mathematical operations.

Question 4. Find the sum of all natural numbers between 250 and 1000 which are exactly divisible by 3.
Answer: The first natural number greater than 250 that is divisible by 3 is 252.
The last natural number less than 1000 that is divisible by 3 is 999.
This forms an AP: \( 252, 255, 258, \dots, 999 \)
Here, first term \( a = 252 \), common difference \( d = 3 \), and last term \( a_n = 999 \).
Using the general term formula to find the number of terms \( n \):
\( a_n = a + (n-1)d \)
\( 999 = 252 + (n-1)3 \)
\( 747 = (n-1)3 \)
\( n-1 = 249 \)
\( n = 250 \)
Now, calculate the sum of these 250 terms:
\( S_n = \frac{n}{2}(a + a_n) \)
\( S_{250} = \frac{250}{2}(252 + 999) \)
\( S_{250} = 125(1251) = 156375 \)
The sum of all such numbers is 156375.
In simple words: The numbers start at 252 and go up to 999. There are 250 such numbers in total, and their combined sum is 156,375.

Exam Tip: Using the formula \( S_n = \frac{n}{2}(a + l) \) is much faster and less error-prone than the longer sum formula when the last term \( l \) is explicitly known.

Question 5. If Sn = sum of first n terms of an AP. Prove that S12 = 3(S8 − S4)
Answer: Let the first term of the AP be \( a \) and the common difference be \( d \).
The sum of the first \( n \) terms is:
\( S_n = \frac{n}{2}[2a + (n-1)d] \)
Let us calculate \( S_4 \), \( S_8 \), and \( S_{12} \):
\( S_4 = \frac{4}{2}[2a + 3d] = 2(2a + 3d) = 4a + 6d \)
\( S_8 = \frac{8}{2}[2a + 7d] = 4(2a + 7d) = 8a + 28d \)
\( S_{12} = \frac{12}{2}[2a + 11d] = 6(2a + 11d) = 12a + 66d \)
Now, look at the right-hand side of the statement:
\( \text{RHS} = 3(S_8 - S_4) \)
\( \text{RHS} = 3[(8a + 28d) - (4a + 6d)] \)
\( \text{RHS} = 3(4a + 22d) \)
\( \text{RHS} = 12a + 66d \)
Comparing this to \( S_{12} \):
\( \text{LHS} = S_{12} = 12a + 66d \)
Since LHS = RHS, we have proven that \( S_{12} = 3(S_8 - S_4) \).
In simple words: By writing down the algebraic formula for the sum of 4, 8, and 12 terms, we show that three times the difference between the sum of 8 terms and 4 terms is exactly equal to the sum of 12 terms.

Exam Tip: Clearly write down expressions for each part (LHS and RHS) separately, simplify them completely, and show that they are identical.

Question 6. For what value of n, the nth term of the series ‘3 + 10 + 17 + ………’ and ’63 + 65 + 67 + ………’ are equal ?
Answer: Let the first sequence be AP1: \( 3, 10, 17, \dots \)
First term \( a_1 = 3 \), common difference \( d_1 = 10 - 3 = 7 \).
Its \( n \)-th term is:
\( T_n = a_1 + (n-1)d_1 = 3 + (n-1)7 = 7n - 4 \)
Let the second sequence be AP2: \( 63, 65, 67, \dots \)
First term \( a_2 = 63 \), common difference \( d_2 = 65 - 63 = 2 \).
Its \( n \)-th term is:
\( T_n' = a_2 + (n-1)d_2 = 63 + (n-1)2 = 2n + 61 \)
We are given that their \( n \)-th terms are equal:
\( T_n = T_n' \)
\( 7n - 4 = 2n + 61 \)
\( 5n = 65 \)
\( n = 13 \)
Therefore, the 13th terms of both sequences are equal.
In simple words: The first pattern starts low but grows fast, while the second starts high but grows slow. They cross each other exactly at the 13th term.

Exam Tip: Form equations for the general terms of both progressions and equate them to find the value of \( n \).

Question 7. Find a, b, c such that the following numbers are in AP : a, 7, b, 23, c
Answer: Let the given AP be \( a, 7, b, 23, c \).
Since the terms are in an Arithmetic Progression, the difference between consecutive terms is equal:
\( 7 - a = b - 7 = 23 - b = c - 23 = d \)
From the equality of the middle differences:
\( b - 7 = 23 - b \)
\( 2b = 30 \implies b = 15 \)
Using \( b = 15 \), we can find the common difference \( d \):
\( d = b - 7 = 15 - 7 = 8 \)
Now, we can solve for \( a \) and \( c \):
\( 7 - a = d \implies 7 - a = 8 \implies a = -1 \)
\( c - 23 = d \implies c - 23 = 8 \implies c = 31 \)
The required values are \( a = -1 \), \( b = 15 \), and \( c = 31 \).<

Question 8. Find the sum of all the 11 terms of an AP whose middle most term is 30.
Answer: In an AP consisting of 11 terms, the middle-most term is the 6th term (\( T_6 \)).
Let \( a \) be the first term and \( d \) be the common difference.
Given:
\( T_6 = a + 5d = 30 \)
The sum of the first 11 terms of the AP is calculated as:
\( S_{11} = \frac{11}{2}[2a + (11-1)d] \)
\( S_{11} = \frac{11}{2}[2a + 10d] \)
Factor out 2 from the bracketed expression:
\( S_{11} = 11(a + 5d) \)
Substitute the value of \( a + 5d = 30 \):
\( S_{11} = 11(30) = 330 \)
The sum of all 11 terms is 330.
In simple words: Since the middle term is 30, the average of all the terms is also 30. Multiplying this average by the 11 terms gives a total sum of 330.

Exam Tip: For any AP with an odd number of terms \( n \), the sum is always equal to \( n \times \text{middle term} \).

Question 9. An AP consists of 21 terms. The sum of the three terms in the middle is 129 and of the last three is 237. Find the AP.
Answer: Let the first term of the AP be \( a \) and the common difference be \( d \).
The AP consists of 21 terms. The middle terms are the 10th, 11th, and 12th terms.
The sum of these three middle terms is 129:
\( T_{10} + T_{11} + T_{12} = 129 \)
\( (a + 9d) + (a + 10d) + (a + 11d) = 129 \)
\( 3a + 30d = 129 \)
\( a + 10d = 43 \quad \text{--- (Equation 1)} \)
The last three terms of the 21-term AP are the 19th, 20th, and 21st terms.
The sum of these last three terms is 237:
\( T_{19} + T_{20} + T_{21} = 237 \)
\( (a + 18d) + (a + 19d) + (a + 20d) = 237 \)
\( 3a + 57d = 237 \)
\( a + 19d = 79 \quad \text{--- (Equation 2)} \)
Subtract Equation 1 from Equation 2:
\( (a + 19d) - (a + 10d) = 79 - 43 \)
\( 9d = 36 \implies d = 4 \)
Substitute \( d = 4 \) into Equation 1:
\( a + 10(4) = 43 \)
\( a + 40 = 43 \implies a = 3 \)
The AP is: \( 3, 7, 11, 15, \dots \)
In simple words: This pattern starts at 3 and grows by 4 each time, giving us the sequence: 3, 7, 11, 15, and so on.

Exam Tip: Clearly identify which specific term positions correspond to "the middle terms" based on the total number of terms in the sequence.

Question 10. The digits of a +ve 3 digit no are in an AP and the sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the numbers.
Answer: Let the hundreds, tens, and units digits of the 3-digit number be represented as \( a-d \), \( a \), and \( a+d \) respectively.
The sum of these digits is 15:
\( (a-d) + a + (a+d) = 15 \)
\( 3a = 15 \implies a = 5 \)
The value of the original 3-digit number is:
\( 100(a-d) + 10a + (a+d) = 111a - 99d \)
The value of the reversed 3-digit number is:
\( 100(a+d) + 10a + (a-d) = 111a + 99d \)
According to the problem, the reversed number is 594 less than the original number:
\( (111a - 99d) - (111a + 99d) = 594 \)
\( -198d = 594 \)
\( d = -3 \)
Now, find the digits of the original number:
Hundreds digit: \( a-d = 5 - (-3) = 8 \)
Tens digit: \( a = 5 \)
Units digit: \( a+d = 5 + (-3) = 2 \)
Therefore, the original number is 852.
In simple words: The digits of the number are 8, 5, and 2, which form the number 852. Reversing its digits gives 258, which is exactly 594 less than 852.

Exam Tip: Remember to write the number using expanded place-value notation (hundreds, tens, units) when translating digits into algebraic equations.

Question 11. Find the sum of \( 7 + 10\frac{1}{2} + 14 + \dots + 84 \).
Answer: The given AP is \( 7, 10\frac{1}{2}, 14, \dots, 84 \).
First term \( a = 7 \).
The common difference \( d \) is:
\( d = 10.5 - 7 = 3.5 = \frac{7}{2} \)
The last term is \( a_n = 84 \).
Let us find the number of terms \( n \):
\( a_n = a + (n-1)d \)
\( 84 = 7 + (n-1)\frac{7}{2} \)
\( 77 = (n-1)\frac{7}{2} \)
\( n-1 = 22 \)
\( n = 23 \)
Now, calculate the sum of these 23 terms:
\( S_{23} = \frac{23}{2}(a + a_n) \)
\( S_{23} = \frac{23}{2}(7 + 84) \)
\( S_{23} = \frac{23}{2}(91) = \frac{2093}{2} = 1046.5 \)
The sum of the series is 1046.5.
In simple words: The numbers start at 7 and go up to 84 by jumps of 3.5. There are 23 numbers in total, and their sum is 1046.5.

Exam Tip: Use decimal fractions or simple improper fractions consistently throughout your calculation to make operations less confusing.

Question 12. The 4th term of an AP is three times the first and the 7th term exceeds twice the third term by 1. Find the 1st term and common difference.
Answer: Let the first term be \( a \) and the common difference be \( d \).
Given that the 4th term is three times the first term:
\( T_4 = 3a \)
\( a + 3d = 3a \)
\( 2a = 3d \implies a = \frac{3d}{2} \quad \text{--- (Equation 1)} \)
Also, the 7th term exceeds twice the third term by 1:
\( T_7 = 2(T_3) + 1 \)
\( a + 6d = 2(a + 2d) + 1 \)
\( a + 6d = 2a + 4d + 1 \)
\( 2d - a = 1 \quad \text{--- (Equation 2)} \)
Substitute Equation 1 into Equation 2:
\( 2d - \frac{3d}{2} = 1 \)
\( \frac{d}{2} = 1 \implies d = 2 \)
Now, find \( a \) using Equation 1:
\( a = \frac{3(2)}{2} = 3 \)
The first term \( a \) is 3 and the common difference \( d \) is 2.
In simple words: The sequence starts with 3 and increases by 2 each time. This makes the pattern: 3, 5, 7, 9, and so on.

Exam Tip: Substitute fractional expressions carefully and simplify denominators immediately to keep your calculations clean.

Question 13. Three numbers whose sum is 21 are in AP. If the product of the first and the third numbers exceeds the second number by 6, find the numbers.
Answer: Let the three numbers in AP be represented as \( a-d \), \( a \), and \( a+d \).
Their sum is 21:
\( (a-d) + a + (a+d) = 21 \)
\( 3a = 21 \implies a = 7 \)
The product of the first and third numbers exceeds the second number by 6:
\( (a-d)(a+d) = a + 6 \)
\( a^2 - d^2 = a + 6 \)
Substitute \( a = 7 \):
\( 7^2 - d^2 = 7 + 6 \)
\( 49 - d^2 = 13 \)
\( d^2 = 36 \implies d = \pm 6 \)
If we take \( d = 6 \), the numbers are:
\( a-d = 7 - 6 = 1 \)
\( a = 7 \)
\( a+d = 7 + 6 = 13 \)
The three numbers are 1, 7, and 13.
In simple words: The three numbers are 1, 7, and 13. They add up to 21, and the product of 1 and 13 (which is 13) is exactly 6 more than the middle number 7.

Exam Tip: If your equation yields two values for \( d \), write down the terms for both positive and negative values of \( d \). The set of numbers remains the same, just in reversed order.

Question 14. Find the sum of all 3-digit natural numbers divisible by 7.
Answer: The first 3-digit natural number divisible by 7 is 105.
The last 3-digit natural number divisible by 7 is 994.
This forms an AP: \( 105, 112, 119, \dots, 994 \)
Here, first term \( a = 105 \), common difference \( d = 7 \), and last term \( a_n = 994 \).
Using the general term formula to find the number of terms \( n \):
\( a_n = a + (n-1)d \)
\( 994 = 105 + (n-1)7 \)
\( 889 = (n-1)7 \)
\( n-1 = 127 \)
\( n = 128 \)
Now, calculate the sum of these 128 terms:
\( S_{128} = \frac{128}{2}(a + a_n) \)
\( S_{128} = 64(105 + 994) \)
\( S_{128} = 64(1099) = 70336 \)
The sum of all such numbers is 70336.
In simple words: The first three-digit multiple is 105 and the last is 994. There are 128 such numbers, and adding them up gives 70,336.

Exam Tip: To find the first or last multiple of a number in a given range, divide the boundary values by the divisor and look at the remainder.

Question 15. In an AP, the first term is 22, nth term is −11 and the sum of first n terms is 66. Find n and d.
Answer: Let \( a = 22 \), \( a_n = -11 \), and \( S_n = 66 \).
Using the sum formula:
\( S_n = \frac{n}{2}(a + a_n) \)
\( 66 = \frac{n}{2}[22 + (-11)] \)
\( 66 = \frac{n}{2}(11) \)
\( n = \frac{66 \times 2}{11} = 12 \)
Now, find the common difference \( d \) using the general term formula:
\( a_n = a + (n-1)d \)
\( -11 = 22 + (12-1)d \)
\( -33 = 11d \)
\( d = -3 \)
Therefore, \( n = 12 \) and \( d = -3 \).
In simple words: This decreasing pattern has 12 numbers in total. It starts at 22 and goes down by 3 at each step to reach -11.

Exam Tip: When given the sum, first term, and last term, always find \( n \) first using the simpler sum formula before solving for \( d \).

LONG ANSWER TYPE QUESTIONS

Question 1. Prove that no matter what the real numbers a and b are, the pattern of numbers with nth term a + nb is always an AP. What is the common difference ? What is the sum of the first 20 terms ?
Answer: Let the \( n \)-th term of the pattern be \( T_n = a + nb \).
The previous term is \( T_{n-1} = a + (n-1)b \).
Let's find the difference between consecutive terms:
\( T_n - T_{n-1} = (a + nb) - [a + (n-1)b] \)
\( T_n - T_{n-1} = a + nb - a - nb + b = b \)
Since the difference between any two consecutive terms is a constant value \( b \) which does not depend on \( n \), the pattern of numbers is always an AP.
The common difference is \( b \).
Now, let's find the sum of the first 20 terms (\( S_{20} \)):
The first term \( T_1 \) is:
\( T_1 = a + 1(b) = a + b \)
Using the sum formula for 20 terms:
\( S_{20} = \frac{20}{2}[2(T_1) + (20-1)d] \)
\( S_{20} = 10[2(a+b) + 19b] \)
\( S_{20} = 10[2a + 2b + 19b] \)
\( S_{20} = 10[2a + 21b] = 20a + 210b \)
Therefore, the common difference is \( b \), and the sum of the first 20 terms is \( 20a + 210b \).
In simple words: Because the gap between any two terms is always the same number \( b \), this pattern is always an AP. The total sum of its first 20 terms is \( 20a + 210b \).

Exam Tip: To prove that a sequence is an AP, always show that the difference \( T_n - T_{n-1} \) simplifies to a constant that is completely free of the variable \( n \).

Question 2. The ratio of the 11th term to the 18th term of an AP is 2 : 3. Find the ratio of the 5th term to the 21st term, and also the ratio of the sum of the first 5 terms to the sum of the first 21 terms.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the AP.
Given that:
\( \frac{T_{11}}{T_{18}} = \frac{2}{3} \)
\( \frac{a + 10d}{a + 17d} = \frac{2}{3} \)
\( 3(a + 10d) = 2(a + 17d) \)
\( 3a + 30d = 2a + 34d \)
\( a = 4d \quad \text{--- (Equation 1)} \)
Now, find the ratio of the 5th term to the 21st term:
\( \frac{T_5}{T_{21}} = \frac{a + 4d}{a + 20d} \)
Substitute \( a = 4d \) into the ratio:
\( \frac{T_5}{T_{21}} = \frac{4d + 4d}{4d + 20d} = \frac{8d}{24d} = \frac{1}{3} \)
So, the ratio of the 5th term to the 21st term is \( 1 : 3 \).
Next, find the ratio of the sum of the first 5 terms to the sum of the first 21 terms:
\( \frac{S_5}{S_{21}} = \frac{\frac{5}{2}[2a + 4d]}{\frac{21}{2}[2a + 20d]} \)
\( \frac{S_5}{S_{21}} = \frac{5(2a + 4d)}{21(2a + 20d)} \)
Substitute \( a = 4d \):
\( \frac{S_5}{S_{21}} = \frac{5[2(4d) + 4d]}{21[2(4d) + 20d]} \)
\( \frac{S_5}{S_{21}} = \frac{5(12d)}{21(28d)} = \frac{60d}{588d} = \frac{5}{49} \)
So, the ratio of the sums is \( 5 : 49 \).
In simple words: The starting number \( a \) is 4 times the jump size \( d \). Using this, we find the 5th-to-21st term ratio is 1:3 and the sum-of-5-to-sum-of-21 ratio is 5:49.

Exam Tip: Expressing \( a \) in terms of \( d \) allows you to completely substitute out variables, leaving a simple fraction that is easy to reduce.

Question 3. If (b − c)2, (c − a)2, (a − b)2 are in AP, show that 1/(b-c), 1/(c-a), 1/(a-b) are also in AP.
Answer: Given that \( (b-c)^2, (c-a)^2, (a-b)^2 \) are in AP.
Therefore, the common difference is constant:
\( (c-a)^2 - (b-c)^2 = (a-b)^2 - (c-a)^2 \)
Using the identity \( x^2 - y^2 = (x-y)(x+y) \):
\( [(c-a) - (b-c)][(c-a) + (b-c)] = [(a-b) - (c-a)][(a-b) + (c-a)] \)
\( (2c - a - b)(b - a) = (2a - b - c)(c - b) \)
Dividing both sides by \( -1 \):
\( (a + b - 2c)(a - b) = (b + c - 2a)(b - c) \quad \text{--- (Equation 1)} \)
Now, let us examine the second sequence \( \frac{1}{b-c}, \frac{1}{c-a}, \frac{1}{a-b} \).
For these terms to be in AP, we must show that:
\( \frac{1}{c-a} - \frac{1}{b-c} = \frac{1}{a-b} - \frac{1}{c-a} \)
Taking LCM on both sides:
\( \frac{(b-c) - (c-a)}{(c-a)(b-c)} = \frac{(c-a) - (a-b)}{(a-b)(c-a)} \)
Since \( c-a \neq 0 \), we can cancel \( c-a \) from the denominators:
\( \frac{a + b - 2c}{b-c} = \frac{b + c - 2a}{a-b} \)
Cross-multiplying yields:
\( (a + b - 2c)(a - b) = (b + c - 2a)(b - c) \)
Since this is exactly the same as Equation 1, which is true, the terms \( \frac{1}{b-c}, \frac{1}{c-a}, \frac{1}{a-b} \) are indeed in AP.
In simple words: By simplifying the condition for both sequences, we show that they both require the exact same algebraic relationship to hold true.

Exam Tip: In algebraic AP proofs, simplify the required condition to see if it reduces directly to the given condition.

Question 4. The student of a school decided to beautify the school on the Annual Day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at interval of every 2 m. The flags are stored at the position of the middle most flag. Raju was given the responsibility of placing the flags. Raju kept his books where the flags were stored. He could carry only one flag at a time. How much distance did she cover in completing this job and returning back to collect her books ? What is the maximum distance she travelled carrying a flag ?
Answer: There are 27 flags, so the middle-most flag is the 14th flag.
The flags are stored at the position of this 14th flag. Let this position be \( x = 0 \).
There are 13 flags to the left of the center and 13 flags to the right of the center, each placed at intervals of 2 m.
The distances of the flags on one side from the center are: \( 2, 4, 6, \dots, 26 \) meters.
Raju carries one flag at a time. For each flag on one side (at distance \( d \)), he must travel to the position and return to the center. So, the distance covered for a flag at distance \( d \) is \( 2d \).
The total distance covered for all 13 flags on one side is:
\( D_{\text{one side}} = 2(2) + 2(4) + 2(6) + \dots + 2(26) \)
\( D_{\text{one side}} = 4(1 + 2 + 3 + \dots + 13) \)
\( D_{\text{one side}} = 4 \left( \frac{13 \times 14}{2} \right) = 4(91) = 364 \text{ meters} \)
Since the layout is symmetrical, the distance covered for the 13 flags on the other side is also 364 meters.
The middle-most flag is already at the center, so it requires 0 meters of travel.
The total distance covered is:
\( D_{\text{total}} = 364 + 364 = 728 \text{ meters} \)
The maximum distance Raju travelled carrying a flag is the distance to the outermost flag, which is \( 13 \times 2 = 26 \) meters.
In simple words: Raju walks out and back for each of the 26 flags placed away from the center. He covers 728 meters in total, and the farthest he carries a single flag is 26 meters.

Exam Tip: Be sure to count both going out and returning back when calculating the total travel distance for round trips.

Question 5. The sum of first 5 terms of an AP and the sum of 1st 7 terms of the same AP is 167. If the sum of the first ten terms of this AP is 235, find the sum of the first 20 terms.
Answer: Let \( a \) be the first term and \( d \) be the common difference.
Given:
\( S_5 + S_7 = 167 \)
\( \frac{5}{2}[2a + 4d] + \frac{7}{2}[2a + 6d] = 167 \)
\( 5(a + 2d) + 7(a + 3d) = 167 \)
\( 5a + 10d + 7a + 21d = 167 \)
\( 12a + 31d = 167 \quad \text{--- (Equation 1)} \)
Also, the sum of the first 10 terms is 235:
\( S_{10} = \frac{10}{2}[2a + 9d] = 235 \)
\( 5(2a + 9d) = 235 \)
\( 2a + 9d = 47 \quad \text{--- (Equation 2)} \)
Multiply Equation 2 by 6 to align coefficients:
\( 12a + 54d = 282 \quad \text{--- (Equation 3)} \)
Subtract Equation 1 from Equation 3:
\( (12a + 54d) - (12a + 31d) = 282 - 167 \)
\( 23d = 115 \implies d = 5 \)
Substitute \( d = 5 \) back into Equation 2:
\( 2a + 9(5) = 47 \)
\( 2a + 45 = 47 \implies 2a = 2 \implies a = 1 \)
Now, find the sum of the first 20 terms (\( S_{20} \)):
\( S_{20} = \frac{20}{2}[2a + (20-1)d] \)
\( S_{20} = 10[2(1) + 19(5)] \)
\( S_{20} = 10[2 + 95] = 10(97) = 970 \)
The sum of the first 20 terms is 970.
In simple words: Solving the equations shows that this sequence starts at 1 and increases by 5 each time. The sum of its first 20 terms is 970.

Exam Tip: Be careful when simplifying terms like \( \frac{5}{2}[2a + 4d] \) — factor out the common factor of 2 to remove the denominator early in the process.

Question 6. Find the sum of all natural numbers less than 1000 and which are neither divisible by 5 nor by 2.
Answer: Natural numbers less than 1000 are \( 1, 2, 3, \dots, 999 \) [6].
The numbers that are neither divisible by 2 nor by 5 are those ending with the digits 1, 3, 7, or 9.
We can split these numbers into four separate APs, each with a common difference of 10:
AP1 (numbers ending in 1): \( 1, 11, 21, \dots, 991 \). Here, \( a=1, d=10, n=100 \).
\( S_1 = \frac{100}{2}(1 + 991) = 50(992) = 49600 \)
AP2 (numbers ending in 3): \( 3, 13, 23, \dots, 993 \). Here, \( a=3, d=10, n=100 \).
\( S_2 = \frac{100}{2}(3 + 993) = 50(996) = 49800 \)
AP3 (numbers ending in 7): \( 7, 17, 27, \dots, 997 \). Here, \( a=7, d=10, n=100 \).
\( S_3 = \frac{100}{2}(7 + 997) = 50(1004) = 50200 \)
AP4 (numbers ending in 9): \( 9, 19, 29, \dots, 999 \). Here, \( a=9, d=10, n=100 \).
\( S_4 = \frac{100}{2}(9 + 999) = 50(1008) = 50400 \)
The total sum of all required numbers is:
\( S_{\text{total}} = S_1 + S_2 + S_3 + S_4 \)
\( S_{\text{total}} = 49600 + 49800 + 50200 + 50400 = 200000 \)
Therefore, the sum is 200,000.
In simple words: The numbers that are not divisible by 2 or 5 end in 1, 3, 7, or 9. Splitting them into patterns and adding them up gives a total of 200,000.

Exam Tip: You can also find this by taking the sum of all natural numbers up to 999, and subtracting the sum of multiples of 2, the sum of multiples of 5, and then adding back the sum of multiples of 10 (which were subtracted twice).

Question 7. 390 plants are to be planted in a garden in a number of rows. There are 40 plants in the 1st row, 38 plants in the second row, 36 plants in 3rd row and so on. In how many rows the 390 plants are planted ? Find the no of plants in the last row also.
Answer: The number of plants in consecutive rows forms an AP: \( 40, 38, 36, \dots \) [7].
Here, first term \( a = 40 \), common difference \( d = -2 \), and the total sum \( S_n = 390 \) [7].
Using the sum formula:
\( S_n = \frac{n}{2}[2a + (n-1)d] \)
\( 390 = \frac{n}{2}[2(40) + (n-1)(-2)] \)
\( 390 = \frac{n}{2}[80 - 2n + 2] \)
\( 390 = n(41 - n) \)
\( 390 = 41n - n^2 \)
\( n^2 - 41n + 390 = 0 \)
Solve this quadratic equation by factoring:
\( n^2 - 15n - 26n + 390 = 0 \)
\( n(n-15) - 26(n-15) = 0 \)
\( (n-15)(n-26) = 0 \)
This gives \( n = 15 \) or \( n = 26 \).
If \( n = 26 \), the number of plants in the 26th row is:
\( T_{26} = a + 25d = 40 + 25(-2) = -10 \), which is impossible because the number of plants cannot be negative.
Therefore, \( n = 15 \) is the correct value.
The number of plants in the last (15th) row is:
\( T_{15} = a + 14d = 40 + 14(-2) = 40 - 28 = 12 \)
The plants are planted in 15 rows, and there are 12 plants in the last row.
In simple words: The plants are arranged in 15 rows. The rows decrease in size, and the very last row has 12 plants.

Exam Tip: When you get two positive integer solutions from a quadratic equation, always check if both make physical sense by calculating the value of the last term.

Question 8. In an AP it is given that tn = 4, d = 2, Sn = −14. Find n and a.
Answer: Given \( t_n = 4 \), \( d = 2 \), and \( S_n = -14 \) [8].
Using the general term formula:
\( t_n = a + (n-1)d = 4 \)
\( a + (n-1)2 = 4 \)
\( a + 2n - 2 = 4 \implies a = 6 - 2n \quad \text{--- (Equation 1)} \)
Using the sum formula:
\( S_n = \frac{n}{2}(a + t_n) = -14 \)
\( \frac{n}{2}(a + 4) = -14 \)
\( n(a + 4) = -28 \quad \text{--- (Equation 2)} \)
Substitute Equation 1 into Equation 2:
\( n(6 - 2n + 4) = -28 \)
\( n(10 - 2n) = -28 \)
\( 10n - 2n^2 = -28 \)
\( 2n^2 - 10n - 28 = 0 \)
Divide by 2:
\( n^2 - 5n - 14 = 0 \)
\( (n - 7)(n + 2) = 0 \)
Since the number of terms \( n \) must be positive, we take \( n = 7 \).
Now, substitute \( n = 7 \) back into Equation 1:
\( a = 6 - 2(7) = 6 - 14 = -8 \)
Therefore, \( n = 7 \) and \( a = -8 \).
In simple words: This pattern starts at -8 and increases by 2 each time. It has 7 terms in total, ending at 4.

Exam Tip: Expressing \( a \) in terms of \( n \) and substituting it into the sum formula is the standard approach for this common exam question.

Question 9. A sum of Rs.700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs.20 less than its preceding prize, find the value of each of the prizes.
Answer: Let the value of the first prize be \( a \).
Since each subsequent prize is Rs. 20 less than the preceding one, the values of the prizes form an AP with a common difference \( d = -20 \) [9].
The total number of prizes is \( n = 7 \), and the total sum is \( S_7 = 700 \) [9].
Using the sum formula:
\( S_n = \frac{n}{2}[2a + (n-1)d] \)
\( 700 = \frac{7}{2}[2a + (7-1)(-20)] \)
\( 100 = \frac{1}{2}[2a - 120] \)
\( 100 = a - 60 \)
\( a = 160 \)
The values of the seven prizes are:
Rs. 160, Rs. 140, Rs. 120, Rs. 100, Rs. 80, Rs. 60, Rs. 40.
In simple words: The highest prize is Rs. 160, and each successive prize is Rs. 20 less, going all the way down to Rs. 40 for the 7th prize.

Exam Tip: Always state the final values of all the terms clearly in your conclusion, rather than stopping after just finding the first term \( a \).

Question 10. Sum of 4th and 8th terms of an AP is 24 and sum of 6th and 10th terms is 44. Find AP.
Answer: Let the first term of the AP be \( a \) and the common difference be \( d \).
Given:
\( T_4 + T_8 = 24 \) [10]
\( (a + 3d) + (a + 7d) = 24 \)
\( 2a + 10d = 24 \)
\( a + 5d = 12 \quad \text{--- (Equation 1)} \)
Also, given:
\( T_6 + T_{10} = 44 \) [10]
\( (a + 5d) + (a + 9d) = 44 \)
\( 2a + 14d = 44 \)
\( a + 7d = 22 \quad \text{--- (Equation 2)} \)
Subtract Equation 1 from Equation 2:
\( (a + 7d) - (a + 5d) = 22 - 12 \)
\( 2d = 10 \implies d = 5 \)
Substitute \( d = 5 \) back into Equation 1:
\( a + 5(5) = 12 \)
\( a + 25 = 12 \implies a = -13 \)
The AP is: \( -13, -8, -3, 2, 7, \dots \)
In simple words: The sequence starts with -13 and grows by adding 5 each time, which gives: -13, -8, -3, 2, 7, and so on.

Exam Tip: When asked to "find the AP", always write down at least the first four terms of the sequence in your final answer.

Question 11. The sum of first six terms of an AP is 42. The ratio of its 10th term to its 30th term is 1 : 3. Calculate the first and the thirteenth term of the AP.
Answer: Let the first term of the AP be \( a \) and the common difference be \( d \).
Given that the sum of the first six terms is 42:
\( S_6 = \frac{6}{2}[2a + 5d] = 42 \) [11]
\( 3(2a + 5d) = 42 \)
\( 2a + 5d = 14 \quad \text{--- (Equation 1)} \)
The ratio of the 10th term to the 30th term is \( 1 : 3 \):
\( \frac{T_{10}}{T_{30}} = \frac{a + 9d}{a + 29d} = \frac{1}{3} \) [11]
\( 3(a + 9d) = a + 29d \)
\( 3a + 27d = a + 29d \)
\( 2a = 2d \implies a = d \quad \text{--- (Equation 2)} \)
Substitute Equation 2 into Equation 1:
\( 2a + 5a = 14 \)
\( 7a = 14 \implies a = 2 \)
Since \( a = d \), we also have \( d = 2 \).
The first term \( a \) is 2 [11].
Now, calculate the 13th term [11]:
\( T_{13} = a + 12d = 2 + 12(2) = 2 + 24 = 26 \)
The first term is 2 and the thirteenth term is 26 [11].
In simple words: The sequence starts at 2 and increases by 2 each time. The 13th number in this sequence is 26.

Exam Tip: Simplify the ratio first to establish a direct relationship between \( a \) and \( d \) before substituting it into the sum equation.

Question 12. The sum of the first five terms of an AP is 55 and sum of the first ten terms of this AP is 235, find the sum of first 20 terms.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the AP.
Given that \( S_5 = 55 \) [12]:
\( \frac{5}{2}[2a + 4d] = 55 \)
\( 5(a + 2d) = 55 \)
\( a + 2d = 11 \quad \text{--- (Equation 1)} \)
Given that \( S_{10} = 235 \) [12]:
\( \frac{10}{2}[2a + 9d] = 235 \)
\( 5(2a + 9d) = 235 \)
\( 2a + 9d = 47 \quad \text{--- (Equation 2)} \)
From Equation 1, we get \( a = 11 - 2d \). Substitute this into Equation 2:
\( 2(11 - 2d) + 9d = 47 \)
\( 22 - 4d + 9d = 47 \)
\( 5d = 25 \implies d = 5 \)
Substitute \( d = 5 \) back into the expression for \( a \):
\( a = 11 - 2(5) = 1 \)
Now, find the sum of the first 20 terms (\( S_{20} \)) [12]:
\( S_{20} = \frac{20}{2}[2a + (20-1)d] \)
\( S_{20} = 10[2(1) + 19(5)] \)
\( S_{20} = 10[2 + 95] = 10(97) = 970 \)
The sum of the first 20 terms is 970 [12].
In simple words: The sequence starts at 1 and increases by 5 each time. The sum of the first 20 terms is 970.

Exam Tip: Be consistent with your algebraic substitutions to avoid sign errors, especially in multi-step equations.

Question 13. The ratio of the 11th term to the 18th term of an AP is 2 : 3. Find the ratio of the 5th term to the 21st term, and also the ratio of the sum of the first five terms to the sum of the first 21 terms.
Answer: Let the first term of the AP be \( a \) and the common difference be \( d \).
Given that:
\( \frac{T_{11}}{T_{18}} = \frac{a + 10d}{a + 17d} = \frac{2}{3} \) [13]
Cross-multiplying yields:
\( 3(a + 10d) = 2(a + 17d) \)
\( 3a + 30d = 2a + 34d \)
\( a = 4d \quad \text{--- (Equation 1)} \)
Now, find the ratio of the 5th term to the 21st term [13]:
\( \frac{T_5}{T_{21}} = \frac{a + 4d}{a + 20d} \)
Substitute \( a = 4d \) into the ratio:
\( \frac{T_5}{T_{21}} = \frac{4d + 4d}{4d + 20d} = \frac{8d}{24d} = \frac{1}{3} \)
So, the ratio of the 5th term to the 21st term is \( 1 : 3 \) [13].
Next, find the ratio of the sum of the first five terms to the sum of the first 21 terms [13]:
\( \frac{S_5}{S_{21}} = \frac{\frac{5}{2}[2a + 4d]}{\frac{21}{2}[2a + 20d]} = \frac{5(2a + 4d)}{21(2a + 20d)} \)
Substitute \( a = 4d \):
\( \frac{S_5}{S_{21}} = \frac{5[2(4d) + 4d]}{21[2(4d) + 20d]} = \frac{5(12d)}{21(28d)} = \frac{60d}{588d} = \frac{5}{49} \)
The ratio of the sums is \( 5 : 49 \) [13].
In simple words: The first term is 4 times the jump size. This helps us find that the 5th to 21st term ratio is 1:3, and the ratio of their sums is 5:49.

Exam Tip: Keep the variable \( d \) in your expressions until the final step, as it will cancel out when you simplify the ratios.

Value Based Questions

Question 1. A sum of Rs. 3150 is to be used to give six cash prizes to students of a school for overall academic performance, punctuality, regularity, cleanliness, confidence and creativity. If each prize is Rs. 50 less than its preceding prize, find the value of each of the prizes.
a) Which value according to you should be awarded with maximum amount? Justify your answer.
b) Can you add more values to the above ones which should be awarded?
Answer: Let the first cash prize be \( a \).
Since each prize is Rs. 50 less than the preceding one, the prizes form an AP with \( d = -50 \) [2].
The total number of prizes is \( n = 6 \) [2] and the total amount is \( S_6 = 3150 \) [2].
Using the sum formula:
\( S_n = \frac{n}{2}[2a + (n-1)d] \)
\( 3150 = \frac{6}{2}[2a + 5(-50)] \)
\( 3150 = 3[2a - 250] \)
\( 1050 = 2a - 250 \)
\( 1300 = 2a \implies a = 650 \)
The values of the six cash prizes are:
Rs. 650, Rs. 600, Rs. 550, Rs. 500, Rs. 450, Rs. 400.

a) Creativity or Academic Performance should be awarded with the highest amount because these values require a lot of dedication, critical thinking, and effort, which can be further fostered with financial encouragement.
b) Yes, we can add other values like honesty, discipline, empathy, leadership, and helpfulness to the ones that should be awarded.
In simple words: The six prizes are Rs. 650, Rs. 600, Rs. 550, Rs. 500, Rs. 450, and Rs. 400. Fostering creativity and academics is highly valued, and honesty or discipline can be added to the list.

Exam Tip: For value-based questions, make sure your justifications are positive, constructive, and clearly written.

Question 2. A person donates money to a trust working for education of children and women in some villages. If the person donates Rs. 5,000 in the first year and his donation increases by Rs. 250 every year, find the amount donated by him in the eighth year and the total amount donated in eight years.
a) What mathematical concept is being used here?
b) Write any two values the person mentioned here possess.
c) Why do you think education of women is necessary for the development of a society?
Answer: Let the donations form an AP where the first term is \( a = 5000 \) and the common difference is \( d = 250 \) [2].
The amount donated in the eighth year is [2]:
\( T_8 = a + 7d = 5000 + 7(250) = 5000 + 1750 = \text{Rs. } 6750 \)
The total amount donated in eight years is [2]:
\( S_8 = \frac{8}{2}[2a + (8-1)d] \)
\( S_8 = 4[2(5000) + 7(250)] \)
\( S_8 = 4[10000 + 1750] = 4(11750) = \text{Rs. } 47000 \)

a) The mathematical concept being used here is Arithmetic Progression (AP) [2].
b) The values possessed by the person are generosity and empathy for the underprivileged.
c) Women's education is vital because educating a woman empowers her economically, improves family health and nutrition, reduces social inequalities, and ensures that future generations are well-educated.
In simple words: The person donates Rs. 6,750 in the eighth year, and a total of Rs. 47,000 over eight years. This uses the concept of AP, showing kindness, and women's education helps society grow.

Exam Tip: In financial AP word problems, always specify the currency symbol (Rs.) clearly next to your final calculated numerical values.

Question 3. A sum of Rs. 700 is to be used to give cash prizes to 5 students of a school on the basis of their over all.academic performance, regularity, creativity, confidence and cleanliness. A child having one value is given a fixed amount. A child having 2 or more values is given Rs. 20 more for each additional value. Find the maximum number of students who can be awarded. Find the value of each cash prize for having only 1 value, 2 values and so on. According to you, is there any other value which should be rewarded? How is rewarding the values helpful?
Answer: Let us assume the 5 students have 1, 2, 3, 4, and 5 values respectively.
Let the prize for having exactly 1 value be \( x \) rupees.
The prize for having 2 values is \( x + 20 \) [3].
The prize for having 3 values is \( x + 40 \).
The prize for having 4 values is \( x + 60 \).
The prize for having 5 values is \( x + 80 \).
These prizes form an AP with first term \( a = x \) and common difference \( d = 20 \).
The total sum of these 5 prizes is Rs. 700 [3]:
\( S_5 = \frac{5}{2}[2x + (5-1)20] = 700 \)
\( 5(x + 40) = 700 \)
\( x + 40 = 140 \implies x = 100 \)
So the cash prizes for different number of values are:
For 1 value: Rs. 100
For 2 values: Rs. 120
For 3 values: Rs. 140
For 4 values: Rs. 160
For 5 values: Rs. 180
The maximum number of students who can be awarded is 5 [3].
Another value that should be rewarded is honesty, kindness, or hard work.
Rewarding values is helpful because it builds positive reinforcement, encouraging children to develop strong moral traits alongside their academics.
In simple words: The maximum number of students awarded is 5. The prizes are Rs. 100, Rs. 120, Rs. 140, Rs. 160, and Rs. 180 depending on the number of values they show. Rewarding these helps build good character.

Exam Tip: When setting up the system of equations, state your assumptions clearly so the examiner can follow your logical steps easily.

Question 4. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for first day, Rs. 250 for 2nd day, Rs. 300 for 3rd day and so on. If the contractor pays Rs. 27750 as penalty, find the number of days for which the construction work is delayed. Does the penalty help in providing quality work? Which trait of contractor’s personality is reflected here?
Answer: The daily penalty amounts form an AP: \( 200, 250, 300, \dots \) [3].
Here, first term \( a = 200 \), common difference \( d = 50 \), and total penalty paid is \( S_n = 27750 \) [3].
Using the sum formula to find the number of delayed days \( n \):
\( S_n = \frac{n}{2}[2a + (n-1)d] \)
\( 27750 = \frac{n}{2}[2(200) + (n-1)50] \)
\( 55500 = n[400 + 50n - 50] \)
\( 55500 = n[350 + 50n] \)
\( 55500 = 350n + 50n^2 \)
Divide the entire equation by 50:
\( n^2 + 7n - 1110 = 0 \)
Solve this quadratic equation by factoring:
\( n^2 + 37n - 30n - 1110 = 0 \)
\( n(n + 37) - 30(n + 37) = 0 \)
\( (n + 37)(n - 30) = 0 \)
Since the number of days \( n \) must be a positive integer, we discard \( n = -37 \) and select \( n = 30 \).
Therefore, the construction work was delayed by 30 days [3].
Yes, the penalty encourages punctuality and timely completion, which prevents rushed, poor-quality work at the very end of the project.
The traits reflected in the contractor's personality are poor time management, lack of proper planning, and unreliability.
In simple words: The construction was delayed by 30 days. The penalty system keeps contractors on time, and this delay shows poor planning on their part.

Exam Tip: Be comfortable with factoring large numbers for quadratic equations by listing factors that sum to the linear coefficient.