CBSE Class 10 Mathematics Arithmetic Progression Worksheet Set I

Question. Show that the sum of all terms of an A.P. whose first term is \( a \), the second term is \( b \) and the last term is \( c \) is equal to \( \frac{(a + c)(b + c - 2a)}{2(b - a)} \)
Answer: Given : First term of A.P. = \( a \)
Second term of A.P. = \( b \)
and, Last term of A.P. = \( c \)
So, Common difference, \( d = b - a \)
We know,
\( l = a + (n - 1)d \)
\( \Rightarrow c = a + (n - 1) (b - a) \)
\( \Rightarrow c - a = (n - 1) (b - a) \)
\( \Rightarrow n - 1 = \frac{c - a}{b - a} \)
\( \Rightarrow n = \frac{c - a}{b - a} + 1 \)
\( \Rightarrow n = \frac{b + c - 2a}{b - a} \)
Now, \( S_n = \frac{n}{2} (a + l) \)
\( \Rightarrow S_n = \frac{(b + c - 2a)(a + c)}{2(b - a)} \)
Hence Proved.

Question. The 8th term of an A.P. is 37 and 12th term is 57. Find the A.P.
Answer: Given : \( a_8 = 37 \) and \( a_{12} = 57 \)
\( \Rightarrow a + 7d = 37 \dots(i) \)
and \( a + 11d = 57 \dots(ii) \)
Subtracting equation (i) from (ii)
\( a + 11d = 57 \)
\( a + 7d = 37 \)
——————
\( 4d = 20 \)
\( \Rightarrow d = \frac{20}{4} = 5 \)
Putting \( d = 5 \) in equation (i), we get
\( a + 7(5) = 37 \)
\( \Rightarrow a = 37 - 35 = 2 \)
\(\therefore\) First term = \( a = 2 \)
Second term = \( a_2 = a + d = 2 + 5 = 7 \)
Third term = \( a_3 = a + 2d = 2 + (2 \times 5) = 12 \)
Thus, the A.P. is 2, 7, 12, ........

Question. Solve the equation : \( 1 + 4 + 7 + 10 + \dots + x = 287 \).
Answer: Given : \( 1 + 4 + 7 + 10 + \dots + x = 287 \)
In this A.P., \( a = 1, d = 3, S = 287 \) and \( l = x \)
We know that,
\( l = a + (n - 1)d \)
\( \Rightarrow x = 1 + (n - 1)3 \)
\( \Rightarrow x = 1 + 3n - 3 \)
\( \Rightarrow x = 3n - 2 \)
\( \Rightarrow n = \frac{x + 2}{3} \)
Now, \( S_n = \frac{n}{2}(a + l) \)
\( \Rightarrow 287 = \frac{(x + 2)(1 + x)}{6} \)
\( \Rightarrow x^2 + 3x + 2 = 1722 \)
\( x^2 + 3x - 1720 = 0 \)
\( x^2 + 43x - 40x - 1720 = 0 \)
\( x(x + 43) - 40 (x + 43) = 0 \)
\( x = - 43 \) or \( x = 40 \)
\(\therefore x = - 43 \) is not possible because it is increasing A.P.
\(\therefore x = 40 \).

Question. In an A.P., if sum of its first \( n \) terms is \( 3n^2 + 5n \) and its kth term is 164, find the value of \( k \).
Answer: Given : \( S_n = 3n^2 + 5n \) and \( a_k = 164 \).
Now, we have \( S_n = 3n^2 + 5n \)
For \( n = 1, S_1 = 3(1)^2 + 5(1) = 8 \)
For \( n = 2, S_2 = 3(2)^2 + 5(2) = 22 \)
We know, \( a_1 = S_1 = 8 \)
and \( S_2 = a_1 + a_2 = 22 \)
\( \Rightarrow a + a + d = 22 \)
\( \Rightarrow 2a + d = 22 \)
\( \Rightarrow 2(8) + d = 22 \)
\( \Rightarrow d = 22 - 16 = 6 \)
Now, \( a_k = 164 \)
\( \Rightarrow a + (k - 1)d = 164 \)
\( \Rightarrow 8 + (k - 1)6 = 164 \)
\( \Rightarrow (k - 1)6 = 164 - 8 = 156 \)
\( \Rightarrow k - 1 = \frac{156}{6} = 26 \)
\( \Rightarrow k = 26 + 1 = 27 \).

Question. If m Times the mth term of an Arithmetic Progression is equal to n times its nth term and \( m \neq n \), show that the \( (m + n)^{th} \) term of the A.P. is zero.
Answer: Let the first term of given A.P. be '\( a \)' and the common difference be '\( d \)' and \( a_p \) denotes \( p^{th} \) term.
Given: \( m(a_m) = n(a_n) \) [\( m \neq n \)]
To show: \( a_{m+n} = 0 \)
\( m(a_m) = n(a_n) \)
\( \Rightarrow m[a + (m - 1)d] = n[a + (n - 1)d] \)
\( \Rightarrow am + md(m - 1) = an + nd(n - 1) \)
\( \Rightarrow am - an + d(m^2 - m) - d(n^2 - n) = 0 \)
\( \Rightarrow a(m - n) + d[m^2 - n^2 - (m - n)] = 0 \)
\( \Rightarrow a(m - n) + d[(m - n)(m + n) - (m - n)] = 0 \)
\( \Rightarrow a(m - n) + d[(m - n)\{m + n - 1\}] = 0 \)
Since \( m \neq n \), dividing by \( (m - n) \):
\( \Rightarrow a + d[m + n - 1] = 0 \)
\( \Rightarrow a + (m + n - 1)d = 0 \)
\( \Rightarrow a_{m+n} = 0 \)
Hence, proved!

Question. The 6th term of an A.P. is – 10 and its 10th term is – 26. Determine the 15th term of the A.P.
Answer: Let the first term be \( a \) and the common difference be \( d \).
Thus, \( T_6 = a + (6 - 1)d = - 10 \)
\( \Rightarrow a + 5d = - 10 \dots(i) \)
and \( T_{10} = a + (10 - 1)d = - 26 \)
\( \Rightarrow a + 9d = - 26 \dots(ii) \)
Subtracting equation (i) from equation (ii), we get
\( 4d = - 16 \)
\( \Rightarrow d = - 4 \)
From equation (i),
\( a + 5d = - 10 \)
\( \Rightarrow a + 5 (- 4) = - 10 \)
\( \Rightarrow a - 20 = - 10 \)
\( \Rightarrow a = - 10 + 20 = 10 \)
So, \( T_{15} = 10 + (15 - 1) (- 4) \)
\( = 10 - 4(14) \)
\( = 10 - 56 \)
\( = - 46 \).

Question. The 17th term of an A.P. is 5 more than twice its 8th term. If the 11th term of the A.P. is 43, then find its nth term.
Answer: Given, \( T_{11} = 43 \)
\( \Rightarrow a + 10d = 43 \dots(i) \)
And \( T_{17} = 2T_8 + 5 \)
\( \Rightarrow a + 16d = 2(a + 7d) + 5 \)
\( \Rightarrow a + 16d = 2a + 14d + 5 \)
\( \Rightarrow a - 2d = - 5 \)
On multiplying the above equation by 5, we get
\( \Rightarrow 5a - 10d = - 25 \dots(ii) \)
On adding equations (i) and (ii), we have
\( 6a = 18 \)
\( \Rightarrow a = 3 \)
Substituting \( a = 3 \) in equation (i), we get
\( 3 + 10d = 43 \)
\( \Rightarrow 10d = 40 \)
\( \Rightarrow d = 4 \)
Now, \( T_n = 3 + (n - 1)4 \)
\( = 3 + 4n - 4 \)
\( = 4n - 1 \).

Question. Find the number of terms of the A.P. \( 18, 15\frac{1}{2}, 13, \dots, - 49\frac{1}{2} \) and find the sum of all its terms.
Answer: The given A.P. is \( 18, 15\frac{1}{2}, 13, \dots, - 49\frac{1}{2} \).
So, \( a = 18 \)
\( d = 15\frac{1}{2} - 18 = \frac{31}{2} - 18 = \frac{31 - 36}{2} = - \frac{5}{2} \)
and \( T_n = - 49\frac{1}{2} = - \frac{99}{2} \)
Now, \( T_n = a + (n - 1)d \)
\( \Rightarrow - \frac{99}{2} = 18 + (n - 1) \left( - \frac{5}{2} \right) \)
\( \Rightarrow - \frac{99}{2} - 18 = - (n - 1) \frac{5}{2} \)
\( \Rightarrow - 135 = - 5(n - 1) \)
\( \Rightarrow n - 1 = 27 \)
\( \Rightarrow n = 28 \).
Thus, the number of terms in the series are 28.
Now, \( S_{28} = \frac{28}{2} \left\{ 2(18) + (28 - 1) \left( - \frac{5}{2} \right) \right\} \)
\( = 14 \left\{ 36 - \frac{27 \times 5}{2} \right\} \)
\( = 14 \left\{ 36 - \frac{135}{2} \right\} \)
\( = 7 \{ 72 - 135 \} \)
\( = 7 (- 63) = - 441 \).

Question. The sum of the three numbers in an A.P. is 21 and their product is 231. Find the numbers.
Answer: Let the three numbers of the A.P. be \( (a - d), a, (a + d) \).
So, the first term \( (A) = a - d \)
and the common difference \( (D) = d \)
We know, \( S_n = \frac{n}{2} \{2A + (n - 1) D\} \)
\( \Rightarrow S_3 = \frac{3}{2} \{2(a - d) + (3 - 1)d\} \)
\( \Rightarrow 21 = \frac{3}{2} \{2a - 2d + 2d\} = 3a \)
\( \Rightarrow a = 7 \)
Now, \( (a - d) a (a + d) = 231 \)
\( \Rightarrow (7 - d) 7 (7 + d) = 231 \)
\( \Rightarrow 49 - d^2 = 33 \)
\( \Rightarrow d^2 = 16 \)
\( \Rightarrow d = \pm 4 \)
Thus, the numbers are either 3, 7, 11 or 11, 7, 3.

Question. Find the sum of the first 24 terms of an A.P., whose nth term is given by \( a_n = 3 + 2n \).
Answer: Given : \( a_n = 3 + 2n \)
So, \( a_1 = 3 + 2(1) = 5 \)
\( a_2 = 3 + 2(2) = 7 \)
We know, \( a_2 = a + d \)
\( \Rightarrow 7 = 5 + d \)
\( \Rightarrow d = 7 - 5 = 2 \)
Now, sum of first 24 terms of A.P. will be given as,
\( S_{24} = \frac{24}{2} [2a + (24 - 1)d] \)
\( = 12 [2 \times 5 + 23 \times 2] \)
\( = 12 [10 + 46] \)
\( = 12 \times 56 = 672 \).

Question. If the 3rd and 9th term of an A.P. are 4 and – 8 respectively, which term of this A.P. is zero ?
Answer: Given : \( a_3 = 4 \) and \( a_9 = - 8 \)
\( \Rightarrow a + 2d = 4 \) and \( a + 8d = - 8 \)
Subtracting the above equations, we get
\( a + 8d = - 8 \)
\( a + 2d = 4 \)
——————
\( 6d = - 12 \)
\( \Rightarrow d = - 2 \)
We have, \( a + 2d = 4 \)
\( \Rightarrow a + 2 ( - 2) = 4 \)
\( \Rightarrow a - 4 = 4 \)
\( \Rightarrow a = 8 \)
Now, let nth term of this A.P. is zero.
Then,
\( a_n = 0 \)
\( \Rightarrow 8 + (n - 1) (- 2) = 0 \)
\( 8 - 2n + 2 = 0 \)
\( 10 - 2n = 0 \)
\( \Rightarrow n = 5 \)
Thus, 5th term of this A.P. is zero.

Question. Find the sum of the odd numbers between 0 and 50.
Answer: Odd numbers between 0 and 50 are 1, 3, 5, ....... 49.
The above series is an A.P. with \( a = 1, d = 3 - 1 = 2 \), and \( a_n = 49 \).
Now, we know \( a_n = a + (n - 1)d \)
\( \Rightarrow 49 = 1 + (n - 1)2 \)
\( \Rightarrow 48 = (n - 1)2 \)
\( \Rightarrow n - 1 = \frac{48}{2} = 24 \)
\( \Rightarrow n = 25 \)
Sum of n terms of an A.P. is given as,
\( S_n = \frac{n}{2} (a + a_n) \)
\(\therefore S_{25} = \frac{25}{2} ( 1 + 49) \)
\( = \frac{25}{2} \times 50 = 625 \).

Question. The sum of the first 15 terms of an A.P. is 750 and its first term is 15. Find its 20th term.
Answer: Given : \( S_{15} = 750 \) and \( a = 15 \).
Now, we know
\( S_n = \frac{n}{2} [2a + (n - 1)d] \)
\(\therefore 750 = \frac{15}{2} [2 \times 15 + (15 - 1)d] \)
\( \Rightarrow 750 = \frac{15}{2} [30 + 14d] \)
\( \Rightarrow 50 = 15 + 7d \)
\( \Rightarrow 7d = 35 \)
\( \Rightarrow d = 5 \)
So, 20th term, \( a_{20} = a + 19d \)
\( = 15 + 19 \times 5 = 110 \).

Question. An A.P. consist of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Answer: Given : \( n = 50, a_3 = 12 \) and \( a_n = 106 \)
Now, \( a_n = 106 \)
\( \Rightarrow a + (n - 1)d = 106 \)
\( \Rightarrow a + (50 - 1)d = 106 \) [\( \because n = 50 \)]
\( \Rightarrow a + 49d = 106 \dots(i) \)
Also, \( a_3 = 12 \)
\( \Rightarrow a + 2d = 12 \dots(ii) \)
Solving equations (i) and (ii), we get
\( a = 8 \) and \( d = 2 \)
So, 29th term, \( a_{29} = a + 28d \)
\( = 8 + (28 \times 2) = 64 \).

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: Number of trees planted by students of class I = \( 3 \times 1 = 3 \)
Number of trees planted by students of class II = \( 3 \times 2 = 6 \)
Number of trees planted by students of class III = \( 3 \times 3 = 9 \)
Number of trees planted by students of class XII = \( 3 \times 12 = 36 \)
Total number of trees planted = \( 3 + 6 + 9 + \dots + 36 \)
The above series is an A.P. with \( a = 3, d = 3 \) and last term \( l = 36 \).
\(\therefore \text{Sum} = \frac{n}{2} (a + l) \)
\( = \frac{12}{2} (3 + 36) = 234 \)
So, number of plants planted by the students are 234.

Question. Distance travelled by competitior to pick up seventh potato
Answer: \( a_7 = a + (7 - 1)d \)
\( a_7 = 10 + 6 \times 6 = 10 + 36 \)
\( = 46 \text{ m} \)

Question. Find \( S_{10} \)
Answer: We know, \( S_n = \frac{n}{2} [2a + (n - 1)d] \)
\( \therefore S_{10} = \frac{10}{2} [2 \times 10 + (10 - 1)6] \)
\( = 5 [20 + 54] \)
\( = 5 \times 74 = 370 \text{ m} \)

Question. Find \( S_{12} \)
Answer: \( S_{12} = \frac{12}{2} [2 \times 10 + (12 - 1)6] \)
\( = 6 [20 + 66] \)
\( = 6 \times 86 = 516 \text{ m} \)

Jack started giving home tuitions from the month of January. He charges Rs 1500 from each students. He spends Rs 1200 on his rent, Rs 2700 on his food, Rs 980 on taxi fare and Rs 1785 on electricity. In the first month, he had 23 students and in each subsequent months, the number of students increases by 3. Based on the given situation, answer the following questions :

Question. Do his saving form an arithmetic progression? If so, write the first two terms of this arithmetic progression.
Answer: Earnings from home tuitions = Rs1500 times 23) = Rs 34,500 \)
Total expenses = \( Rs (1200 + 2700 + 980 + 1785) = Rs 6,665 \)
Jack's savings in the 1st month = Rs (34,500 - 6,665) = Rs 27,835 \)
Jack's saving in the 2nd month = Rs (27,835 + 3 \times 1500) = Rs (27,835 + 4,500) = Rs 32,335 \)
Jack's saving in the 3rd month = Rs (27,835 + 9,000) = Rs 36,835 \)
In each subsequent months, his savings increases by a fixed amount. Therefore, his savings form an arithmetic progression.
The arithmetic progression is \( 27,835, 32,335, \dots \)

Question. How much did Jack save in the month of September?
Answer: Jack's savings in September will be the 9th term of this arithmetic progression.
9th term = \( 27,835 + 8(4,500) = 27,835 + 36,000 = Rs 63,835 \)
So, he saves \( Rs 63,835 \) in the month of September.

Question. How much money Jack saves till 11th month?
Answer: The money Jack saves till 11th month will be the sum of 11 terms of this arithmetic progression.
\( S_{11} = \frac{11}{2} [2(27,835) + 10(4,500)] = Rs 5,53,685 \)
So, he saves \( Rs 5,53,685 \) till 11th month.

Long Answer Type Questions

Question. Pradeep repays the total loan of \( Rs 1,18,000 \) by paying every month starting with the first instalment of \( Rs 1000 \). He increases the instalment by \( Rs 100 \) every month. What amount will he pay as the last instalment of loan?
Answer: Ist instalment = \( Rs 1000 \)
IInd instalment = \( Rs (1000 + 100) = Rs 1100 \)
IIIrd instalment = \( Rs (1100 + 100) = Rs 1200 \)
Let the number of instalments be \( n \).
\(\therefore \text{Sum} = 1000 + 1100 + 1200 + \dots + \text{nth term} \)
These terms are in A.P. with \( a = 1000, d = 100 \).
\(\therefore \text{Sum} = \frac{n}{2} [2 \times 1000 + (n - 1) 100] \)
\( \Rightarrow \frac{n}{2} (2000 + 100n - 100) = 118000 \)
\( \Rightarrow n(1900 + 100n) = 236000 \)
\( \Rightarrow 100n^2 + 1900n - 236000 = 0 \)
\( \Rightarrow n^2 + 19n - 2360 = 0 \)
\( \Rightarrow n^2 + 59n - 40n - 2360 = 0 \)
\( \Rightarrow (n + 59) (n - 40) = 0 \)
\( \Rightarrow n = 40 \text{ or } n = - 59 \) (Rejected)
\(\therefore\) Last instalment, \( a_{40} = a + 39d \)
\( = 1000 + 39 \times 100 = Rs 4900 \)

Question. Which term of the Arithmetic Progression \( - 7, - 12, - 17, - 22, \dots \) will be \( - 82 \)? Is \( - 100 \) any term of the A.P.? Give reason for your answer.
Answer: \( - 7, - 12, - 17, - 22, \dots \)
Here \( a = - 7, d = - 12 - (- 7) = - 12 + 7 = - 5 \)
Let \( T_n = - 82 \)
\(\therefore T_n = a + (n - 1) d \)
\( - 82 = - 7 + (n - 1) (- 5) \)
\( - 82 = - 7 - 5n + 5 \)
\( - 82 = - 2 - 5n \)
\( - 82 + 2 = - 5n \)
\( - 80 = - 5n \)
\( n = 16 \)
Therefore, 16th term will be \( - 82 \).
Let \( T_n = - 100 \)
Again, \( T_n = a + (n - 1) d \)
\( - 100 = - 7 + (n - 1) (- 5) \)
\( - 100 = - 7 - 5n + 5 \)
\( - 100 = - 2 - 5n \)
\( - 100 + 2 = - 5n \)
\( - 98 = - 5n \)
\( n = \frac{98}{5} \)
But the number of terms can not be in fraction.
So, \( - 100 \) can not be the term of this A.P.

Question. If the sum of first four terms of an A.P. is 40 and that of first 14 terms is 280. Find the sum of its first \( n \) terms.
Answer: Given, \( S_4 = 40 \) and \( S_{14} = 280 \)
If \( a \) be the first term and \( d \) be the common difference of an A.P.
Then, Sum of \( n \) terms \( (S_n) = \frac{n}{2} [2a + (n - 1)d] \)
\( S_4 = \frac{4}{2} [2a + 3d] = 40 \)
\( 2(2a + 3d) = 40 \Rightarrow 2a + 3d = 20 \dots(i) \)
Also, Sum of first 14 terms = 280
\( S_{14} = \frac{14}{2} [2a + 13d] = 280 \)
\( 7(2a + 13d) = 280 \Rightarrow 2a + 13d = 40 \dots(ii) \)
On solving equation (i) and (ii), we get \( a = 7, d = 2 \)
Now, sum of \( n \) terms \( S_n = \frac{n}{2} [2(7) + (n - 1)2] \)
\( S_n = \frac{n}{2} [14 + 2n - 2] \)
\( = \frac{n}{2} [12 + 2n] = n [6 + n] = n^2 + 6n \)
Hence, the sum of first \( n \) terms is \( n^2 + 6n \).

Question. An A.P. consist of 37 terms. The sum of the three middle most terms is 225 and the sum of last three terms is 429. Find the A.P.
Answer: Given : \( n = 37 \)
Since, \( n = 37 \) i.e., odd
\(\therefore \text{Middle most term} = \frac{n + 1}{2} = \frac{37 + 1}{2} = 19\text{th} \)
So, three middle-most terms are 18th, 19th, 20th.
Now, according to the question,
\( a_{18} + a_{19} + a_{20} = 225 \)
\( (a + 17d) + (a + 18d) + (a + 19d) = 225 \)
\( 3a + 54d = 225 \)
\( \Rightarrow a + 18d = 75 \dots(i) \)
Also, \( a_{35} + a_{36} + a_{37} = 429 \)
\( (a + 34d) + (a + 35d) + (a + 36d) = 429 \)
\( 3a + 105d = 429 \)
\( \Rightarrow a + 35d = 143 \dots(ii) \)
Solving equations (i) and (ii), we get \( a = 3 \) and \( d = 4 \)
Thus, the A.P. is \( 3, 7, 11, 15, \dots \)

Question. If the ratio of the sum of first \( n \) terms of two A.P.'s is \( (7n + 1) : (4n + 27) \), find the ratio of their mth terms.
Answer: Let the sum of first \( n \) terms of two A.P.’s be \( S_n \) and \( S'_n \).
Then, \( \frac{S_n}{S'_n} = \frac{\frac{n}{2} [2a + (n - 1)d]}{\frac{n}{2} [2a' + (n - 1)d']} = \frac{7n + 1}{4n + 27} \)
\( \Rightarrow \frac{a + \left( \frac{n-1}{2} \right)d}{a' + \left( \frac{n-1}{2} \right)d'} = \frac{7n + 1}{4n + 27} \dots(i) \)
Also, let mth term of two A.P’s be \( T_m \) and \( T'_m \).
So, \( \frac{T_m}{T'_m} = \frac{a + (m - 1)d}{a' + (m - 1)d'} \)
Replacing \( \frac{n - 1}{2} \) by \( m - 1 \) in equation (i), we get
\( n - 1 = 2(m - 1) \Rightarrow n = 2m - 2 + 1 = 2m - 1 \)
\( \frac{T_m}{T'_m} = \frac{7(2m - 1) + 1}{4(2m - 1) + 27} = \frac{14m - 7 + 1}{8m - 4 + 27} = \frac{14m - 6}{8m + 23} \)
\(\therefore\) Ratio of mth term of two A.P.’s is \( (14m - 6) : (8m + 23) \).

Question. If the ratio of the 11th term of an A.P. to its 18th term is \( 2 : 3 \), find the ratio of the sum of the first five terms to the sum of its first 10 terms.
Answer: Given : \( a_{11} : a_{18} = 2 : 3 \)
\( \Rightarrow \frac{a + 10d}{a + 17d} = \frac{2}{3} \)
\( \Rightarrow 3(a + 10d) = 2(a + 17d) \)
\( \Rightarrow 3a + 30d = 2a + 34d \)
\( \Rightarrow a = 4d \dots(i) \)
Now, \( \frac{S_5}{S_{10}} = \frac{\frac{5}{2} [2a + 4d]}{\frac{10}{2} [2a + 9d]} = \frac{2a + 4d}{2(2a + 9d)} = \frac{2(4d) + 4d}{2(2(4d) + 9d)} \)
[Using (i)]
\( = \frac{8d + 4d}{2(8d + 9d)} = \frac{12d}{2(17d)} = \frac{6d}{17d} = \frac{6}{17} \)
Thus, \( S_5 : S_{10} = 6 : 17 \)

Question. If mth term of an A.P. is \( \frac{1}{n} \) and nth term is \( \frac{1}{m} \), then find the sum of its first \( mn \) terms.
Answer: Let \( a \) and \( d \) be the first term and common difference respectively of the given A. P.
Then \( \frac{1}{n} = a + (m - 1) d \dots(i) \)
and \( \frac{1}{m} = a + (n - 1) d \dots(ii) \)
By subtracting eqn. (ii) from eqn. (i)
\( \frac{1}{n} - \frac{1}{m} = (m - n) d \)
\( \Rightarrow \frac{m - n}{mn} = (m - n) d \)
\( \Rightarrow d = \frac{1}{mn} \)
Putting \( d = \frac{1}{mn} \) in equation (i), we get
\( \frac{1}{n} = a + (m - 1) \frac{1}{mn} \)
\( \Rightarrow \frac{1}{n} = a + \frac{1}{n} - \frac{1}{mn} \)
\( \Rightarrow a = \frac{1}{mn} \)
Now, sum of first mn terms \( S_{mn} = \frac{mn}{2} [2a + (mn - 1) d] \)
\( = \frac{mn}{2} \left[ \frac{2}{mn} + (mn - 1) \frac{1}{mn} \right] = \frac{mn}{2} \left[ \frac{2 + mn - 1}{mn} \right] = \frac{1 + mn}{2} \)

Question. Find the sum of the two middle most terms of the A.P. \( -\frac{4}{3}, -1, -\frac{2}{3}, \dots, 4\frac{1}{3} \).
Answer: Given A.P. is, \( -\frac{4}{3}, -1, -\frac{2}{3}, \dots, \frac{13}{3} \)
Here, \( a = -\frac{4}{3}, d = -1 - (-\frac{4}{3}) = \frac{1}{3} \text{ and } a_n = \frac{13}{3} \)
We know, \( a_n = a + (n - 1)d \)
\( \Rightarrow \frac{13}{3} = -\frac{4}{3} + (n - 1) \frac{1}{3} \)
\( \Rightarrow 13 = - 4 + n - 1 \Rightarrow n = 18 \)
The two middle most terms of the A.P. = \( \left( \frac{n}{2} \right)\text{th}, \left( \frac{n}{2} + 1 \right)\text{th} = \text{9th term, 10th term} \)
So, \( a_9 + a_{10} = a + 8d + a + 9d = 2a + 17d \)
\( = 2(-\frac{4}{3}) + 17(\frac{1}{3}) = -\frac{8}{3} + \frac{17}{3} = \frac{9}{3} = 3 \)

Question. Find the sum of all 3-digit numbers which are multiples of 7.
Answer: 3-digit numbers which are multiple of 7 are \( 105, 112, 119, 126, \dots, 994 \).
The above series is an A.P. So, \( a = 105, d = 7 \) and \( t_n = 994 \)
We know that, \( t_n = a + (n - 1)d \Rightarrow 994 = 105 + (n - 1)7 \Rightarrow 889 = (n - 1)7 \Rightarrow 127 = n - 1 \Rightarrow n = 128 \)
\(\therefore S_{128} = \frac{128}{2} [2(105) + (128 - 1)(7)] = 64 [210 + 889] = 64 \times 1099 = 70336 \).

Question. Find the sum of all three-digit natural numbers, which are divisible by 13.
Answer: Three-digit natural numbers which are divisible by 13 are \( 104, 117, 130, 143, \dots, 988 \).
The above series is an A.P. So, \( a = 104, d = 13 \) and \( t_n = 988 \)
We know that, \( t_n = a + (n - 1)d \Rightarrow 988 = 104 + (n - 1)13 \Rightarrow 76 = 8 + n - 1 \Rightarrow n = 69 \)
\(\therefore S_{69} = \frac{69}{2} [2(104) + (69 - 1)(13)] = 69 [104 + (34)(13)] = 69 [104 + 442] = 69 \times 546 = 37674 \).

Question. Find the number of terms of the A.P. \( 63, 60, 57, \dots \) so that their sum is 693. Explain the double answer.
Answer: Given A.P. is \( 63, 60, 57, \dots \). So, \( a = 63, d = - 3 \), and \( S_n = 693 \)
\( S_n = \frac{n}{2} [2(63) + (n - 1)(- 3)] = 693 \)
\( \Rightarrow n [126 - 3n + 3] = 1386 \)
\( \Rightarrow n [129 - 3n] = 1386 \Rightarrow 3n^2 - 129n + 1386 = 0 \Rightarrow n^2 - 43n + 462 = 0 \)
\( \Rightarrow n^2 - 22n - 21n + 462 = 0 \Rightarrow n(n - 22) - 21(n - 22) = 0 \Rightarrow (n - 22) (n - 21) = 0 \)
\( \Rightarrow n = 21 \text{ or } 22 \)
The double answer is because of the zero. It does not add to the total sum but is definitely a number in the series (22nd term).
\( t_{22} = 63 + (22 - 1) (- 3) = 63 - 63 = 0 \).
i.e., 22th term is zero so sum of 21 terms as well as 22 terms is 693 as 0 will not contribute to the sum.

Question. Find the sum of the following : \( \left(1 - \frac{1}{n}\right) + \left(1 - \frac{2}{n}\right) + \left(1 - \frac{3}{n}\right) + \dots \) upto n terms.
Answer: We have \( \left(1 - \frac{1}{n}\right) + \left(1 - \frac{2}{n}\right) + \left(1 - \frac{3}{n}\right) + \dots + \left(1 - \frac{n}{n}\right) \)
\( = (1 + 1 + 1 + \dots + 1) - \left( \frac{1}{n} + \frac{2}{n} + \frac{3}{n} + \dots + \frac{n}{n} \right) \)
\( = n - \frac{1}{n} [1 + 2 + 3 + \dots + n] = n - \frac{1}{n} \left[ \frac{n(n + 1)}{2} \right] \)
\( = n - \frac{n + 1}{2} = \frac{2n - n - 1}{2} = \frac{n - 1}{2} \)

Question. If the pth term of an A.P. is \( \frac{1}{q} \) and qth term is \( \frac{1}{p} \), prove that the sum of first \( pq \) terms of the A.P. is \( \left( \frac{pq + 1}{2} \right) \).
Answer: Let \( a \) be the first term and \( d \) is common difference.
Then, \( a_p = \frac{1}{q} \Rightarrow a + ( p - 1) d = \frac{1}{q} \dots(i) \)
and \( a_q = \frac{1}{p} \Rightarrow a + (q - 1) d = \frac{1}{p} \dots(ii) \)
Subtracting equation (ii) from equation (i):
\( (p - q)d = \frac{1}{q} - \frac{1}{p} = \frac{p - q}{pq} \Rightarrow d = \frac{1}{pq} \)
Putting the value of \( d \) in equation (i), we get \( a = \frac{1}{pq} \).
\(\therefore S_{pq} = \frac{pq}{2} [2a + ( pq - 1)d] = \frac{pq}{2} \left[ \frac{2}{pq} + \frac{pq - 1}{pq} \right] = \frac{pq}{2} \left[ \frac{2 + pq - 1}{pq} \right] = \frac{pq + 1}{2} \).
Hence Proved.

Question. If the sum of first m terms of an A.P. is the same as the sum of its first n terms, show that the sum of its first \( (m + n) \) terms is zero.
Answer: Let \( a \) be first term and \( d \) be common difference.
Then, \( S_m = S_n \)
\( \Rightarrow \frac{m}{2} [2a + (m - 1)d] = \frac{n}{2} [2a + (n - 1)d] \)
\( \Rightarrow 2am + m(m - 1)d = 2an + n(n - 1)d \)
\( \Rightarrow 2a(m - n) + [m^2 - m - n^2 + n]d = 0 \)
\( \Rightarrow 2a(m - n) + [(m - n)(m + n) - (m - n)]d = 0 \)
Dividing by \( (m - n) \):
\( 2a + (m + n - 1)d = 0 \)
Now, \( S_{m+n} = \frac{m + n}{2} [2a + (m + n - 1)d] = \frac{m + n}{2} [0] = 0 \).

Question. Show that the sum of an A.P. whose first term is \( a \), second term \( b \) and the last term \( c \), is equal to \( \frac{(a + c)(b + c - 2a)}{2(b - a)} \).
Answer: Given : \( a_1 = a; a_2 = b \) and \( a_n = c \)
Now, \( a_2 = a_1 + d \)
\( \Rightarrow b = a + d \)
\( \Rightarrow d = b - a \dots(i) \)
Also, \( a_n = a_1 + (n - 1)d \)
\( \Rightarrow c = a + (n - 1)d \)
\( \Rightarrow c = a + (n - 1) (b - a) \) [Using (i)]
\( \Rightarrow (n - 1) (b - a) = c - a \)
\( \Rightarrow (n - 1) = \frac{c - a}{b - a} \)
\( \Rightarrow n = \frac{c - a}{b - a} + 1 \)
\( \Rightarrow n = \frac{c - a + b - a}{b - a} = \frac{b + c - 2a}{b - a} \dots(ii) \)
We know, sum of \( n \) terms of an A.P. is given as,
\( S_n = \frac{n}{2} [a_1 + a_n] \)
\( = \frac{(b + c - 2a)}{2(b - a)} (a + c) \)
\( = \frac{(a + c)(b + c - 2a)}{2(b - a)} \)
Hence Proved.

Question. If \( a^2, b^2, c^2 \) are in A.P., prove that \( \frac{a}{b + c}, \frac{b}{c + a}, \frac{c}{a + b} \) are in A.P.
Answer: Given, \( a^2, b^2, c^2 \) are in A.P.
\( \therefore b^2 - a^2 = c^2 - b^2 \dots(i) \)
Now, \( \frac{b}{c + a} - \frac{a}{b + c} = \frac{b^2 + bc - ac - a^2}{(a + c)(b + c)} = \frac{(b^2 - a^2) + c(b - a)}{(a + c)(b + c)} = \frac{(b - a)(b + a) + c(b - a)}{(a + c)(b + c)} = \frac{(b - a)(a + b + c)}{(a + c)(b + c)} \dots(ii) \)
Also, \( \frac{c}{a + b} - \frac{b}{c + a} = \frac{c^2 + ac - ab - b^2}{(a + b)(c + a)} = \frac{(c^2 - b^2) + a(c - b)}{(a + b)(c + a)} = \frac{(c - b)(c + b) + a(c - b)}{(a + b)(c + a)} = \frac{(c - b)(a + b + c)}{(a + b)(c + a)} \)
Multiplying numerator and denominator by \( b + c \):
\( = \frac{(c - b)(c + b)(a + b + c)}{(a + b)(b + c)(c + a)} = \frac{(c^2 - b^2)(a + b + c)}{(a + b)(b + c)(c + a)} \)
Using (i):
\( = \frac{(b^2 - a^2)(a + b + c)}{(a + b)(b + c)(c + a)} = \frac{(b - a)(b + a)(a + b + c)}{(a + b)(b + c)(c + a)} = \frac{(b - a)(a + b + c)}{(b + c)(c + a)} \dots(iii) \)
From equations (ii) and (iii), we have
\( \frac{b}{c + a} - \frac{a}{b + c} = \frac{c}{a + b} - \frac{b}{c + a} \)
i.e., \( \frac{a}{b + c}, \frac{b}{c + a}, \frac{c}{a + b} \) are in A.P. Hence Proved.

Question. Find the sum of the integers lying between 1 and 100 (both inclusive) and divisible by 3, 5 or 7.
Answer: Integers divisible by 3 from 1 to 100 are 3, 6, 9,.....99, i.e., total 33 in number.
Integers divisible by 5 are, 5, 10, 15,........100 i.e., total 20 in number.
Integers divisible by 7 are 7, 14,..., 98, i.e., total 14 in number.
Integers divisible by both 3 and 5 are 15, 30........90, i.e., total 6 in number.
Integers divisible by both 3 and 7 are 21, 42, 63 and 84. i.e., total 4 in number.
and Integers divisible by both 5 and 7 are 35, 70 i.e., total 2 in number.
So, sum of numbers divisible by 3, 5 or 7 is
\( = \frac{33}{2} (3 + 99) + \frac{20}{2} (5 + 100) + \frac{14}{2} (7 + 98) - \frac{6}{2} (15 + 90) - \frac{4}{2} (21 + 84) - \frac{2}{2} (35 + 70) \)
\( = \frac{33}{2} \times 102 + \frac{20}{2} \times 105 + \frac{14}{2} \times 105 - \frac{6}{2} \times 105 - \frac{4}{2} \times 105 - \frac{2}{2} \times 105 \)
\( = 33 \times 51 + 105 (10 + 7 - 3 - 2 - 1) \)
\( = 1683 + 105 \times 11 \)
\( = 1683 + 1155 = 2838 \)

Case Study

A man has to take the debt of Rs 3,600 to pay the fees of his child's school. He decided to repay that debt in annual installments. The number of installments will be 40 in numbers. These installments are increasing in a definite amount which forms an A.P. When 30 of installments were paid, he died leaving one-third of the debt unpaid?

Question. If the man unpaid \( \frac{1}{3} \) of his debt, how much amount was paid?
(a) 1200
(b) 2400
(c) 1400
(d) 2000
Answer: (b) Unpaid amount \( = \frac{1}{3} \times 3600 = 1200 \). Then, Paid amount \( = 3600 - 1200 = 2,400 \).

Question. What is the formula for finding the nth term of an A.P.?
(a) \( A + (n - 1)D \)
(b) \( a + nd \)
(c) \( D + (A - 1)n \)
(d) \( A + D \)
Answer: (a)

Question. Find the amount paid in the first installment.
(a) Rs 45
(b) Rs 51
(c) Rs 40
(d) Rs 100
Answer: (b) Here, \( S_n = \frac{n}{2} [2a + (n - 1)d] \). Then, \( 3600 = \frac{40}{2} [2a + (40 - 1)d] \Rightarrow 180 = 2a + 39d \dots(i) \). And \( 2400 = \frac{30}{2} [2a + (30 - 1)d] \Rightarrow 160 = 2a + 29d \dots(ii) \). On solving eq. (i) and (ii), we get \( d = 2 \) and \( a = 51 \).

Question. Find the common difference of the A.P.
(a) Rs 10
(b) Rs 5
(c) Rs 8
(d) Rs 2
Answer: (d) On solving the equations in question above, we get \( d = Rs 2 \).

Question. If the amount left after 30 installments is to be paid in 15 equal installments as decided by two parties mutually. How much amount to be paid in each month?
(a) Rs 60
(b) Rs 80
(c) Rs 100
(d) Rs 90
Answer: (b) Amount left unpaid \( = Rs 1200 \). \( \therefore 15 \text{ equal installments} = \frac{1200}{15} = 80 \). So, Rs 80 will be paid in 15 equal installments.

A company manufactures TV sets. The company is increasing the production of its T.V. sets uniformly by seeing the growth in the demand. If the company produces 8000 sets in the 6th year of manufacturing and 11,300 T.V. sets in its 9th year of manufacturing.

Question. Find the production of T.V. sets in the first year of its manufacturing.
(a) 2000
(b) 2500
(c) 1500
(d) 1800
Answer: (b) Since, the production of T.V. sets are increasing every year. So, it forms an A.P. Here, \( a_6 = 8000 \) and \( a_9 = 11,300 \). Let, first term be '\( a \)' and common difference be '\( d \)'. Then, \( a + 5d = 8,000 \) and \( a + 8d = 11,300 \). Subtracting gives \( 3d = 3,300 \Rightarrow d = 1,100 \). Then \( a = 2,500 \). \(\therefore\) First year production is 2,500 T.V. sets.

Question. What will be the T.V. sets production in the 2nd year?
(a) 3500
(b) 3000
(c) 3800
(d) 3600
Answer: (d) Here, first year production \( = 2,500 \) and production is increasing by 1,100 every year. Then, second year production \( = 2,500 + 1100 = 3,600 \).

Question. After, how many years production will be 5,800?
(a) 6
(b) 5
(c) 4
(d) 3
Answer: (c) Let, after '\( n \)' years production will be 5,800. Then, \( a_n = a + (n - 1)d \Rightarrow 5,800 = 2,500 + (n - 1) \times 1100 \Rightarrow 3300 = (n - 1) \times 1100 \Rightarrow 3 = n - 1 \Rightarrow n = 4 \).

Question. What is the formula for calculating the sum of 'n' terms?
(a) \( \frac{n}{2} [2a + l] \)
(b) \( \frac{n}{2} [a + l] \)
(c) \( a + (n - 1)d \)
(d) \( n[an + 2a] \)
Answer: (b)

Question. In arithmetic progression, the difference between the consecutive terms is same, this constant gap is called :
(a) difference
(b) common difference
(c) gap
(d) constant
Answer: (b)

Assertion and Reasoning Based Questions

DIRECTIONS : In the following questions, a statement 1 is followed by statement 2. Mark the correct choice as :
(A) If both statement 1 and statement 2 are true and statement 2 is the correct explanation of statement 1.
(B) If both statement 1 and statement 2 are true, but statement 2 is not the correct explanation of statement 1.
(C) If statement 1 is true, but statement 2 is false.
(D) If statement 1 is false, but statement 2 is true.

Question. Statement 1 : The given A.P. is \( \frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \dots \). The sum of its 11 terms is \( \frac{33}{20} \). Statement 2 : The sum of 'n' terms of an A.P. is given by \( S_n = \frac{n}{2} [2a + (n - 1)d] \).
Answer: (D) In Statement 1 : \( a = \frac{1}{15}, d = \frac{1}{60} \) and \( n = 11 \). \( S_{11} = \frac{11}{2} [2 \times \frac{1}{15} + (11 - 1) \times \frac{1}{60}] = \frac{11}{2} [\frac{2}{15} + \frac{1}{6}] = \frac{11}{2} [\frac{4 + 5}{30}] = \frac{11}{2} \times \frac{9}{30} = \frac{11}{2} \times \frac{3}{10} = \frac{33}{20} \). Snippet calcs: \( a=1/15, d=1/60 \dots S_{11} = 11/2 [2/15 + 10/60] = 11/2 [4/30 + 5/30] = 11/2 [9/30] = 33/20 \). Wait, OCR calculation in snippet: \( \frac{11}{2} \times \frac{7}{30} = \frac{77}{60} \), thus statement 1 is not true. Statement 2 is true.

Question. Statement 1 : Between 10 and 250, there are 60 multiples of 4. Statement 2 : The formula for calculating number of terms in an A.P. when first and last terms are given, is \( T_n = a + (n - 1)d \).
Answer: (A) In statement 1 : \( T_n = 12 \), \( d = 4 \) and \( T_n = 248 \). We know that \( T_n = a + (n - 1)d \Rightarrow 248 = 12 + (n - 1) \times 4 \Rightarrow 248 = 4n - 4 + 12 \Rightarrow 248 = 4n + 8 \Rightarrow 240 = 4n \Rightarrow n = 60 \). Statement 1 is true and statement 2 is correct explanation.

Archives

Question. Find the 21st term of the A.P. \( -4 \frac{1}{2}, -3, -1 \frac{1}{2}, \dots \)
Answer: \( 25 \frac{1}{2} \)

Question. How many two digits numbers are divisible by 3?
Answer: 30

Question. Find the 9th term from the end (towards the first term) of the A.P. 5, 9, 13, ..., 185.
Answer: 153

Question. If \( S_n \), the sum of first n terms of an A.P. is given by \( S_n = 3n^2 - 4n \), find the nth term.
Answer: \( 6n - 7 \)

Question. Which term of the A.P. 8, 14, 20, 26, .... will by 72 more than its 41st term?
Answer: 53rd term

Question. In an A.P., if \( S_5 + S_7 = 167 \) and \( S_{10} = 235 \), then find the A.P., where \( S_n \) denotes the sum of its first n terms.
Answer: 1, 6, 11, ...

Question. The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.
Answer: Descriptive proof confirmed in text.

Question. Find the middle term of the A.P. 213, 205, 197, ....., 37.
Answer: 125

Question. If the pth term of an A.P. is q and qth term is p, prove that its nth term is \( (p + q - n) \).
Answer: Descriptive proof confirmed in text.

Question. If the sum of the first n terms of an A.P. \( \frac{1}{2} (3n^2 + 7n) \), then find its nth term. Hence write its 20th term.
Answer: \( 3n + 2, 62 \)

Question. The sum of four consecutive numbers in an A.P. is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7 : 15. Find the numbers.
Answer: 2, 6, 10, 14 or 14, 10, 6, 2

Question. Find how many integers between 200 and 500 are divisible by 8.
Answer: 37