OP Malhotra Class 10 Maths Solutions Chapter 9 Arithmetic and Geometric Progression Exercise 9 (A)

S Chand Class 10 ICSE Maths Solutions Chapter 9 Arithmetic and Geometric Progression Ex 9(a)

 

Question 1. Which of the following situations does the list of numbers involved make an arithmetic progression and why?
(i) the amount of air present in a balloon when \( \frac { 1 }{ 5 } \)th of the air remaining in the by a vacuum pump at a time.
(ii) the cost of digging a well after every metre of digging, if it costs Rs. 125 for the first metre and increases by Rs. 65 for every subsequent metre.
Answer:
(i) The given situation does not form an arithmetic progression. When air is removed from a balloon at a constant *fraction* of the remaining air, the amounts form a geometric progression, not an arithmetic one. This is because the amount removed changes each time based on the current quantity.
(ii) Yes, this situation forms an arithmetic progression. The cost of digging for the first metre is Rs. 125, and then it increases by a fixed amount of Rs. 65 for every additional metre. This constant increase means the difference between consecutive terms remains the same.
In simple words: The first part is not an AP because the amount of air removed changes each time. The second part is an AP because the cost goes up by the same fixed amount (Rs. 65) for each extra metre dug.

🎯 Exam Tip: An arithmetic progression (AP) is a sequence where the difference between consecutive terms is constant. Always check if a fixed amount is added or subtracted to determine if a sequence is an AP.

 

Question 2. Write the first five terms of an AP. in which
(i) a = 10, d = 7
(ii) a = 55, d = -4
(iii) a = \( - 1 \), d = \( \frac { 1 }{ 2 } \)
(iv) a = - 3.25, d = – 0.25
Answer:
(i) Given first term \( a = 10 \) and common difference \( d = 7 \).
The first five terms are:
\( a_1 = a = 10 \)
\( a_2 = a + d = 10 + 7 = 17 \)
\( a_3 = a + 2d = 10 + 2(7) = 10 + 14 = 24 \)
\( a_4 = a + 3d = 10 + 3(7) = 10 + 21 = 31 \)
\( a_5 = a + 4d = 10 + 4(7) = 10 + 28 = 38 \)
Thus, the AP is 10, 17, 24, 31, 38.

(ii) Given first term \( a = 55 \) and common difference \( d = -4 \).
The first five terms are:
\( a_1 = a = 55 \)
\( a_2 = a + d = 55 + (-4) = 51 \)
\( a_3 = a + 2d = 55 + 2(-4) = 55 - 8 = 47 \)
\( a_4 = a + 3d = 55 + 3(-4) = 55 - 12 = 43 \)
\( a_5 = a + 4d = 55 + 4(-4) = 55 - 16 = 39 \)
Thus, the AP is 55, 51, 47, 43, 39.

(iii) Given first term \( a = -1 \) and common difference \( d = \frac { 1 }{ 2 } \).
The first five terms are:
\( a_1 = a = -1 \)
\( a_2 = a + d = -1 + \frac { 1 }{ 2 } = -\frac { 1 }{ 2 } \)
\( a_3 = a + 2d = -1 + 2(\frac { 1 }{ 2 }) = -1 + 1 = 0 \)
\( a_4 = a + 3d = -1 + 3(\frac { 1 }{ 2 }) = -1 + \frac { 3 }{ 2 } = \frac { 1 }{ 2 } \)
\( a_5 = a + 4d = -1 + 4(\frac { 1 }{ 2 }) = -1 + 2 = 1 \)
Thus, the AP is \( -1, -\frac { 1 }{ 2 }, 0, \frac { 1 }{ 2 }, 1 \).

(iv) Given first term \( a = -3.25 \) and common difference \( d = -0.25 \).
The first five terms are:
\( a_1 = a = -3.25 \)
\( a_2 = a + d = -3.25 + (-0.25) = -3.50 \)
\( a_3 = a + 2d = -3.25 + 2(-0.25) = -3.25 - 0.50 = -3.75 \)
\( a_4 = a + 3d = -3.25 + 3(-0.25) = -3.25 - 0.75 = -4.00 \)
\( a_5 = a + 4d = -3.25 + 4(-0.25) = -3.25 - 1.00 = -4.25 \)
Thus, the AP is -3.25, -3.50, -3.75, -4.00, -4.25.
In simple words: To find the terms of an arithmetic progression (AP), start with the first term 'a' and keep adding the common difference 'd' to get the next term. For example, the second term is a+d, the third term is a+2d, and so on.

🎯 Exam Tip: Remember that a common difference 'd' can be positive, negative, or even zero. Pay careful attention to the signs when adding or subtracting to avoid calculation errors.

 

Question 3. Choose the correct answer in the following:
(i) the list of numbers -12, -9, -6, -3, 0,3 is
(a) not an AP
(b) an AP with d = 3
(c) an AP with d = 0
(d) an AP with d- 3
(ii) In an AP if a = 3, d = 0 and n = 7, then an will be
(a) 4
(b) 1
(c) 3
(d) 2
Answer:
(i) Let's find the difference between consecutive terms:
\( -9 - (-12) = -9 + 12 = 3 \)
\( -6 - (-9) = -6 + 9 = 3 \)
\( -3 - (-6) = -3 + 6 = 3 \)
\( 0 - (-3) = 0 + 3 = 3 \)
\( 3 - 0 = 3 \)
Since the common difference \( d = 3 \) is constant, the list of numbers is an AP with \( d = 3 \).
(b) an AP with d = 3

(ii) The formula for the nth term of an AP is \( a_n = a + (n - 1)d \).
Given \( a = 3, d = 0, n = 7 \).
Substitute these values into the formula:
\( a_7 = 3 + (7 - 1)(0) \)
\( a_7 = 3 + (6)(0) \)
\( a_7 = 3 + 0 \)
\( a_7 = 3 \)
(c) 3
In simple words: For part (i), we check if the numbers increase by the same amount each time. Since they all go up by 3, it's an AP with a common difference of 3. For part (ii), if the common difference is 0, it means all terms are the same as the first term, so the 7th term is still 3.

🎯 Exam Tip: For MCQs involving APs, always test the common difference by subtracting consecutive terms. When \( d = 0 \), every term in the AP is the same as the first term.

 

Question 4. What is the:
(i) 25th term of the A.P.: \( - 5, - \frac { 5 }{ 2 }, 0, \frac { 5 }{ 2 }, ... \)
(ii) 30th term of the A.P.: 27, 22, 17, ...
(iii) 10th term of the A.P.: -0.1, -0.2, -0.3,...
Answer:
(i) For the AP: \( - 5, - \frac { 5 }{ 2 }, 0, \frac { 5 }{ 2 }, ... \)
The first term \( a = -5 \).
The common difference \( d = - \frac { 5 }{ 2 } - (-5) = - \frac { 5 }{ 2 } + 5 = - \frac { 5 }{ 2 } + \frac { 10 }{ 2 } = \frac { 5 }{ 2 } \).
We need to find the 25th term, so \( n = 25 \).
Using the formula \( a_n = a + (n - 1)d \):
\( a_{25} = -5 + (25 - 1)(\frac { 5 }{ 2 }) \)
\( a_{25} = -5 + (24)(\frac { 5 }{ 2 }) \)
\( a_{25} = -5 + 12 \times 5 \)
\( a_{25} = -5 + 60 \)
\( a_{25} = 55 \)

(ii) For the AP: 27, 22, 17, ...
The first term \( a = 27 \).
The common difference \( d = 22 - 27 = -5 \).
We need to find the 30th term, so \( n = 30 \).
Using the formula \( a_n = a + (n - 1)d \):
\( a_{30} = 27 + (30 - 1)(-5) \)
\( a_{30} = 27 + (29)(-5) \)
\( a_{30} = 27 - 145 \)
\( a_{30} = -118 \)

(iii) For the AP: -0.1, -0.2, -0.3,...
The first term \( a = -0.1 \).
The common difference \( d = -0.2 - (-0.1) = -0.2 + 0.1 = -0.1 \).
We need to find the 10th term, so \( n = 10 \).
Using the formula \( a_n = a + (n - 1)d \):
\( a_{10} = -0.1 + (10 - 1)(-0.1) \)
\( a_{10} = -0.1 + (9)(-0.1) \)
\( a_{10} = -0.1 - 0.9 \)
\( a_{10} = -1 \)
In simple words: To find any term in an AP, you need the first term (a) and the common difference (d). The formula \( a_n = a + (n-1)d \) helps you calculate the 'nth' term by adding 'd' to 'a' \( (n-1) \) times.

🎯 Exam Tip: Be careful with negative numbers and fractions when calculating the common difference or applying the formula. A common mistake is getting the sign wrong for 'd'.

 

Question 5. If k, 2k – 1 and 2k + 1 are the three consecutive terms of an AP, then find the value of k.
Answer: If k, \( 2k - 1 \), and \( 2k + 1 \) are consecutive terms of an AP, then the common difference between them must be the same. This means the second term minus the first term should equal the third term minus the second term.
First common difference: \( d_1 = (2k - 1) - k = k - 1 \)
Second common difference: \( d_2 = (2k + 1) - (2k - 1) = 2k + 1 - 2k + 1 = 2 \)
Since \( d_1 = d_2 \):
\( k - 1 = 2 \)
\( k = 2 + 1 \)
\( k = 3 \)
In simple words: In an AP, the difference between any two terms next to each other is always the same. So, we set the difference between the first two terms equal to the difference between the next two terms to find 'k'.

🎯 Exam Tip: The key property of an AP is that the common difference is constant. Use this property \( (a_2 - a_1 = a_3 - a_2) \) to solve problems involving unknown terms.

 

Question 6. What is the common difference of the A.P. \( \frac{1}{3 q}, \frac{1-6 q}{3 q}, \frac{1-12 q}{3 q} \).
Answer: Given the AP terms are \( \frac{1}{3 q}, \frac{1-6 q}{3 q}, \frac{1-12 q}{3 q} \).
The common difference 'd' can be found by subtracting any term from its succeeding term.
Let's use the first two terms:
\( d = \frac{1-6 q}{3 q} - \frac{1}{3 q} \)
Since the denominators are already the same, we can combine the numerators:
\( d = \frac{(1 - 6q) - 1}{3 q} \)
\( d = \frac{1 - 6q - 1}{3 q} \)
\( d = \frac{-6 q}{3 q} \)
We can cancel out \( q \) (assuming \( q \ne 0 \)) and simplify the numbers:
\( d = -2 \)
In simple words: To find the common difference, subtract the first term from the second term. Since the bottoms of the fractions are the same, just subtract the top parts, and then simplify the result.

🎯 Exam Tip: When terms are fractions, finding a common denominator (if not already present) is crucial before subtracting to find the common difference. Always simplify the final expression if possible.

 

Question 7. Which term of AP 5, 13, 21, ... is 181?
Answer: Given the AP: 5, 13, 21, ...
First term \( a = 5 \).
Common difference \( d = 13 - 5 = 8 \).
Let the nth term \( a_n \) be 181.
Using the formula \( a_n = a + (n - 1)d \):
\( 181 = 5 + (n - 1)8 \)
Subtract 5 from both sides:
\( 181 - 5 = (n - 1)8 \)
\( 176 = (n - 1)8 \)
Divide by 8:
\( \frac { 176 }{ 8 } = n - 1 \)
\( 22 = n - 1 \)
Add 1 to both sides:
\( n = 22 + 1 \)
\( n = 23 \)
So, 181 is the 23rd term of the AP. Finding 'n' is a common type of problem in APs.
In simple words: We know the start, the step size (common difference), and the target number. We use a formula to figure out how many steps it takes to reach that target number from the start.

🎯 Exam Tip: Clearly identify 'a', 'd', and 'a_n' before plugging them into the formula \( a_n = a + (n-1)d \). Be careful with the algebraic rearrangement to solve for 'n'.

 

Question 8. Which term of the A.P. 3, 10, 17,... will be 84 more than its 13th term?
Answer: Given the AP: 3, 10, 17,...
First term \( a = 3 \).
Common difference \( d = 10 - 3 = 7 \).
First, let's find the 13th term, \( a_{13} \).
Using \( a_n = a + (n - 1)d \):
\( a_{13} = 3 + (13 - 1)7 \)
\( a_{13} = 3 + (12)7 \)
\( a_{13} = 3 + 84 \)
\( a_{13} = 87 \)
Now, we need to find which term, let's say \( a_n \), is 84 more than \( a_{13} \).
So, \( a_n = a_{13} + 84 \)
\( a_n = 87 + 84 \)
\( a_n = 171 \)
Now we use the formula \( a_n = a + (n - 1)d \) again to find 'n' for \( a_n = 171 \).
\( 171 = 3 + (n - 1)7 \)
Subtract 3 from both sides:
\( 168 = (n - 1)7 \)
Divide by 7:
\( \frac { 168 }{ 7 } = n - 1 \)
\( 24 = n - 1 \)
Add 1 to both sides:
\( n = 24 + 1 \)
\( n = 25 \)
So, the 25th term will be 84 more than its 13th term. This involves calculating two terms of the AP.
In simple words: First, we find the 13th number in the pattern. Then, we add 84 to that number to get our new target. Finally, we find out which position that new target number is in the original pattern.

🎯 Exam Tip: Break down complex problems into smaller, manageable steps. First find the specific term, then apply the condition (e.g., "84 more than"), and finally solve for the unknown term number 'n'.

 

Question 9. Find the AP if the 6th term of the A.P. is 19 and the 16th term is 15 more than the 11th term.
Answer: Let the first term of the AP be 'a' and the common difference be 'd'.
The nth term of an AP is given by \( a_n = a + (n - 1)d \).
Given that the 6th term is 19:
\( a_6 = a + (6 - 1)d \)
\( 19 = a + 5d \) ... (i)
Given that the 16th term is 15 more than the 11th term:
\( a_{16} = a_{11} + 15 \)
Using the formula for \( a_{16} \) and \( a_{11} \):
\( a + (16 - 1)d = a + (11 - 1)d + 15 \)
\( a + 15d = a + 10d + 15 \)
Subtract 'a' from both sides:
\( 15d = 10d + 15 \)
Subtract \( 10d \) from both sides:
\( 15d - 10d = 15 \)
\( 5d = 15 \)
Divide by 5:
\( d = \frac { 15 }{ 5 } \)
\( d = 3 \)
Now, substitute the value of \( d = 3 \) back into equation (i):
\( 19 = a + 5(3) \)
\( 19 = a + 15 \)
Subtract 15 from both sides:
\( a = 19 - 15 \)
\( a = 4 \)
So, the first term is 4 and the common difference is 3. We can now find the AP.
The AP is \( a, a+d, a+2d, a+3d, ... \)
AP: 4, \( 4+3 \), \( 4+2(3) \), \( 4+3(3) \), ...
AP: 4, 7, 10, 13, 16, ...
In simple words: We are given clues about two terms of the AP. We write these clues as equations using 'a' (first term) and 'd' (common difference). By solving these equations together, we find 'a' and 'd', which then lets us write out the AP.

🎯 Exam Tip: Problems involving multiple terms of an AP usually require setting up a system of linear equations using the general term formula \( a_n = a + (n-1)d \). Solving these equations simultaneously will give you 'a' and 'd'.

 

Question 10. The sum of the 2nd and 7th terms of an A.P. is 30. If its 15th term is 1 less than twice its 8th term, find the AP.
Answer: Let the first term of the AP be 'a' and the common difference be 'd'.
The general term of an AP is \( a_n = a + (n - 1)d \).
Given that the sum of the 2nd and 7th terms is 30:
\( a_2 + a_7 = 30 \)
\( (a + (2 - 1)d) + (a + (7 - 1)d) = 30 \)
\( (a + d) + (a + 6d) = 30 \)
\( 2a + 7d = 30 \) ... (i)
Given that the 15th term is 1 less than twice its 8th term:
\( a_{15} = 2a_8 - 1 \)
\( a + (15 - 1)d = 2(a + (8 - 1)d) - 1 \)
\( a + 14d = 2(a + 7d) - 1 \)
\( a + 14d = 2a + 14d - 1 \)
Subtract \( 14d \) from both sides:
\( a = 2a - 1 \)
Subtract 'a' from both sides:
\( 0 = a - 1 \)
\( a = 1 \)
Now substitute \( a = 1 \) into equation (i):
\( 2(1) + 7d = 30 \)
\( 2 + 7d = 30 \)
Subtract 2 from both sides:
\( 7d = 30 - 2 \)
\( 7d = 28 \)
Divide by 7:
\( d = \frac { 28 }{ 7 } \)
\( d = 4 \)
So, the first term \( a = 1 \) and the common difference \( d = 4 \).
The AP is \( a, a+d, a+2d, a+3d, ... \)
AP: 1, \( 1+4 \), \( 1+2(4) \), \( 1+3(4) \), ...
AP: 1, 5, 9, 13, 17, ...
In simple words: We used the given information to create two equations with 'a' and 'd'. First, we set up an equation for the sum of two terms. Then, we set up another equation based on how the 15th term relates to the 8th term. Solving these helped us find 'a' and 'd', which are needed to write the AP.

🎯 Exam Tip: Carefully translate the verbal descriptions into algebraic equations. "1 less than twice its 8th term" means \( 2a_8 - 1 \), not \( 2(a_8 - 1) \). Precision in forming equations is vital.

 

Question 11. The fourth term of an A.P. is 11. The sum of the fifth and seventh terms of the A.P. is 34, find its common difference.
Answer: Let the first term of the AP be 'a' and the common difference be 'd'.
The general term of an AP is \( a_n = a + (n - 1)d \).
Given that the fourth term (\( a_4 \)) is 11:
\( a_4 = a + (4 - 1)d \)
\( a + 3d = 11 \) ... (i)
Given that the sum of the fifth (\( a_5 \)) and seventh (\( a_7 \)) terms is 34:
\( a_5 + a_7 = 34 \)
\( (a + (5 - 1)d) + (a + (7 - 1)d) = 34 \)
\( (a + 4d) + (a + 6d) = 34 \)
\( 2a + 10d = 34 \)
We can divide the entire equation by 2 to simplify:
\( a + 5d = 17 \) ... (ii)
Now we have a system of two linear equations:
1) \( a + 3d = 11 \)
2) \( a + 5d = 17 \)
Subtract equation (i) from equation (ii):
\( (a + 5d) - (a + 3d) = 17 - 11 \)
\( a + 5d - a - 3d = 6 \)
\( 2d = 6 \)
Divide by 2:
\( d = \frac { 6 }{ 2 } \)
\( d = 3 \)
The common difference of the AP is 3. We didn't even need to find 'a' here, as the question only asked for 'd'.
In simple words: We used the given information about the 4th term and the sum of the 5th and 7th terms to create two equations. By solving these two equations together, we could directly find the common difference 'd'.

🎯 Exam Tip: When asked for only one unknown (like 'd' in this case), look for opportunities to eliminate the other unknown ('a') through subtraction or substitution, which can sometimes save steps.

 

Question 12. Find the middle term of the A.P. 213,205, 197, ..., 37.
Answer: Given the AP: 213, 205, 197, ..., 37.
First term \( a = 213 \).
Common difference \( d = 205 - 213 = -8 \).
Last term \( a_n = 37 \).
First, we need to find the number of terms 'n' in this AP.
Using the formula \( a_n = a + (n - 1)d \):
\( 37 = 213 + (n - 1)(-8) \)
\( 37 = 213 - 8n + 8 \)
\( 37 = 221 - 8n \)
Subtract 221 from both sides:
\( 37 - 221 = -8n \)
\( -184 = -8n \)
Divide by -8:
\( n = \frac { -184 }{ -8 } \)
\( n = 23 \)
Since 'n' is an odd number (23), there is exactly one middle term. The position of the middle term is given by \( \frac { n + 1 }{ 2 } \).
Middle term position \( = \frac { 23 + 1 }{ 2 } = \frac { 24 }{ 2 } = 12 \).
So, we need to find the 12th term (\( a_{12} \)).
\( a_{12} = a + (12 - 1)d \)
\( a_{12} = 213 + (11)(-8) \)
\( a_{12} = 213 - 88 \)
\( a_{12} = 125 \)
The middle term of the AP is 125. Finding the number of terms first is a crucial step.
In simple words: First, we count how many numbers are in the list. Since there's an odd number of terms, there's a single middle number. We find its position by adding 1 to the total count and dividing by 2. Then, we use the AP formula to find the value of the number at that middle position.

🎯 Exam Tip: To find the middle term(s) of an AP, always first determine the total number of terms ('n'). If 'n' is odd, there's one middle term at position \( \frac{n+1}{2} \). If 'n' is even, there are two middle terms at positions \( \frac{n}{2} \) and \( \frac{n}{2} + 1 \).

 

Question 13. If the 3rd and 9th terms of an A.P. are 4 and – 8 respectively, then which term of this A.P. is zero?
Answer: Let the first term of the AP be 'a' and the common difference be 'd'.
The general term of an AP is \( a_n = a + (n - 1)d \).
Given that the 3rd term (\( a_3 \)) is 4:
\( a_3 = a + (3 - 1)d \)
\( a + 2d = 4 \) ... (i)
Given that the 9th term (\( a_9 \)) is -8:
\( a_9 = a + (9 - 1)d \)
\( a + 8d = -8 \) ... (ii)
Now we have a system of two linear equations:
1) \( a + 2d = 4 \)
2) \( a + 8d = -8 \)
Subtract equation (i) from equation (ii):
\( (a + 8d) - (a + 2d) = -8 - 4 \)
\( a + 8d - a - 2d = -12 \)
\( 6d = -12 \)
Divide by 6:
\( d = \frac { -12 }{ 6 } \)
\( d = -2 \)
Now substitute the value of \( d = -2 \) back into equation (i):
\( a + 2(-2) = 4 \)
\( a - 4 = 4 \)
Add 4 to both sides:
\( a = 4 + 4 \)
\( a = 8 \)
So, the first term \( a = 8 \) and the common difference \( d = -2 \).
Now we need to find which term 'n' has a value of 0, i.e., \( a_n = 0 \).
Using the formula \( a_n = a + (n - 1)d \):
\( 0 = 8 + (n - 1)(-2) \)
Subtract 8 from both sides:
\( -8 = (n - 1)(-2) \)
Divide by -2:
\( \frac { -8 }{ -2 } = n - 1 \)
\( 4 = n - 1 \)
Add 1 to both sides:
\( n = 4 + 1 \)
\( n = 5 \)
Therefore, the 5th term of the AP is zero. This problem combines finding 'a' and 'd' with finding 'n'.
In simple words: First, we used the 3rd and 9th terms to figure out the starting number ('a') and how much the numbers change each time ('d'). Then, we used these to find out which term in the sequence would eventually become zero.

🎯 Exam Tip: A negative common difference means the terms of the AP are decreasing. Always set up two equations for 'a' and 'd' when given two terms, then solve them simultaneously. Finding 'n' when \( a_n = 0 \) is a standard application.

 

Question 14. The 8th term of an A.P. is zero. Prove that its 38th term is three times its 18th term.
Answer: Let the first term of the AP be 'a' and the common difference be 'd'.
The general term of an AP is \( a_n = a + (n - 1)d \).
Given that the 8th term (\( a_8 \)) is 0:
\( a_8 = a + (8 - 1)d \)
\( 0 = a + 7d \)
From this, we can express 'a' in terms of 'd':
\( a = -7d \) ... (i)
Now, let's find the 18th term (\( a_{18} \)):
\( a_{18} = a + (18 - 1)d \)
\( a_{18} = a + 17d \)
Substitute \( a = -7d \) from (i) into the expression for \( a_{18} \):
\( a_{18} = -7d + 17d \)
\( a_{18} = 10d \) ... (ii)
Next, let's find the 38th term (\( a_{38} \)):
\( a_{38} = a + (38 - 1)d \)
\( a_{38} = a + 37d \)
Substitute \( a = -7d \) from (i) into the expression for \( a_{38} \):
\( a_{38} = -7d + 37d \)
\( a_{38} = 30d \) ... (iii)
Now, we need to prove that \( a_{38} = 3 \times a_{18} \).
From (ii), \( a_{18} = 10d \).
From (iii), \( a_{38} = 30d \).
We can write \( 30d \) as \( 3 \times 10d \).
\( a_{38} = 3 \times (10d) \)
\( a_{38} = 3 \times a_{18} \)
Hence, it is proven that the 38th term is three times its 18th term. This type of proof highlights the linear relationship between terms in an AP.
In simple words: We are told that the 8th number in the pattern is zero. This helps us find a link between the first number 'a' and the common difference 'd'. Then, we use this link to show that the 38th number is three times bigger than the 18th number in the same pattern.

🎯 Exam Tip: For proof questions, start by expressing all given terms and terms to be proven in terms of 'a' and 'd'. Use any given conditions to simplify the expressions (like \( a = -7d \) here) and then substitute to show the required relationship.

 

Question 15. Which term of the A.P. 120, 116, 112,... is its first negative term?
Answer: Given the AP: 120, 116, 112,...
First term \( a = 120 \).
Common difference \( d = 116 - 120 = -4 \).
We need to find the first term \( a_n \) that is negative, which means \( a_n < 0 \).
Using the formula \( a_n = a + (n - 1)d \):
\( a + (n - 1)d < 0 \)
\( 120 + (n - 1)(-4) < 0 \)
\( 120 - 4n + 4 < 0 \)
\( 124 - 4n < 0 \)
Now, we need to solve this inequality for 'n'. Add \( 4n \) to both sides:
\( 124 < 4n \)
Divide by 4:
\( \frac { 124 }{ 4 } < n \)
\( 31 < n \)
This means 'n' must be greater than 31. The smallest integer value for 'n' that is greater than 31 is 32.
So, the 32nd term is the first negative term. This involves solving an inequality.
In simple words: We want to find when the numbers in the pattern first drop below zero. We set up an equation that says the 'nth' term is less than zero. Solving this helps us find the smallest term number 'n' that makes the term negative.

🎯 Exam Tip: To find the first negative term, set \( a_n < 0 \) and solve for 'n' as an inequality. Remember that 'n' must be a whole number, so if \( n > X \), then 'n' is the smallest integer greater than X.

 

Question 16. How many 3-digit natural numbers are divisible by 7?
Answer: Three-digit natural numbers range from 100 to 999.
We need to find the numbers in this range that are divisible by 7.
First, find the smallest 3-digit number divisible by 7:
Divide 100 by 7: \( 100 \div 7 = 14 \) with a remainder of 2.
So, \( 100 - 2 = 98 \) is divisible by 7. The next number will be \( 98 + 7 = 105 \).
So, the first term of our AP is \( a = 105 \).
Next, find the largest 3-digit number divisible by 7:
Divide 999 by 7: \( 999 \div 7 = 142 \) with a remainder of 5.
So, \( 999 - 5 = 994 \) is divisible by 7.
So, the last term of our AP is \( a_n = 994 \).
The common difference \( d \) for numbers divisible by 7 is 7.
Now, use the formula \( a_n = a + (n - 1)d \) to find the number of terms 'n':
\( 994 = 105 + (n - 1)7 \)
Subtract 105 from both sides:
\( 994 - 105 = (n - 1)7 \)
\( 889 = (n - 1)7 \)
Divide by 7:
\( \frac { 889 }{ 7 } = n - 1 \)
\( 127 = n - 1 \)
Add 1 to both sides:
\( n = 127 + 1 \)
\( n = 128 \)
Therefore, there are 128 three-digit natural numbers divisible by 7. This is a practical application of arithmetic progressions.
In simple words: We first found the smallest three-digit number that 7 can divide evenly. Then, we found the largest three-digit number that 7 can divide evenly. These two numbers become the start and end of our number pattern. Since the step size is 7, we used a formula to count how many numbers are in this pattern.

🎯 Exam Tip: To solve problems about numbers divisible by a certain integer within a range, first find the smallest and largest numbers in that range that meet the divisibility condition. These will be your 'a' and 'a_n' for the AP, with 'd' being the divisor itself.

 

Question 17. Two APs have the same common difference. The first term of one of these is – 1 and that of the other is – 8. What is the difference between their 4th terms?
Answer: Let the first AP be denoted by \( AP_1 \) and the second AP by \( AP_2 \).
Let the common difference for both APs be 'd'.
For \( AP_1 \):
First term \( a_1 = -1 \).
The 4th term of \( AP_1 \) is \( A_4 = a_1 + (4 - 1)d = -1 + 3d \).
For \( AP_2 \):
First term \( a_2 = -8 \).
The 4th term of \( AP_2 \) is \( B_4 = a_2 + (4 - 1)d = -8 + 3d \).
We need to find the difference between their 4th terms, which is \( A_4 - B_4 \).
\( A_4 - B_4 = (-1 + 3d) - (-8 + 3d) \)
\( A_4 - B_4 = -1 + 3d + 8 - 3d \)
\( A_4 - B_4 = -1 + 8 \)
\( A_4 - B_4 = 7 \)
The difference between their 4th terms is 7. Interestingly, when the common differences are the same, the difference between any corresponding terms will also be constant and equal to the difference between their first terms.
In simple words: We have two lists of numbers that grow at the same pace. The first number of one list is -1, and the first number of the other is -8. We need to find the difference between the fourth numbers in each list. Because they grow at the same pace, the difference between their fourth numbers will be the same as the difference between their first numbers.

🎯 Exam Tip: If two APs have the same common difference 'd', then the difference between their nth terms \( (a_n - b_n) \) will always be equal to the difference between their first terms \( (a_1 - b_1) \), regardless of 'n'. This can save calculation time.

 

Question 18. If an AP, ratio of the 4th and 9th terms is 1 : 3, find the ratio of 12 term and 5th term?
Answer: Let the first term of the AP be 'a' and the common difference be 'd'.
The general term of an AP is \( a_n = a + (n - 1)d \).
Given that the ratio of the 4th term (\( a_4 \)) and 9th term (\( a_9 \)) is 1 : 3.
\( \frac { a_4 }{ a_9 } = \frac { 1 }{ 3 } \)
Substitute the formula for \( a_4 \) and \( a_9 \):
\( \frac { a + (4 - 1)d }{ a + (9 - 1)d } = \frac { 1 }{ 3 } \)
\( \frac { a + 3d }{ a + 8d } = \frac { 1 }{ 3 } \)
Cross-multiply:
\( 3(a + 3d) = 1(a + 8d) \)
\( 3a + 9d = a + 8d \)
Rearrange the terms to find a relationship between 'a' and 'd':
\( 3a - a = 8d - 9d \)
\( 2a = -d \)
\( d = -2a \) ... (i)
Now, we need to find the ratio of the 12th term (\( a_{12} \)) and the 5th term (\( a_5 \)).
\( a_{12} = a + (12 - 1)d = a + 11d \)
\( a_5 = a + (5 - 1)d = a + 4d \)
Substitute \( d = -2a \) from (i) into these expressions:
For \( a_{12} \):
\( a_{12} = a + 11(-2a) \)
\( a_{12} = a - 22a \)
\( a_{12} = -21a \)
For \( a_5 \):
\( a_5 = a + 4(-2a) \)
\( a_5 = a - 8a \)
\( a_5 = -7a \)
Now, find the ratio \( \frac { a_{12} }{ a_5 } \):
\( \frac { a_{12} }{ a_5 } = \frac { -21a }{ -7a } \)
Assuming \( a \ne 0 \), we can cancel 'a' from numerator and denominator:
\( \frac { a_{12} }{ a_5 } = \frac { -21 }{ -7 } \)
\( \frac { a_{12} }{ a_5 } = 3 \)
So, the ratio of the 12th term and 5th term is 3 : 1. This problem demonstrates solving for relationships between 'a' and 'd'.
In simple words: We are given how the 4th and 9th numbers in the pattern compare to each other. We use this information to find a link between the starting number and the step size. Then, we use that link to compare the 12th and 5th numbers in the pattern.

🎯 Exam Tip: When given ratios of terms, always set up equations relating 'a' and 'd'. Express one variable in terms of the other (e.g., \( d = -2a \)) and then substitute this relationship into the expressions for the terms whose ratio you need to find.

 

Question 19. Find the 7th term from the end of the A.P. 7, 10, 13, ..., 184.
Answer: Given the AP: 7, 10, 13, ..., 184.
First term \( a = 7 \).
Common difference \( d = 10 - 7 = 3 \).
Last term \( a_n = 184 \).
First, we need to find the total number of terms 'n' in the AP.
Using the formula \( a_n = a + (n - 1)d \):
\( 184 = 7 + (n - 1)3 \)
Subtract 7 from both sides:
\( 177 = (n - 1)3 \)
Divide by 3:
\( \frac { 177 }{ 3 } = n - 1 \)
\( 59 = n - 1 \)
Add 1 to both sides:
\( n = 59 + 1 \)
\( n = 60 \)
There are 60 terms in this AP. Now, we need to find the 7th term from the end.
The 7th term from the end of an AP with 'n' terms is the \( (n - 7 + 1) \)th term from the beginning.
Position from beginning \( = 60 - 7 + 1 = 54 \).
So, we need to find the 54th term (\( a_{54} \)).
Using the formula \( a_n = a + (n - 1)d \):
\( a_{54} = 7 + (54 - 1)3 \)
\( a_{54} = 7 + (53)3 \)
\( a_{54} = 7 + 159 \)
\( a_{54} = 166 \)
Therefore, the 7th term from the end of the AP is 166. This method works by converting the "from the end" position to a "from the beginning" position.
In simple words: First, we count how many numbers are in the whole list. Then, to find the 7th number from the end, we figure out which number it is when counted from the beginning of the list. Finally, we use the AP formula to find the value of that number.

🎯 Exam Tip: To find the \( k \)th term from the end of an AP with 'n' terms, calculate it as the \( (n - k + 1) \)th term from the beginning. Alternatively, you can reverse the AP and find the \( k \)th term of the new AP (where the last term becomes the first, and 'd' becomes '-d').

 

Question 20. The four angles of a quadrilateral form an A.P. If the sum of the first three angles is twice the fourth angle, then find all the angles.
Answer: Let the four angles of the quadrilateral in AP be \( a - 3d, a - d, a + d, a + 3d \). (Using this form simplifies calculations as the common difference is 2d).
The sum of the angles of a quadrilateral is \( 360^\circ \).
So, \( (a - 3d) + (a - d) + (a + d) + (a + 3d) = 360^\circ \)
\( 4a = 360^\circ \)
\( a = \frac { 360^\circ }{ 4 } \)
\( a = 90^\circ \)
The angles are \( 90^\circ - 3d, 90^\circ - d, 90^\circ + d, 90^\circ + 3d \).

Given that the sum of the first three angles is twice the fourth angle:
\( (a - 3d) + (a - d) + (a + d) = 2(a + 3d) \)
\( 3a - 3d = 2a + 6d \)
Substitute \( a = 90^\circ \):
\( 3(90^\circ) - 3d = 2(90^\circ) + 6d \)
\( 270^\circ - 3d = 180^\circ + 6d \)
Rearrange the terms:
\( 270^\circ - 180^\circ = 6d + 3d \)
\( 90^\circ = 9d \)
\( d = \frac { 90^\circ }{ 9 } \)
\( d = 10^\circ \)
Now we have \( a = 90^\circ \) and \( d = 10^\circ \). The common difference between consecutive terms in the list \( a-3d, a-d, a+d, a+3d \) is \( (a-d) - (a-3d) = 2d \). So if 'd' is \( 10^\circ \), the step is \( 20^\circ \).
The four angles are:
First angle: \( a - 3d = 90^\circ - 3(10^\circ) = 90^\circ - 30^\circ = 60^\circ \)
Second angle: \( a - d = 90^\circ - 10^\circ = 80^\circ \)
Third angle: \( a + d = 90^\circ + 10^\circ = 100^\circ \)
Fourth angle: \( a + 3d = 90^\circ + 3(10^\circ) = 90^\circ + 30^\circ = 120^\circ \)
The angles are \( 60^\circ, 80^\circ, 100^\circ, 120^\circ \). These angles satisfy both conditions.
In simple words: We know the four angles form an AP and add up to \( 360^\circ \). We also have a special rule about the sum of the first three angles compared to the fourth. Using these two facts, we set up equations to find 'a' and 'd', which then gives us the values of all four angles.

🎯 Exam Tip: When dealing with an odd number of terms in an AP, take the middle term as 'a' and 'd' as the common difference. For an even number of terms (like 4 here), taking the terms as \( a-3d, a-d, a+d, a+3d \) (with common difference \( 2d \)) can simplify calculations by making the sum equal to \( 4a \).

 

Question 21. The sum of three numbers in A.P. is – 3 and their product is 8. Find the numbers.
Answer: Let the three numbers in AP be \( a - d, a, a + d \). This form simplifies the sum.
Given that the sum of the three numbers is -3:
\( (a - d) + a + (a + d) = -3 \)
\( 3a = -3 \)
Divide by 3:
\( a = \frac { -3 }{ 3 } \)
\( a = -1 \)
So, the numbers are \( -1 - d, -1, -1 + d \).
Given that their product is 8:
\( (-1 - d)(-1)(-1 + d) = 8 \)
Notice that \( (-1 - d)(-1 + d) \) is of the form \( (x - y)(x + y) = x^2 - y^2 \), where \( x = -1 \) and \( y = d \). Also, the middle \( -1 \) means we can just write \( (-1 - d)(-1 + d) \times (-1) = 8 \).
\( ((-1)^2 - d^2)(-1) = 8 \)
\( (1 - d^2)(-1) = 8 \)
Multiply both sides by -1:
\( 1 - d^2 = -8 \)
Add \( d^2 \) to both sides and add 8 to both sides:
\( 1 + 8 = d^2 \)
\( 9 = d^2 \)
Take the square root of both sides:
\( d = \pm \sqrt{9} \)
\( d = \pm 3 \)
Now we have two possible values for 'd': 3 and -3.

Case 1: If \( d = 3 \)
The numbers are: \( a - d = -1 - 3 = -4 \)
\( a = -1 \)
\( a + d = -1 + 3 = 2 \)
The numbers are -4, -1, 2.

Case 2: If \( d = -3 \)
The numbers are: \( a - d = -1 - (-3) = -1 + 3 = 2 \)
\( a = -1 \)
\( a + d = -1 + (-3) = -1 - 3 = -4 \)
The numbers are 2, -1, -4.
In both cases, the set of numbers is \(\{ -4, -1, 2 \} \). Finding such numbers demonstrates the power of variable choice.
In simple words: We guessed the numbers as 'a-d', 'a', and 'a+d' so their sum would be simple. From the sum, we found 'a'. Then, we used their product to find 'd'. Since 'd' could be positive or negative, we got two possible sets of numbers, which are actually the same set in a different order.

🎯 Exam Tip: When choosing terms for an AP, for an odd number of terms (like 3 or 5), represent them as \( a-d, a, a+d \) or \( a-2d, a-d, a, a+d, a+2d \) respectively. This makes the sum simplify directly to 'na', where 'n' is the number of terms, which is very useful.

 

Question 22. Ram prasad saved Rs. 10 in the first week of the year and then increased his weekly savings by Rs. 2.75. If in the nth week, his savings become Rs. 59.50, find n.
Answer: This problem represents an arithmetic progression.
First week's saving (first term) \( a = Rs. 10 \).
Weekly increase in savings (common difference) \( d = Rs. 2.75 \).
Savings in the nth week (nth term) \( a_n = Rs. 59.50 \).
We need to find 'n', the number of weeks.
Using the formula for the nth term of an AP: \( a_n = a + (n - 1)d \)
\( 59.50 = 10 + (n - 1)(2.75) \)
Subtract 10 from both sides:
\( 59.50 - 10 = (n - 1)(2.75) \)
\( 49.50 = (n - 1)(2.75) \)
Divide by 2.75:
\( \frac { 49.50 }{ 2.75 } = n - 1 \)
To perform the division, we can multiply numerator and denominator by 100 to remove decimals:
\( \frac { 4950 }{ 275 } = n - 1 \)
\( 18 = n - 1 \)
Add 1 to both sides:
\( n = 18 + 1 \)
\( n = 19 \)
So, Ram Prasad's savings will become Rs. 59.50 in the 19th week. This problem shows how APs apply to real-life situations like savings.
In simple words: Ram started with Rs. 10 and added Rs. 2.75 each week. We want to know after how many weeks his total savings reached Rs. 59.50. We used the AP formula to solve for the number of weeks, 'n'.

🎯 Exam Tip: Real-world problems often translate directly into AP concepts. Carefully identify the initial value (first term 'a'), the consistent change (common difference 'd'), and the target value (nth term 'a_n') before applying the formula.

 

Question 23. For an A.P. show that \( T_p + T_{p+2q} = 2T_{p+q} \).
Answer: Let the first term of the AP be 'a' and the common difference be 'd'.
The general term of an AP is \( T_n = a + (n - 1)d \).
We need to show that \( T_p + T_{p+2q} = 2T_{p+q} \).

First, let's write out each term using the general formula:
\( T_p = a + (p - 1)d \)
\( T_{p+2q} = a + (p + 2q - 1)d \)
\( T_{p+q} = a + (p + q - 1)d \)

Now, let's consider the Left Hand Side (LHS) of the equation:
LHS \( = T_p + T_{p+2q} \)
LHS \( = [a + (p - 1)d] + [a + (p + 2q - 1)d] \)
LHS \( = a + pd - d + a + pd + 2qd - d \)
LHS \( = 2a + 2pd + 2qd - 2d \)
Factor out 2 from the entire expression:
LHS \( = 2[a + pd + qd - d] \)
LHS \( = 2[a + (p + q - 1)d] \) ... (i)

Next, let's consider the Right Hand Side (RHS) of the equation:
RHS \( = 2T_{p+q} \)
RHS \( = 2[a + (p + q - 1)d] \) ... (ii)

By comparing (i) and (ii), we can see that LHS = RHS.
Therefore, \( T_p + T_{p+2q} = 2T_{p+q} \) is proven. This property shows the average of terms in an AP. For any three terms \( T_A, T_B, T_C \) where their indices \( A, B, C \) are themselves in AP (e.g., \( p, p+q, p+2q \)), then \( T_B \) is the arithmetic mean of \( T_A \) and \( T_C \).
In simple words: We used the general rule for finding any term in an AP. Then, we plugged in the specific term positions (p, p+q, p+2q) into this rule. By adding the first and third terms (LHS) and comparing it to twice the middle term (RHS), we showed that both sides are exactly the same, proving the statement.

🎯 Exam Tip: For proofs involving AP terms, always start by defining the first term 'a' and common difference 'd'. Substitute the general term formula \( T_n = a + (n-1)d \) into both sides of the equation and simplify to show they are equal. This property is also known as the Arithmetic Mean property.

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