Maharashtra Board Class 10 Maths Chapter 3 Arithmetic Progression Set 3 Solutions

Question 1. Choose the correct alternative answer for each of the following sub questions.
(i) The sequence – 10,- 6,- 2, 2, ...
(A) is an A.P. Reason d = 16
(B) is an A.P. Reason d = 4
(C) is an A.P. Reason d = - 4
(D) is not an A.P.
Answer: (B) is an A.P. Reason d = 4
In simple words: An arithmetic progression (A.P.) has a constant difference between consecutive terms. In this sequence, adding 4 to each term gives the next term, so the common difference (d) is 4.

🎯 Exam Tip: To find the common difference (d) of an A.P., subtract any term from its succeeding term. Ensure this difference is consistent across the sequence.

Question 1. Choose the correct alternative answer for each of the following sub questions.
(ii) First four terms of an A.P. are ..., whose first term is -2 and common difference is -2.
(A) -2, 0, 2, 4
(B) -2, 4, 8, 16
(C) -2, -4, -6, -8
(D) -2,-4, -8, -16
Answer: (C) -2, -4, -6, -8
In simple words: In an A.P., each term is found by adding the common difference to the previous term. Starting with -2 and a common difference of -2, the terms are -2, (-2)+(-2)=-4, (-4)+(-2)=-6, and so on.

🎯 Exam Tip: The n-th term of an A.P. is given by \(t_n = a + (n-1)d\). You can generate terms by repeatedly adding 'd' to the previous term, starting from 'a'.

Question 1. Choose the correct alternative answer for each of the following sub questions.
(iii) What is the sum of the first 30 natural numbers?
(A) 464
(B) 465
(C) 462
(D) 461
Answer: (B) 465
Hints:
First 30 natural numbers are
1, 2, 3, ..., 30
The above sequence is an A.P.
\( t_1 = 1, t_{30} = 30 \)
\( S_{30} = \frac{30}{2} (1+30) \)
\( = 15 \times 31 = 465 \)
In simple words: The sum of the first 'n' natural numbers can be found using the formula \(S_n = \frac{n(n+1)}{2}\). For the first 30 numbers, this calculates to 465.

🎯 Exam Tip: Remember the formula for the sum of the first 'n' natural numbers, \(S_n = \frac{n(n+1)}{2}\), as it's a common shortcut in AP problems.

Question 1. Choose the correct alternative answer for each of the following sub questions.
(iv) For an given A.P. \( t_7 = 4, d = - 4 \), then a = .......
(A) 6
(B) 7
(C) 20
(D) 28
Answer: (D) 28
In simple words: Using the formula \(t_n = a + (n-1)d\), we can substitute the given values \(t_7 = 4\), \(n = 7\), and \(d = -4\) to solve for the first term 'a'. This calculation yields \(a = 28\).

🎯 Exam Tip: Always write down the given information and the formula you intend to use. This helps in systematically substituting values and solving for the unknown variable.

Question 1. Choose the correct alternative answer for each of the following sub questions.
(v) For an given A.P. \( a = 3.5, d = 0, n = 101 \), then \( t_n \) = ....
(A) 0
(B) 3.5
(c) 103.5
(D) 104.5
Answer: (B) 3.5
In simple words: If the common difference \(d\) is 0, then all terms in the A.P. are the same as the first term \(a\). Therefore, the \(n^{th}\) term \(t_n\) will also be equal to \(a\), which is 3.5.

🎯 Exam Tip: A common difference of zero means the sequence is constant. Hence, every term, including the \(n^{th}\) term, will be equal to the first term.

Question 1. Choose the correct alternative answer for each of the following sub questions.
(vi) In an A.P. first two terms are - 3, 4, then 21st term is ....
(A) -143
(B) 143
(C) 137
(D) 17
Answer: (C) 137
In simple words: From the first two terms, -3 and 4, we find the first term \(a = -3\) and common difference \(d = 4 - (-3) = 7\). Then, use the formula \(t_n = a + (n-1)d\) to find the 21st term.

🎯 Exam Tip: Always calculate the common difference accurately. If the terms are given, \(d = t_2 - t_1\). Then apply the \(n^{th}\) term formula precisely.

Question 1. Choose the correct alternative answer for each of the following sub questions.
(vii) If for any A.P. \( d = 5 \), then \( t_{18} - t_{13} \) = ....
(A) 5
(B) 20
(C) 25
(D) 30
Answer: (C) 25
Hints:
\( t_{18}-t_{13} = a + (18-1)d-[a+ (13-1)d] \)
\( = a+17d-a-12d \)
\( = 5d = 5 \times 5 = 25 \)
In simple words: The difference between any two terms \(t_m\) and \(t_n\) in an A.P. is \((m-n)d\). Here, \((18-13)d = 5d\). Given \(d=5\), the difference is \(5 \times 5 = 25\).

🎯 Exam Tip: The difference \(t_m - t_n\) directly relates to \((m-n)\) times the common difference 'd'. This shortcut saves time in calculations.

Question 1. Choose the correct alternative answer for each of the following sub questions.
(viii) Sum of first five multiples of 3 is ...
(A) 45
(B) 55
(C) 15<
(D) 75
Answer: (A) 45
In simple words: The first five multiples of 3 are 3, 6, 9, 12, 15. This is an A.P. with \(a=3\) and \(d=3\). The sum of these terms is \(3+6+9+12+15 = 45\).

🎯 Exam Tip: For small 'n', directly listing and summing the terms is often quicker than using the \(S_n\) formula. For larger 'n', the formula \(S_n = \frac{n}{2} [2a + (n-1)d]\) is essential.

Question 1. Choose the correct alternative answer for each of the following sub questions.
(ix) 15, 10, 5, ... In this A.P. sum of first 10 terms is...
(A) -75
(B) -125
(C) 75
(D) 125
Answer: (A) -75
In simple words: The given A.P. has \(a=15\) and \(d = 10-15 = -5\). To find the sum of the first 10 terms, use the formula \(S_n = \frac{n}{2} [2a + (n-1)d]\). Substituting \(n=10\), \(a=15\), and \(d=-5\), we get \(S_{10} = \frac{10}{2} [2(15) + (10-1)(-5)] = 5 [30 + 9(-5)] = 5 [30 - 45] = 5(-15) = -75\).

🎯 Exam Tip: Be careful with negative common differences when calculating sums. Pay attention to the signs in the formula's terms.

Question 1. Choose the correct alternative answer for each of the following sub questions.
(x) In an A.P. 1st term is 1 and the last term is 20. The sum of all terms is 399, then \(n\) = ....
(A) 42
(B) 38
(C) 21
(D) 19
Answer: (B) 38
Hints:
\( S_n = \frac{n}{2} (\text{first term + last term}) \)
\( 399 = \frac{n}{2} (1+20) \)
\( 399 \times 2 = 21n \)
\( n = \frac{798}{21} = 38 \)
In simple words: Given the first term, last term, and sum, we can use the sum formula \(S_n = \frac{n}{2} (a + t_n)\) to solve for 'n', the number of terms. Plugging in \(S_n = 399\), \(a = 1\), and \(t_n = 20\), we find \(n = 38\).

🎯 Exam Tip: When the first and last terms are known, the formula \(S_n = \frac{n}{2} (a + l)\) is generally simpler to use than the one involving \(2a + (n-1)d\).

Question 2. Find the fourth term from the end in an
Α.Ρ.: -11, -8, -5, ..., 49.
Solution:
The given A.P. is
-11,-8,-5, ..., 49
Reversing the A.P., we get 49, ..., -5, -8, -11
Here, \( a = 49 \), \( d = -11 -(-8) = -11 + 8 = -3 \)
Since, \( t_n = a + (n - 1)d \)
\( \implies t_4 = 49 + (4 - 1)(-3) \)
\( = 49 + (3) (-3) \)
\( = 49-9 \)
\( = 40 \)
Fourth term from the end in the given A.P. is 40.
In simple words: To find a term from the end of an A.P., reverse the sequence. The last term becomes the new first term, and the common difference changes its sign. Then, apply the standard \(n^{th}\) term formula to the reversed A.P.

🎯 Exam Tip: Reversing the A.P. simplifies finding terms from the end. Remember to correctly identify the new first term and common difference for the reversed sequence.

Question 3. In an A.P. the \(10^{th}\) term is 46, sum of the \(5^{th}\) and \(7^{th}\) term is 52. Find the A.P.
Solution:
For an A.P., let a be the first term and d be the common difference.
\( t_{10} = 46, t_5 + t_7 = 52 \)...[Given]
Since, \( t_n = a + (n - 1)d \)
\( \implies t_{10} = a + (10 - 1)d \)
\( \implies 46 = a + 9d \)
i. e. \( a + 9d = 46 \)...(i)
Also, \( t_5 + t_7 = 52 \)
\( \implies a + (5 - 1)d + a + (7 - 1)d = 52 \)
\( \implies a + 4d + a + 6d = 52 \)
\( \implies 2a + 10d = 52 \)
\( \implies 2 (a + 5d) = 52 \)
\( \implies a + 5d = \frac{52}{2} \)
\( \implies a + 5d = 26 \)...(ii)
Subtracting equation (ii) from (i), we get

\( \implies d = \frac{20}{4} = 5 \)
Substituting \( d = 5 \) in equation (ii), we get
\( a + 5(5) = 26 \)
\( \implies a + 25 = 26 \)
\( \implies a = 26 - 25 = 1 \)
\( t_1 = a = 1 \)
\( t_2 = t_1 + d = 1 + 5 = 6 \)
\( t_3 = t_2 + d = 6 + 5 = 11 \)
\( t_4 = t_3 + d = 11 + 5 = 16 \)
The required A.P. is 1,6,11,16,....
In simple words: We used the \(n^{th}\) term formula to set up two linear equations based on the given conditions. Solving these simultaneous equations helped us find the first term 'a' and common difference 'd'. Once 'a' and 'd' are known, the A.P. can be generated.

🎯 Exam Tip: Many A.P. problems reduce to solving simultaneous linear equations for 'a' and 'd'. Clearly label your equations and perform algebraic operations carefully to avoid errors.

Question 4. The A.P. in which \(4^{th}\) term is -15 and \(9^{th}\) term is -30. Find the sum of the first 10 numbers.
Solution:
\( t_4 = -15, t_9 = - 30 \)...[Given]
Since, \( t_n = a + (n - 1)d \)
\( \implies t_4 = a + (4 - 1)d \)
\( \implies - 15 = a + 3d \)
i. e. \( a + 3d = -15 \)...(i)
Also, \( t_9 = a + (9 - 1)d \)
\( \implies -30 = a + 8d \)
i.e. \( a + 8d = -30 \)...(ii)
Subtracting equation (i) from (ii), we get

\( \implies d = \frac{-15}{5} = -3 \)
Substituting \( d = -3 \) in equation (i), we get
\( a + 3 (-3)=-15 \)
\( \implies a-9=-15 \)
\( \implies a=-15+9=-6 \)
\( S_n = \frac{n}{2} [2a + (n-1)d] \)
\( S_{10} = \frac{10}{2} [2(-6)+(10-1) (-3)] \)
\( = 5 (-12+9 \times -3) \)
\( = 5 (-12-27) \)
\( = 5 \times (-39) \)
\( S_{10} = -195 \)
The sum of the first 10 numbers is -195.
In simple words: First, we determined the first term 'a' and common difference 'd' using the given 4th and 9th terms by solving simultaneous equations. Then, with 'a' and 'd', we applied the sum formula \(S_n = \frac{n}{2} [2a + (n-1)d]\) to find the sum of the first 10 terms.

🎯 Exam Tip: Clearly show the steps for finding 'a' and 'd'. Remember that a negative common difference often leads to decreasing terms and potentially a negative sum, which is mathematically valid.

Question 5. Two given A.P.'s are 9, 7, 5, ... and 24, 21, 18, ... If nth term of both the progressions are equal then find the value of n and n,h term.
Solution:
The first A.P. is 9, 7, 5,...
Here, \( a = 9, d = 7- 9 = -2 \)
\( \implies n^{th} \text{ term} = a + (n - 1)d \)
\( = 9 + (n - 1) (-2) \)
\( = 9 - 2n + 2 \)
\( = 11 - 2n \)
The second A.P. is 24, 21, 18, ...
Here, \( a = 24, d = 21 - 24 = - 3 \)
\( \implies n^{th} \text{ term} = a + (n - 1)d \)
\( = 24 + (n - 1) (-3) \)
\( = 24 - 3n + 3 \)
\( = 27 - 3n \)
Since, the \(n^{th}\) terms of the two A.P.'s are equal.
\( \implies 11 - 2n = 27 - 3n \)
\( \implies 3n - 2n = 27 - 11 \)
\( \implies n = 16 \)
\( \implies t_{16} = 9 + (16 - 1) (-2) \)
\( = 9 + 15 \times (-2) \)
\( = 9-30 \)
\( \implies t_{16} = -21 \)
The values of n and \(n^{th}\) term are 16 and -21 respectively.
In simple words: We found the formula for the \(n^{th}\) term for each A.P. separately. By equating these two expressions, we solved for 'n'. Then, we substituted the value of 'n' back into either \(n^{th}\) term formula to find the actual \(n^{th}\) term value.

🎯 Exam Tip: When comparing terms from different A.P.s, always clearly distinguish their respective 'a' and 'd' values. Setting their \(t_n\) expressions equal is a standard method to find the term number where they match.

Question 6. If sum of \(3^{rd}\) and \(8^{th}\) terms of an A.P. is 7 and sum of \(7^{th}\) and \(14^{th}\) terms is -3, then find the \(10^{th}\) term.
Solution:
for an A.P., let a be the first term and d be the common difference.
According to the first condition,
\( t_3 + t_8 = 7 \)
\( \implies a + (3 - 1) d + a + (8 - 1)d = 7 \)
\( [\because t_n = a + (n - 1)d] \)
\( \implies a + 2d + a + 7d = 7 \)
\( \implies 2a + 9d = 7 \)...(i)
According to the second condition,
\( t_7 + t_{14} = -3 \)
\( \implies a + (7 - 1)d + a + (14 - 1 )d = -3 \)
\( \implies a + 6d + a + 13d = -3 \)
\( \implies 2a + 19 d = - 3 \)...(ii)
Subtracting equation (i) from (ii), we get

\( \implies d = \frac{-10}{10} = -1 \)
Substituting \( d = -1 \) in equation (i), we get
\( 2a +9(-1) = 7 \)
\( \implies 2a - 9 = 7 \)
\( \implies 2a = 7+9 \)
\( \implies 2a = 16 \)
\( \implies a = \frac{16}{2} = 8 \)
\( \implies t_{10} = 8+ (10-1) (-1) \)
\( = 8 + 9 \times (-1) \)
\( = 8-9 \)
\( \implies t_{10} = -1 \)
\(10^{th}\) term of the A.P. is -1.
In simple words: We formed two linear equations in 'a' and 'd' using the given sums of terms. Solving these equations simultaneously gave us the first term 'a' and common difference 'd'. Finally, we used these values to find the \(10^{th}\) term.

🎯 Exam Tip: Be meticulous with algebraic signs, especially when solving simultaneous equations involving negative numbers. A small error in subtraction or multiplication can propagate through the entire solution.

Question 7. In an A.P. the first term is -5 and last term is 45. If sum of all numbers in the A.P. is 120, then how many terms are there? What is the common difference?
Solution:
Let the number of terms in the A.P. be n and the common difference be d.
Then, \( a = -5, t_n = 45, S_n = 120 \)
Since, \( t_n = a + (n - 1)d \)
\( \implies 45 = -5 + (n - 1)d \)
\( \implies 45 + 5 = (n - 1)d \)
\( \implies (n - 1)d = 50 \)...(i)
\( S_n = \frac{n}{2} [2a + (n-1)d] \)
\( \implies 120 = \frac{n}{2} [2 (-5) + (n - 1)d] \)
\( \implies 120 = \frac{n}{2} (-10+50) \)...[From (i)]
\( \implies 120 = \frac{n}{2} \times 40 \)
\( \implies 120 = 20n \)
\( \implies n = \frac{120}{20} = 6 \)
Substituting \( n = 6 \) in equation (i), we get
\( (6-1)d = 50 \)
\( \implies 5d = 50 \)
\( \implies d = \frac{50}{5} = 10 \)
There are 6 terms in the A.P. and the common difference is 10.
Alternate Method:
Let the number of terms in the A.P. be n.
Then, \( t_1 = a = -5, t_n = 45, S_n = 120 \)
\( S_n = \frac{n}{2} (t_1+t_n) \)
\( \implies 120 = \frac{n}{2} (-5+45) \)
\( \implies 120 = \frac{n}{2} \times 40 \)
\( \implies 120 = 20n \)
\( \implies n = \frac{120}{20} = 6 \)
Since, \( t_n = a + (n-1)d \)
\( \implies 45 = -5+(6-1)d \)
\( \implies 45+5=5d \)
\( \implies 50 = 5d \)
\( \implies d = \frac{50}{5} = 10 \)
There are 6 terms in the A.P. and the common difference is 10.
In simple words: We used two A.P. formulas: the \(n^{th}\) term formula (\(t_n\)) and the sum of \(n\) terms formula (\(S_n\)). By substituting the given first term, last term, and sum, we first found 'n' using the \(S_n\) formula involving \(t_n\), then used 'n' in the \(t_n\) formula to find 'd'.

🎯 Exam Tip: When given \(a\), \(t_n\) (or \(l\)), and \(S_n\), it's often more efficient to use the sum formula \(S_n = \frac{n}{2}(a+l)\) to find 'n' first, and then use \(t_n = a + (n-1)d\) to find 'd'.

Question 8. Sum of 1 to n natural numbers is 36, then find the value of n.
Solution:
The natural numbers from 1 to n are
1,2, 3, ......, n.
The above sequence is an A.P.
\( a = 1, d = 2 - 1 = 1 \)
\( S_n = 36 \)...[Given]
Now, \( S_n = \frac{n}{2} [2a + (n - 1)d] \)
\( \implies 36 = \frac{n}{2} [2(1) + (n - 1)(1)] \)
\( \implies 36 = \frac{n}{2} (2 + n - 1) \)
\( \implies 36 \times 2 = n (n + 1) \)
\( \implies 72 = n (n + 1) \)
\( \implies 72 = n^2 + n \)
\( \implies n^2 + n - 72 = 0 \)
\( \implies n^2 + 9n - 8n - 72 = 0 \)
\( \implies n(n + 9) - 8 (n + 9) = 0 \)
\( \implies (n + 9) (n - 8) = 0 \)
\( \implies n + 9 = 0 \text{ or } n - 8 = 0 \)
\( \implies n = -9 \text{ or } n = 8 \)
But, n cannot be negative.
\( \implies n = 8 \)
The value of n is 8.
In simple words: We used the formula for the sum of an A.P., setting \(a=1\), \(d=1\), and \(S_n=36\). This resulted in a quadratic equation for 'n'. Solving the quadratic equation gave two possible values for 'n', but since 'n' must be positive, we chose the positive integer solution.

🎯 Exam Tip: When solving for 'n' (number of terms), remember that 'n' must always be a positive integer. Discard any negative or fractional solutions for 'n'.

Question 9. Divide 207 in three parts, such that all parts are in A.P. and product of two smaller parts will be 4623.
Solution:
Let the three parts of 207 that are in A.P. be
\( a - d, a, a + d \)
According to the first condition,
\( (a - d) + a + (a + d) = 207 \)
\( \implies 3a = 207 \)
\( \implies a = \frac{207}{3} \)
\( \implies a = 69 \)...(i)
According to the second condition,
\( (a - d) \times a = 4623 \)
\( \implies (69 - d) \times 69 = 4623 \)...[From (i)]
\( \implies 69 - d = \frac{4623}{69} \)
\( \implies 69 - d = 67 \)
\( \implies d = 69-67 \)
\( \implies d = 2 \)
\( \implies a - d = 69 - 2 = 67 \)
\( a = 69 \)
\( a + d = 69 + 2 = 71 \)
The three parts of 207 that are in A.P. are 67, 69 and 71.
In simple words: We represented the three parts in A.P. as \(a-d\), \(a\), and \(a+d\). Using the sum condition, we found 'a'. Then, using the product condition for the two smaller parts \((a-d)\) and \(a\), we solved for 'd'. Finally, we calculated the three parts.

🎯 Exam Tip: For A.P. problems involving three terms, representing them as \(a-d\), \(a\), \(a+d\) often simplifies calculations, especially when sums are involved, as 'd' cancels out.

Question 10. There are 37 terms in an A.P., the sum of three terms placed exactly at the middle is 225 and the sum of last three terms is 429. Write the A.P.
Solution:
Since, there are 37 terms in the A.P.
The middle term = \( \frac{37+1}{2} \text{ term} \)
\( = 19^{th} \text{ term} \)
\( \implies 18^{th}, 19^{th} \text{ and } 20^{th} \) terms are placed exactly in
the middle of the sequence.
According to the first condition,
\( t_{18}+t_{19}+t_{20} = 225 \)
\( \implies a+(18-1)d+a+(19-1)d+a+(20-1)d = 225 \)
\( [\because t_n = a + (n-1)d] \)
\( \implies (a + 17d) + (a + 18d) + (a + 19d) = 225 \)
\( \implies 3a + 54d = 225 \)...(i)
According to the second condition,
\( t_{35}+t_{36}+t_{37} = 429 \)
\( \implies a+(35-1)d+a+(36-1)d+a+(37-1)d = 429 \)
\( \implies (a+34d) + (a+35d) + (a+36d) = 429 \)
\( \implies 3a + 105d = 429 \)...(ii)
Subtracting equation (i) from (ii), we get

\( \implies d = \frac{204}{51} = 4 \)
Substituting \( d = 4 \) in equation (i), we get
\( 3a + 54(4) = 225 \)
\( \implies 3a + 216 = 225 \)
\( \implies 3a = 225 - 216 \)
\( \implies 3a = 9 \)
\( \implies a = \frac{9}{3} = 3 \)
The required A. P. is
\( a, a + d, a + 2d, a + 3d, ...., a + (n - 1)d \)
i.e. \( 3, 3 + 4, 3 + 2 \times 4, 3 + 3 \times 4,..., 3 + (37 - 1)4 \)
i.e. 3, 7,11,15, ...,147
In simple words: First, we identified the middle terms and the last three terms. Then, we used the sum condition for these terms to create two linear equations involving 'a' and 'd'. Solving these equations simultaneously gave us the first term and common difference, allowing us to write out the A.P.

🎯 Exam Tip: Carefully identify the indices of the terms involved in sums (e.g., \(t_{18}, t_{19}, t_{20}\)). This is crucial for correctly setting up the equations in terms of 'a' and 'd'.

Question 11. If first term of an A.P. is a, second term is b and last term is c, then show that sum of all
terms is \( \frac{(a+c)(b+c-2a)}{2(b-a)} \)
Solution:
(i) \( t_1 = a, t_2 = b, t_n = c \)...[Given]
(ii) \( d = t_2 - t_1 = b-a \)
\( t_n = a+(n-1)d \)
\( \implies c=a+(n-1) (b - a) \)
\( \implies c-a=(n-1) (b-a) \)
\( \implies \frac{c-a}{b-a} = n-1 \)
\( \implies \frac{c-a}{b-a} +1=n \)
\( \implies n = \frac{c-a+b-a}{b-a} \)
\( \implies n = \frac{b+c-2a}{b-a} \)...(i)
(iii) \( S_n = \frac{n}{2} (t_1+t_n) \)
\( \implies S_n = \frac{1}{2} \left( \frac{b+c-2a}{b-a} \right) (a+c) \)...[From (i)]
\( \implies S_n = \frac{(a+c)(b+c-2a)}{2(b-a)} \)
In simple words: We used the given terms to find the common difference 'd' and then expressed the number of terms 'n' in terms of a, b, and c. Finally, we substituted 'n' into the sum formula \(S_n = \frac{n}{2}(a+c)\) to derive the required expression for the sum.

🎯 Exam Tip: This is a proof-based question. Ensure each step is logically derived from the previous one, clearly stating the formula used. Algebraic manipulation must be precise to arrive at the desired result.

Question 12. If the sum of first p terms of an A.P. is equal to the sum of first q terms then show that the sum of its first (p + q) terms is zero, (p ≠ q)
Solution:
For an A.P., let a be the first term and d be the common difference.
The sum of first n terms of an A.P. is given by
\( S_n = \frac{n}{2} [2a + (n - 1)d] \)
According to the given condition,
\( S_p = S_q \)
\( \implies \frac{p}{2} [2a + (p-1)d] = \frac{q}{2} [2a+(q-1)d] \)
\( \implies p [2a+(p-1)d] = q [2a+(q-1)d] \)
\( \implies 2ap + p (p-1)d = 2aq + q (q-1)d \)
\( \implies 2ap + p^2d - pd = 2aq + q^2d - qd \)
\( \implies 2ap + p^2d - pd - 2aq - q^2d + qd = 0 \)
\( \implies (2ap-2aq) + (p^2d - q^2d) - (pd - qd) = 0 \)
\( \implies 2a(p-q) + d(p^2 - q^2) - d(p - q) = 0 \)
\( \implies 2a(p-q) + d(p+q)(p-q) - d(p - q) = 0 \)
\( \implies (p-q)[2a + d(p + q) - d] = 0 \)
\( \implies (p-q) [2a+(p + q -1)d] = 0 \)
But \( p \neq q \)
\( \implies 2a+(p+q-1)d = 0 \)...(i)
Sum of first \( (p + q) \) terms,
\( S_{p+q} = \frac{p+q}{2} [2a+(p+q-1)d] \)
\( \implies S_{p+q} = \frac{p+q}{2} (0) \)...[From (i)]
\( \implies S_{p+q}=0 \)
The sum of the first \( (p + q) \) terms is zero
In simple words: By equating the formulas for \(S_p\) and \(S_q\), we derived a key relationship: \(2a+(p+q-1)d = 0\). Recognizing that the expression for \(S_{p+q}\) contains exactly this relationship, we substituted 0 into it, proving that \(S_{p+q}\) is zero.

🎯 Exam Tip: For proof-based questions, begin by clearly stating the given conditions and relevant formulas. Manipulate the equations algebraically to reveal the required relationship, paying close attention to factorization and common terms.

Question 13. If m times the \(m^{th}\) term of an A.P. is equal to n times \(n^{th}\) term, then show that the \( (m + n)^{th} \) term of the A.P. is zero.
Solution:
According to the given condition,
\( mt_m = nt_n \)
\( \implies m[a + (m - 1)d] = n[a + (n - 1)d] \)
\( \implies ma + md(m - 1) = na + nd(n-1) \)
\( \implies ma + m^2d - md = na + n^2d - nd \)
\( \implies ma + m^2d - md - na - n^2d + nd = 0 \)
\( \implies (ma - na) + (m^2d - n^2d) - (md - nd) = 0 \)
\( \implies a(m - n) + d(m^2 - n^2) - d(m - n) = 0 \)
\( \implies a(m - n) + d(m + n) (m - n) - d(m - n) = 0 \)
\( \implies (m - n)[a + (m + n - 1) d] = 0 \)
\( \implies [a+ (m + n - 1)d] = 0 \)...[Dividing both sides by \( (m - n) \)]
\( \implies t_{m+n} = 0 \)
The \( (m + n)^{th} \) term of the A.P. is zero.
In simple words: By setting \(m \cdot t_m = n \cdot t_n\) and expanding using the \(n^{th}\) term formula, we rearranged the terms to factor out \((m-n)\). Since \(m \neq n\), we could divide by \((m-n)\), leaving an expression that is precisely the formula for \(t_{m+n}\), which equals zero.

🎯 Exam Tip: This is a classic A.P. proof. Pay attention to the algebraic manipulation, especially factoring out common terms like \((m-n)\) and recognizing the final expression as the \( (m+n)^{th} \) term formula.

Question 14. Rs.1000 is invested at 10 percent simple interest. Check at the end of every year if the total interest amount is in A.P. If this is an A.P. then find interest amount after 20 years. For this complete the following activity.
Solution:
\( \text{Simple interest} = \frac{P \times R \times N}{100} \)
Simple interest after 1 year \( = \frac{1000 \times 10 \times 1}{100} = \text{Rs.}100 \)
Simple interest after 2 years \( = \frac{1000 \times 10 \times 2}{100} = \text{Rs.}200 \)
Simple interest after 3 years \( = \frac{1000 \times 10 \times 3}{100} = \text{Rs.}300 \)
According to this the simple interest for 4, 5, 6
years will be Rs.400, Rs.500, Rs.600
respectively.
From this \( d = 200-100 = 100 \), and \( a = 100 \)
Amount of simple interest after 20 years
\( t_n = a + (n-1) d \)
\( \implies t_{20} = 100+ (20-1) 100 \)
\( = 100 +19 \times 100 = 100+ 1900 \)
\( \implies t_{20} = 2000 \)
Amount of simple interest after 20 years = Rs.2000
In simple words: Simple interest accumulated each year forms an A.P. because the interest earned is constant per year, which acts as the common difference. We calculate the interest for the first few years to establish the 'a' and 'd' values, then use the \(n^{th}\) term formula to find the interest after 20 years.

🎯 Exam Tip: Simple interest problems often translate directly into A.P.s where the principal interest amount acts as 'a' (or \(t_1\)) and the annual simple interest itself acts as 'd'. Compound interest, however, forms a Geometric Progression.