CBSE Class 10 Maths HOTs Arithmetic Progressions Set 15

Question. Find the number of two-digit numbers which are divisible by 6.
Answer: AP: 12, 18, 24, ..., 96.
\( a = 12, d = 6, a_n = 96 \).
\( 96 = 12 + (n-1)6 \implies 84 = (n-1)6 \implies 14 = n-1 \implies n = 15 \).

Question. In an AP, the first term is 12 and the common difference is 6. If the last term of the AP is 252, find its middle term. 
Answer: \( a = 12, d = 6, a_n = 252 \).
\( 252 = 12 + (n-1)6 \implies 240 = (n-1)6 \implies n-1 = 40 \implies n = 41 \).
Middle term is \( \frac{41+1}{2} = 21^{st} \) term.
\( a_{21} = 12 + 20(6) = 12 + 120 = 132 \).

Question. Find the number of three-digit natural numbers which are divisible by 11. 
Answer: Three digit numbers: 100 to 999. Divisible by 11: 110, 121, ..., 990.
\( a = 110, d = 11, a_n = 990 \).
\( 990 = 110 + (n-1)11 \implies 880 = (n-1)11 \implies n-1 = 80 \implies n = 81 \).

Question. The 8th term of an arithmetic progression is zero. Prove that its 38th term is triple of its 18th term.
Answer: \( a_8 = 0 \implies a + 7d = 0 \implies a = -7d \).
\( a_{18} = a + 17d = -7d + 17d = 10d \).
\( a_{38} = a + 37d = -7d + 37d = 30d \).
\( a_{38} = 3 \times a_{18} \). Hence proved.

Question. The 19th term of an AP is equal to three times its sixth term. If its 9th term is 19, find the AP. 
Answer: \( a_{19} = 3 \times a_6 \implies a + 18d = 3(a + 5d) \implies a + 18d = 3a + 15d \implies 2a = 3d \implies a = 1.5d \).
\( a_9 = 19 \implies a + 8d = 19 \implies 1.5d + 8d = 19 \implies 9.5d = 19 \implies d = 2 \).
\( a = 1.5(2) = 3 \). AP: 3, 5, 7, 9, ...

Question. Which term of the sequence \( 17, 16 \frac{1}{5}, 15 \frac{2}{5}, 14 \frac{3}{5}, ... \) is the first negative term?
Answer: \( a = 17, d = 16 \frac{1}{5} - 17 = 81/5 - 85/5 = -4/5 \).
\( a_n < 0 \implies 17 + (n-1)(-4/5) < 0 \implies 85 - 4(n-1) < 0 \implies 85 - 4n + 4 < 0 \implies 89 < 4n \implies n > 22.25 \).
The first negative term is the \( 23^{rd} \) term.

Question. The eighth term of an AP is half its second term and the eleventh term exceeds one-third of its fourth term by 1. Find the 15th term.
Answer: \( a_8 = \frac{1}{2} a_2 \implies a+7d = \frac{1}{2}(a+d) \implies 2a+14d = a+d \implies a = -13d \).
\( a_{11} = \frac{1}{3} a_4 + 1 \implies a+10d = \frac{1}{3}(a+3d) + 1 \implies 3a+30d = a+3d+3 \implies 2a+27d = 3 \).
Substitute \( a = -13d \): \( 2(-13d) + 27d = 3 \implies -26d + 27d = 3 \implies d = 3 \).
\( a = -13(3) = -39 \).
\( a_{15} = a+14d = -39 + 14(3) = -39 + 42 = 3 \).

Question. If \( a_n = \frac{n(n - 3)}{n + 4} \), then find 18th term of this sequence.
Answer: \( a_{18} = \frac{18(18-3)}{18+4} = \frac{18 \times 15}{22} = \frac{9 \times 15}{11} = \frac{135}{11} \).

Question. If the numbers m, n, p, q, r form an A.P., find the value of m – 4n + 6p – 4q + r.
Answer: Let common difference be d. Then \( n = m+d, p = m+2d, q = m+3d, r = m+4d \).
\( m - 4(m+d) + 6(m+2d) - 4(m+3d) + (m+4d) \)
\( = m - 4m - 4d + 6m + 12d - 4m - 12d + m + 4d \)
\( = (1-4+6-4+1)m + (-4+12-12+4)d = 0m + 0d = 0 \).

Question. If m times the mth term of an AP is equal to n times its nth term, find the (m + n)th term of the AP.
Answer: \( m \cdot a_m = n \cdot a_n \implies m(a + (m-1)d) = n(a + (n-1)d) \)
\( \implies am + m(m-1)d = an + n(n-1)d \)
\( \implies a(m-n) + (m^2 - m - n^2 + n)d = 0 \)
\( \implies a(m-n) + [(m-n)(m+n) - (m-n)]d = 0 \)
\( \implies (m-n)[a + (m+n-1)d] = 0 \).
Since \( m \neq n \), \( a + (m+n-1)d = 0 \).
Therefore, \( a_{m+n} = 0 \).

Question. If \( a_1, a_2, a_3, … \) are in A.P. then show that \( \frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \frac{1}{\sqrt{a_3} + \sqrt{a_4}} ... + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}} = \frac{n - 1}{\sqrt{a_1} + \sqrt{a_n}} \).
Answer: Rationalizing each term:
\( \frac{\sqrt{a_2} - \sqrt{a_1}}{a_2 - a_1} + \frac{\sqrt{a_3} - \sqrt{a_2}}{a_3 - a_2} + ... + \frac{\sqrt{a_n} - \sqrt{a_{n-1}}}{a_n - a_{n-1}} \)
\( \because a_2 - a_1 = a_3 - a_2 = ... = d \)
\( = \frac{1}{d} [ \sqrt{a_2} - \sqrt{a_1} + \sqrt{a_3} - \sqrt{a_2} + ... + \sqrt{a_n} - \sqrt{a_{n-1}} ] \)
\( = \frac{\sqrt{a_n} - \sqrt{a_1}}{d} \).
Multiply numerator and denominator by \( \sqrt{a_n} + \sqrt{a_1} \):
\( = \frac{a_n - a_1}{d(\sqrt{a_n} + \sqrt{a_1})} = \frac{(a_1 + (n-1)d) - a_1}{d(\sqrt{a_n} + \sqrt{a_1})} = \frac{(n-1)d}{d(\sqrt{a_n} + \sqrt{a_1})} = \frac{n - 1}{\sqrt{a_1} + \sqrt{a_n}} \). Hence proved.

Question. If the nth term of an AP is \( (2n + 1) \), then the sum of its first three terms is
Sol. \( a_1 = 2 \times 1 + 1 = 3, \)
\( a_2 = 2 \times 2 + 1 = 5, a_3 = 2 \times 3 + 1 = 7 \)
\( \therefore \) Sum = \( 3 + 5 + 7 = 15 \)
(a) \( 6n + 3 \)
(b) 15
(c) 12
(d) 21
Answer: (b) 15

Question. An AP consists of 31 terms. If its 16th term is m, then sum of all the terms of this AP is
Sol. \( S_{31} = \frac{31}{2} (2a + 30d) \)
\( a_{16} = a + 15d = m \)

\( \implies \) \( S_{31} = \frac{31}{2} \times 2(a + 15d) \)

\( \implies \) \( S_{31} = 31m \)
(a) 16m
(b) 47m
(c) 31m
(d) 52m
Answer: (c) 31m

Question. If the first term of an AP is 2 and common difference is 4, then sum of its first 40 terms is
Sol. \( S_{40} = \frac{40}{2} (2 \times 2 + 39 \times 4) = 40 \times (80) = 3200 \)
(a) 1600
(b) 3200
(c) 6400
(d) 19000
Answer: (a) 1600

Question. 7th term of an AP is 40. The sum of its first 13 terms is
Sol. \( a + 6d = 40 \)
\( S_{13} = \frac{13}{2} [2a + 12d] = 13(a + 6d) \)
\( = 13 \times 40 = 520. \)
(a) 260
(b) 540
(c) 520
(d) 560
Answer: (c) 520

Question. If the sum of first m terms of an AP is \( 2m^2 + 3m \), then its second term is
Sol. \( S_m = 2m^2 + 3m \)
\( \therefore S_1 = 2 \times 1^2 + 3 \times 1 = 5 = a_1 \)
and \( S_2 = 2 \times 2^2 + 3 \times 2 = 14 \)

\( \implies \) \( a_1 + a_2 = 14 \)

\( \implies \) \( 5 + a_2 = 14 \)

\( \implies \) \( a_2 = 9 \)
(a) 6
(b) 7
(c) 9
(d) 11
Answer: (c) 9

Question. If the sum of first p terms of an AP is \( ap^2 + bp \), then its common difference is
Sol. \( S_p = ap^2 + bp \)

\( \implies \) \( S_1 = a \times 1^2 + b \times 1 = a + b = a_1 \)
and \( S_2 = a \times 2^2 + b \times 2 = 4a + 2b \)

\( \implies \) \( a_1 + a_2 = 4a + 2b \)

\( \implies \) \( a + b + a_2 = 4a + 2b \)

\( \implies \) \( a_2 = 3a + b \)
Now, \( d = a_2 - a_1 \)
\( = (3a + b) - (a + b) = 2a. \)
(a) a
(b) 2a
(c) 3a
(d) 4a
Answer: (b) 2a

Question. Find the sum of first 8 multiples of 3. 
Answer: Sum of first 8 multiples of 3:
Forms an AP, \( a = 3, d = 3, n = 8 \).
\( \text{Sum} = S_n = \frac{n}{2} [2a + (n - 1)d] \)
\( S_8 = \frac{8}{2} [2(3) + (8 - 1)3] = 4 [6 + 21] = 4 \times 27 = 108 \).
The sum of the first 8 multiples of 3 is 108.

Question. Find the common difference of an AP whose first term is 4, the last term is 49 and the sum of all its terms is 265.
Answer: Given; \( a = 4, l = 49 \) and \( S_n = 265 \)
\( \therefore 265 = \frac{n}{2} (4 + 49) \)

\( \implies \) \( 530 = 53n \)

\( \implies \) \( n = 10 \)
\( \therefore l = a_{10} = a + 9d \)

\( \implies \) \( 49 = 4 + 9d \)

\( \implies \) \( 9d = 45 \)

\( \implies \) \( d = 5 \)
\( \therefore \) Common difference = 5

Question. Find the sum: \( 2 + 4 + 6 + ... + 200 \)
Answer: Here, \( a = 2, d = 4 - 2 = 2 \)
\( a_n = a + (n - 1)d \)

\( \implies \) \( 200 = 2 + (n - 1)2 \)

\( \implies \) \( n = 100 \)
\( \therefore \) There are 100 terms in the given AP.
Now \( S_n = \frac{n}{2} [a + l] \)

\( \implies \) \( S_{100} = \frac{100}{2} [2 + 200] = 10100 \).

Question. 100% Scoring Tips: If question is "solve the equation for x. \( 1 + 6 + 11 + ... + x = 148 \)"
Answer: Then, solution is given as "Sum of first n terms is given as 148
\( \therefore 148 = \frac{n}{2} [2 \times 1 + (n - 1)5] \)

\( \implies \) \( n = 8 \)
Here, we have taken x as \( a_n \)
\( \therefore x = a_8 \)

\( \implies \) \( x = 1 + 7 \times 5 = 36 \)"
Note: But don't solve as "Sum of first x terms is given as 148. \( \therefore 148 = \frac{x}{2} [2 \times 1 + (n - 1)5] \implies x = 8 \)". This is a wrong solution. Here, you must think x as \( a_n \).

Question. If \( S_n \) the sum of first n terms of an AP is given by \( S_n = 3n^2 - 4n \), find the nth term.  
Answer: \( a_1 = S_1 = 3 - 4 = -1 \)
\( a_2 = S_2 - S_1 = [3(2)^2 - 4(2)] - (-1) = 5 \)
\( \therefore d = a_2 - a_1 = 6 \)
Hence \( a_n = -1 + (n - 1) \times 6 = 6n - 7 \)
Alternate method:
\( S_n = 3n^2 - 4n \)
\( \therefore S_{n-1} = 3(n - 1)^2 - 4(n - 1) = 3n^2 - 10n + 7 \)
Hence \( a_n = S_n - S_{n-1} = (3n^2 - 4n) - (3n^2 - 10n + 7) = 6n - 7 \)

Question. Find the sum of all the natural numbers less than 100 which are divisible by 6.
Answer: Natural numbers less than 100 and divisible by 6 are 6, 12, 18, ..., 96.
\( a = 6, d = 6, n = 16, a_n = 96 \)
\( \therefore \text{Sum} = \frac{16}{2} [6 + 96] = 816 \)

Question. The sum of three numbers of an AP is 27 and their product is 405. Find the numbers.
Answer: Let three numbers in AP be \( a - d, a \) and \( a + d \)
\( \therefore (a - d) + a + (a + d) = 27 \)

\( \implies \) \( 3a = 27 \)

\( \implies \) \( a = 9 \)
Also \( (a - d) (a) (a + d) = 405 \)

\( \implies \) \( (9 - d) (9) (9 + d) = 405 \)

\( \implies \) \( (9 - d) (9 + d) = 45 \)

\( \implies \) \( 81 - d^2 = 45 \)

\( \implies \) \( d^2 = 36 \)

\( \implies \) \( d = 6, -6 \)
When \( d = 6 \), numbers are 3, 9, 15
When \( d = -6 \), numbers are 15, 9, 3

Question. An A.P. consists of 25 terms. If 13th term is 50, find the sum of all terms of the A.P.
Answer: Let first term of the A.P. = a and common difference = d
A.T.Q. \( a_{13} = 50 \)

\( \implies \) \( a + 12d = 50 \) ...(i)
Now, \( S_{25} = \frac{25}{2} (a_1 + a_{25}) = \frac{25}{2} (a + a + 24d) \)
\( = \frac{25}{2} (2a + 24d) = \frac{25}{2} \times 2(a + 12d) = 25(a + 12d) \)
\( = 25 \times 50 \) [using (i)]
\( = 1250 \)

Question. Find the sum of n terms of an A.P. whose first term is x, the last term is y and the common difference is 1.
Answer: Here \( S_n = \frac{n}{2} (\text{Ist term} + \text{last term}) \)
\( S_n = \frac{n}{2} (x + y) \) ...(i)
Let \( a_n \) is the last term

\( \implies \) \( a + (n - 1)d = y \)

\( \implies \) \( x + (n - 1) \times 1 = y \)
\( n - 1 = y - x \)
\( n = y - x + 1 \)
Substituting in eq. (i), we get
\( S_n = \frac{(y - x + 1)}{2} (x + y) \)
\( = \frac{1}{2} (y - x + 1) (x + y) \)

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: The sum of 1st n terms is \( 3n^2 + 5n \)
\( S_n = 3n^2 + 5n \)
Change n to n – 1
\( S_{n - 1} = 3(n - 1)^2 + 5(n - 1) \)
\( = 3n^2 + 3 - 6n + 5n - 5 = 3n^2 - n - 2 \)
nth term = \( S_n - S_{n - 1} \)
\( = (3n^2 + 5n) - (3n^2 - n - 2) = 6n + 2 \)
So, kth term = \( 6k + 2 \)
\( \therefore 164 = 6k + 2 \)
\( 162 = 6k \implies k = 27 \)

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 term. 
Answer: \( a_m = \frac{1}{n} \) and \( a_n = \frac{1}{m} \)

\( \implies \) \( a + (m - 1)d = \frac{1}{n} \) and \( a + (n - 1)d = \frac{1}{m} \)
\( a + (m - 1)d = \frac{1}{n} \) ...(i)
\( a + (n - 1)d = \frac{1}{m} \) ...(ii)
Subtracting (ii) from (i);
\( (m - 1)d - (n - 1)d = \frac{1}{n} - \frac{1}{m} \)

\( \implies \) \( md - nd = \frac{m - n}{mn} \)

\( \implies \) \( (m - n)d = \frac{m - n}{mn} \)

\( \implies \) \( d = \frac{1}{mn} \)
From (i) \( a + (m - 1) \cdot \frac{1}{mn} = \frac{1}{n} \)

\( \implies \) \( a = \frac{1}{n} - \frac{(m - 1)}{mn} = \frac{m - m + 1}{mn} \)
\( a = \frac{1}{mn} \)
\( \therefore S_{mn} = \frac{mn}{2} [2 \cdot \frac{1}{mn} + (mn - 1) \cdot \frac{1}{mn}] \)
\( = \frac{mn}{2} [\frac{2}{mn} + 1 - \frac{1}{mn}] \)
\( = \frac{mn}{2} [\frac{1}{mn} + 1] = \frac{mn}{2} \times \frac{1 + mn}{mn} \)
\( = \frac{1 + mn}{2} = \frac{1}{2}(1 + mn) \)

Question. The sum of n terms of an AP is \( 5n^2 - 3n \). Find the AP and also its 10th term.
Answer: Sum of n terms of given AP, \( S_n = 5n^2 - 3n \)
\( \therefore \) Sum of (n – 1) terms of given AP,
\( S_{n - 1} = 5(n - 1)^2 - 3(n - 1) \)

\( \implies \) \( S_{n - 1} = 5(n^2 - 2n + 1) - 3n + 3 = 5n^2 - 13n + 8 \)
\( \therefore \) nth term of AP = \( a_n = S_n - S_{n - 1} \)
\( = (5n^2 - 3n) - (5n^2 - 13n + 8) \)

\( \implies \) \( a_n = 10n - 8 \)
\( \therefore \) 1st term of AP = \( 10 \times 1 - 8 = 2 \)
2nd term of AP = \( 10 \times 2 - 8 = 12 \)
and 3rd term of AP = \( 10 \times 3 - 8 = 22 \)
\( \therefore \) Required AP is 2, 12, 22, ...
\( \therefore a_{10} = 10 \times 10 - 8 = 92 \)

Question. The first term of an A.P. is 5, the last term is 45 and the sum of all its term is 400. Find the number of terms and the common difference of the A.P.  
Answer: Let ‘a’ be the first term and ‘d’ be the common difference of the given AP.
Let \( a_n \) be the last term and \( S_n \) be the sum of n terms
So \( a = 5, a_n = 45, S_n = 400 \)
We need to find the values of ‘n’ and ‘d’
We know that \( S_n = \frac{n}{2} (a + a_n) \) ...(i)
Substituting the values in (i)
\( 400 = \frac{n}{2} (5 + 45) \)

\( \implies \) \( \frac{400 \times 2}{50} = n \)

\( \implies \) \( n = 16 \)
Also, \( a_n = a + (n - 1)d \)

\( \implies \) \( 45 = 5 + (16 - 1)d \)

\( \implies \) \( 45 - 5 = 15d \)

\( \implies \) \( d = \frac{40}{15} \)

\( \implies \) \( d = \frac{8}{3} \)
Hence, there are 16 terms and common difference is \( \frac{8}{3} \).

Question. Show that the sum of first n even natural numbers is equal to \( (1 + \frac{1}{n}) \) times the sum of the first n odd natural numbers.
Answer: Let \( S_1 \) be the sum of first n even natural numbers.
Then \( S_1 = 2 + 4 + 6 + ...... + 2n \)

\( \implies \) \( S_1 = \frac{n}{2} [2 \times 2 + (n - 1) 2] \)

\( \implies \) \( S_1 = \frac{n}{2} [4 + 2n - 2] = n (n + 1) \) ...(i)
Let \( S_2 \) be the sum of first n odd natural numbers.
\( S_2 = 1 + 3 + 5 + ...... + (2n - 1) \)
Then \( S_2 = \frac{n}{2} [2 \times 1 + (n - 1) 2] = n^2 \)
\( S_1 = (1 + \frac{1}{n}) S_2 \) [From (i)]

Question. If there are \( (2n + 1) \) terms in an A.P., prove that the ratio of the sum of odd terms and the sum of even terms is \( (n + 1) : n \).
Answer: Let Ist term of the A.P. = a and common difference = d
Let \( S_1 \) = Sum of odd terms

\( \implies \) \( S_1 = a_1 + a_3 + a_5 + ..... + a_{2n + 1} \)

\( \implies \) \( S_1 = \frac{n+1}{2} (a_1 + a_{2n + 1}) \) (no. of terms = n + 1)

\( \implies \) \( S_1 = \frac{n+1}{2} (a + a + 2nd) = (n + 1)(a + nd) \) ...(i)
\( S_2 \) = sum of even terms

\( \implies \) \( S_2 = a_2 + a_4 + a_6 + ..... + a_{2n} \)
\( S_2 = \frac{n}{2} [(a_2 + a_{2n}) \text{ (no. of terms = n)} \)
\( S_2 = \frac{n}{2} [(a + d) + (a + (2n - 1)d)] \)
\( = \frac{n}{2} (2a + 2nd) \)

\( \implies \) \( S_2 = n(a + nd) \) ...(ii)
Now, \( S_1 : S_2 = (n + 1) (a + nd): n(a + nd) \)
\( = (n + 1) : n \)

Question. The sum of first 16 terms of an A.P. is 120. Find second term of another A.P. whose first term is \( \frac{2}{15} \) times of first terms of given A.P. and common difference same as common difference of given A.P.
Answer: Let Ist term of given A.P. = a and common difference = d.
A.T.Q. \( S_{16} = 120 \)

\( \implies \) \( \frac{16}{2} (2a + 15d) = 120 \)

\( \implies \) \( 2a + 15d = 15 \)

\( \implies \) \( 15d = 15 - 2a \)

\( \implies \) \( d = 1 - \frac{2}{15}a \) ...(i)
Now, first term of IInd A.P. = A

\( \implies \) \( A = \frac{2}{15} a \) (given) ...(ii)
Common difference = d
\( \therefore \) Second term of this A.P. = A + d
\( = \frac{2}{15} a + 1 - \frac{2}{15} a \) [using (i) and (ii)] = 1

Question. The sum of four consecutive numbers in an AP 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: Let the four consecutive numbers of the AP are \( a - 3d, a - d, a + d \) and \( a + 3d \) respectively.
ATQ, \( a - 3d + a - d + a + d + a + 3d = 32 \)

\( \implies \) \( 4a = 32 \implies a = 8 \)
Also, \( \frac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac{7}{15} \)

\( \implies \) \( \frac{a^2 - 9d^2}{a^2 - d^2} = \frac{7}{15} \)

\( \implies \) \( 15(a^2 - 9d^2) = 7(a^2 - d^2) \)

\( \implies \) \( 15a^2 - 135d^2 = 7a^2 - 7d^2 \)

\( \implies \) \( 15a^2 - 7a^2 = - 7d^2 + 135d^2 \)

\( \implies \) \( 8a^2 = 128d^2 \)

\( \implies \) \( 8 \times 8^2 = 128d^2 \) (\( \because a = 8 \))

\( \implies \) \( 128d^2 = 8 \times 64 \)
\( \therefore d^2 = \frac{8 \times 64}{128} = 4 \)
\( \therefore d = \pm \sqrt{4} = \pm 2 \)
When \( a = 8 \) and \( d = 2 \)
\( a - 3d = 8 - 3 \times 2 = 8 - 6 = 2 \)
\( a - d = 8 - 2 = 6 \)
\( a + d = 8 + 2 = 10 \)
\( a + 3d = 8 + 3 \times 2 = 14 \)
When \( d = - 2 \) and \( a = 8 \)
\( a - 3d = 8 - 3 \times (-2) = 8 + 6 = 14 \)
\( a - d = 8 - (-2) = 8 + 2 = 10 \)
\( a + d = 8 + (-2) = 8 - 2 = 6 \)
\( a + 3d = 8 + 3 \times (-2) = 8 - 6 = 2 \)
Therefore, four consecutive terms of the AP's are 2, 6, 10, 14 or 14, 10, 6, 2 respectively.

Question. If the ratio of the sum of the first n terms of two A.P.s is (7n + 1) : (4n + 27), then find the ratio of their 9th terms. 
Answer: Let \( a, d \) and \( A, D \) be the \( 1^{st} \) term and common difference of the 2 APs respectively.
Then,
\( \frac{\frac{n}{2} [ 2a + (n - 1)d ]}{\frac{n}{2} [ 2A + (n - 1)D ]} = \frac{7n + 1}{4n + 27} \)

\( \implies \) \( \frac{2a + (n - 1)d}{2A + (n - 1)D} = \frac{7n + 1}{4n + 27} \)
Replacing \( n \) by 17 in both LHS and RHS,
\( \frac{2a + (17 - 1)d}{2A + (17 - 1)D} = \frac{7(17) + 1}{4(17) + 27} \)

\( \implies \) \( \frac{2a + 16d}{2A + 16D} = \frac{119 + 1}{68 + 27} \)

\( \implies \) \( \frac{2(a + 8d)}{2(A + 8D)} = \frac{120}{95} \)
as \( a + (n - 1)d = a_n \),
\( \frac{a_9}{A_9} = \frac{24}{19} \)
\( \therefore \) ratio of \( 9^{th} \) terms is 24 : 19.

Question. If the ratio of the 11th term of an AP 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: Let first term of an AP be a and common difference be d.
\( \frac{T_{11}}{T_{18}} = \frac{2}{3} \)
\( \frac{a + 10d}{a + 17d} = \frac{2}{3} \)
\( 3a + 30d = 2a + 34d \)
\( a = 4d \) ...(i)
so, \( \frac{S_5}{S_{10}} = \frac{\frac{5}{2} (2a + 4d)}{\frac{10}{2} (2a + 9d)} \)
\( \frac{S_5}{S_{10}} = \frac{1}{2} [ \frac{2a + 4d}{2a + 9d} ] \)
Put \( a = 4d \)
\( \frac{S_5}{S_{10}} = \frac{1}{2} [ \frac{2 \times 4d + 4d}{2 \times 4d + 9d} ] = \frac{1}{2} [ \frac{12d}{17d} ] \)
\( \frac{S_5}{S_{10}} = \frac{6}{17} \)
So, the ratio is 6:17.

Question. Find the sum of first 24 terms of an A.P. whose nth term is given by \( a_n = 3 + 2n \). [AI 2017 (C)]
Answer: nth term of the given AP is \( a_n = 3 + 2n \)
1st term = \( a_1 = 3 + 2 = 5 \)
2nd term = \( a_2 = 3 + 4 = 7 \)
3rd term = \( a_3 = 3 + 6 = 9 \)
So, AP is 5, 7, 9, ...
Sum of 24 terms of the AP = \( \frac{n}{2} [2a + (n - 1)d] \)
\( = \frac{24}{2} [2 \times 5 + (24 - 1)2] \)
\( = 12(10 + 46) = 12 \times 56 = 672 \)

Question. The sum of the first 7 terms of an AP is 63 and that of its next 7 terms is 161. Find the AP.   
Answer: Let a be first term, d be common difference of an AP
\( S_7 = 63 \) and \( S_{14} = S_7 + 161 = 63 + 161 = 224 \)
\( \frac{7}{2} [2a + 6d] = 63 \implies a + 3d = 9 \) ...(i)
\( \frac{14}{2} [2a + 13d] = 224 \implies 2a + 13d = 32 \) ...(ii)
Solving (i) and (ii), we get
\( a = 3, d = 2 \)
\( \therefore \) AP is 3, 5, 7, 9, ...

Question. Find the common difference of an AP whose first term is 5 and the sum of its first four terms is half the sum of the next four terms. 
Answer: Given; Ist term of the AP, \( a = 5 \). Let common difference be d.
Then A.T.Q.,
\( (a_1 + a_2 + a_3 + a_4) = \frac{1}{2} (a_5 + a_6 + a_7 + a_8) \)

\( \implies \) \( [a + (a + d) + (a + 2d) + (a + 3d)] = \frac{1}{2} [(a + 4d) + (a + 5d) + (a + 6d) + (a + 7d)] \)

\( \implies \) \( (4a + 6d) = \frac{1}{2} (4a + 22d) \)

\( \implies \) \( 2 \times (4 \times 5 + 6d) = (4 \times 5 + 22d) \) (\( \because a = 5 \))

\( \implies \) \( 40 + 12d = 20 + 22d \)

\( \implies \) \( 10d = 20 \implies d = 2 \)

Question. A manufacturer of laptop produced 6000 units in 3rd year and 7000 units in the 7th year. Assuming that production increases uniformly by a fixed number every year, find (i) the production in the 1st year, (ii) the production in the 5th year, (iii) the total production in 7 years.
Answer: (i) Let production in Ist year be a units and increase in production (every year) be d units.
\( \because \) Increase in production is constant, therefore unit produced every year forms an AP
Now, \( a_3 = 6000 \)

\( \implies \) \( a + 2d = 6000 \)

\( \implies \) \( a = 6000 - 2d \) ...(i)
and \( a_7 = 7000 \)

\( \implies \) \( a + 6d = 7000 \)

\( \implies \) \( (6000 - 2d) + 6d = 7000 \)

\( \implies \) \( 4d = 1000 \) [using eq. (i)]

\( \implies \) \( d = 250 \)
When \( d = 250 \), eq. (i) becomes
\( a = 6000 - 2 \times 250 = 5500 \)
\( \therefore \) Production in Ist year = 5500
(ii) Production in fifth year
\( a_5 = a + 4d = 5500 + 4 \times 250 = 6500 \)
(iii) Total production in 7 years
\( = \frac{7}{2} (a_1 + a_7) = \frac{7}{2} (5500 + 7000) = 43750 \)