CBSE Class 10 Arithmetic Progressions Sure Shot Questions Set F

Case Study I:
Your friend Veer wants to participate in a 200 m race. Presently, he can run 200 m in 51 seconds and during each day practice it takes him 2 seconds less. He wants to do in 31 seconds.

Question. Which of the following terms are in AP for the given situation?
(a) 51, 53, 55, \dots (b) 51, 49, 47, \dots (c) -51, -53, -55, \dots (d) 51, 55, 59, \dots

Question. What is the minimum number of days he needs to practice till his goal is achieved?
(a) 10 (b) 12 (c) 11 (d) 9

Question. Which of the following term is not in the AP of the above given situation?
(a) 41 (b) 30 (c) 37 (d) 39

Question. If \( n^{th} \) term of an AP is given by \( a_n = 2n + 3 \) then common difference of an AP is
(a) 2 (b) 3 (c) 5 (d) 1

Question. The value of \( x \), for which \( 2x, x + 10, 3x + 2 \) are three consecutive terms of an AP is
(a) 6 (b) - 6 (c) 18 (d) -18
Answer: 1.(b), 2.(c), 3.(b), 4.(a), 5.(a)

Case Study II:

India is competitive manufacturing location due to the low cost of manpower and strong technical and engineering capabilities contributing to higher quality production runs. The production of TV sets in a factory increases uniformly by a fixed number every year. It produced 16000 sets in \( 6^{th} \) year and 22600 in \( 9^{th} \) year.

Question. The production during first year is
(a) 3000 TV sets (b) 5000 TV sets (c) 7000 TV sets (d) 10000 TV sets

Question. The production during \( 8^{th} \) year is
(a) 10500 (b) 11900 (c) 12500 (d) 20400

Question. The production during first 3 years is
(a) 12800 (b) 19300 (c) 21600 (d) 25200

Question. In which year, the production is 29,200?
(a) \( 10^{th} \) year (b) \( 12^{th} \) year (c) \( 15^{th} \) year (d) \( 18^{th} \) year

Question. he difference of the production during \( 7^{th} \) year and \( 4^{th} \) year is
(a) 6600 (b) 6800 (c) 5400 (d) 7200
Answer: 1.(b), 2.(d), 3.(c), 4.(b), 5.(a)

3. Sum of First \( n \) Terms of an AP

• If first term of an AP be \( a \) and its common difference is \( d \), then the sum \( S_n \) of the first \( n \) terms of an AP is given by \( S_n = \frac{n}{2} [2a + (n - 1) d] \) or, \( S_n = \frac{n}{2} (a + a_n) \) where \( a_n = n^{th} \) term of the AP.
• If \( l \) is the last term of an AP of \( n \) terms, then the sum of all ‘\( n \)’ terms can also be given by \( S_n = \frac{n}{2} (a + l) \).
• The sum of first \( n \) positive integers is given by \( S_n = \frac{n(n + 1)}{2} \).
• If \( S_n \) is the sum of the first \( n \) terms of an AP, then its \( n^{th} \) term is given by \( a_n = S_n - S_{n - 1} \), i.e., the \( n^{th} \) term of an AP is the difference of the sum to first \( n \) terms and the sum to first \( (n - 1) \) terms of it.

Question. Find the sum of the given AP: \( - 5 + (- 8) + (- 11) + \dots + (- 230) \).
Answer: We have, \( a = - 5 \) and \( d = - 8 + 5 = - 3 \)
So, \( a_n = a + (n - 1)d \Rightarrow - 230 = - 5 + (n - 1) (- 3) \Rightarrow - 230 = - 5 - 3n + 3 \)
\( \Rightarrow - 230 + 2 = - 3n \Rightarrow - 228 = - 3n \Rightarrow n = \frac{228}{3} = 76 \)
\( \therefore S_n = \frac{n}{2} (a + a_n) \Rightarrow S_{76} = \frac{76}{2} [- 5 - 230] = 38 [- 235] = - 8930 \)

Question. Find the sum of the AP: \( 7 + 10\frac{1}{2} + 14 + \dots + 84 \)
Answer: Let \( a \) be the first term, \( d \) the common difference and \( a_n \) the last term of given AP.
We have, \( a = 7, d = 10\frac{1}{2} - 7 = \frac{21}{2} - 7 = \frac{21 - 14}{2} = \frac{7}{2} \) and \( a_n = 84 \)
Now, \( a_n = a + (n - 1) d \Rightarrow 84 = 7 + (n - 1) \times \frac{7}{2} \Rightarrow 77 = (n - 1) \times \frac{7}{2} \)
\( \Rightarrow 11 \times 2 = (n - 1) \Rightarrow 22 = n - 1 \Rightarrow n = 22 + 1 = 23 \)
Now, \( S_n = \frac{n}{2} [2a + (n - 1)d] \Rightarrow S_{23} = \frac{23}{2} [2 \times 7 + (23 - 1) \times \frac{7}{2}] \)
\( \Rightarrow S_{23} = \frac{23}{2} [14 + 22 \times \frac{7}{2}] = \frac{23}{2} [14 + 77] = \frac{23}{2} \times 91 = \frac{2093}{2} = 1046\frac{1}{2} \)

Question. How many terms of the AP: 9, 17, 25, \dots must be taken to give a sum of 636?
Answer: Let sum of \( n \) terms be 636. Then, \( S_n = 636, a = 9, d = 17 - 9 = 8 \)
\( \Rightarrow \frac{n}{2} [2a + (n - 1)d] = 636 \Rightarrow \frac{n}{2} [2 \times 9 + (n - 1) \times 8] = 636 \)
\( \Rightarrow \frac{n}{2} \times 2 [9 + (n - 1) \times 4] = 636 \Rightarrow n(9 + 4n - 4) = 636 \)
\( \Rightarrow n[5 + 4n] = 636 \Rightarrow 5n + 4n^2 = 636 \Rightarrow 4n^2 + 5n - 636 = 0 \)
\( \therefore n = \frac{-5 \pm \sqrt{(5)^2 - 4 \times 4 \times (-636)}}{2 \times 4} = \frac{-5 \pm \sqrt{25 + 10176}}{8} = \frac{-5 \pm \sqrt{10201}}{8} = \frac{-5 \pm 101}{8} \)
\( = \frac{96}{8}, -\frac{106}{8} = 12, -\frac{53}{4} \)
But \( n \neq -\frac{53}{4} \), So, \( n = 12 \). Thus, the sum of 12 terms of the given AP is 636.

Question. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first \( n \) terms.
Answer: We have, \( S_7 = 49 \Rightarrow 49 = \frac{7}{2} [2a + (7 - 1) \times d] \Rightarrow 7 \times 2 = [2a + 6d] \)
\( \Rightarrow 14 = 2a + 6d \Rightarrow a + 3d = 7 \dots(i) \)
and \( S_{17} = 289 \Rightarrow 289 = \frac{17}{2} [2a + (17 - 1)d] \Rightarrow 2a + 16d = \frac{289 \times 2}{17} = 34 \)
\( \Rightarrow a + 8d = 17 \dots(ii) \)
Now subtracting equation (i) from (ii), we get \( 5d = 10 \Rightarrow d = 2 \)
Putting the value of \( d \) in equation (i), we get \( a + 3 \times 2 = 7 \Rightarrow a = 7 - 6 = 1 \)
Here \( a = 1 \) and \( d = 2 \). Now, \( S_n = \frac{n}{2} [2a + (n - 1)d] \)
\( = \frac{n}{2} [2 \times 1 + (n - 1) \times 2] = \frac{n}{2} [2 + 2n - 2] = \frac{n}{2} \times 2n = n^2 \)

Question. The sum of the first 7 terms of an AP is 63 and the sum of its next 7 terms is 161. Find the \( 28^{th} \) term of this AP.
Answer: We have, \( S_n = \frac{n}{2} [2a + (n - 1)d] \)
\( \therefore S_7 = \frac{7}{2} [2a + (7 - 1)d] \Rightarrow S_7 = \frac{7}{2} [2a + 6d] \Rightarrow 63 = 7a + 21d \dots(i) \)
Also, \( S_{14} = \frac{14}{2} [2a + 13d] \Rightarrow S_{14} = 14a + 91d \)
But according to question, \( S_{1-7} + S_{8-14} = S_{14} \Rightarrow 63 + 161 = 14a + 91d \Rightarrow 224 = 14a + 91d \)
\( \Rightarrow 2a + 13d = 32 \Rightarrow 2\left(\frac{63 - 21d}{7}\right) + 13d = 32 \dots(ii) \)
\( \Rightarrow 126 - 42d + 91d = 224 \Rightarrow 49d = 98 \Rightarrow d = 2 \)
\( \therefore a = \frac{63 - 21 \times 2}{7} = \frac{63 - 42}{7} = \frac{21}{7} = 3 \)
Thus, \( a_{28} = a + 27d = 3 + 27 \times 2 \Rightarrow a_{28} = 3 + 54 = 57 \)

Question. Find the sum of the integers between 100 and 200 that are: (i) divisible by 9 (ii) not divisible by 9
Answer: (i) Numbers divisible by 9 between 100 and 200 are 108, 117, 126, \dots, 198.
Here, \( a = 108, d = 9, a_n = 198 \). \( a_n = a + (n - 1) d \Rightarrow 198 = 108 + (n - 1)9 \)
\( \Rightarrow 198 = 108 + 9n - 9 \Rightarrow 198 = 99 + 9n \Rightarrow n = 11 \)
Thus, \( S_{11} = \frac{11}{2} [2 \times 108 + 10 \times 9] = \frac{11}{2} [216 + 90] = 1683 \)
(ii) Numbers between 100 and 200 are 101, 102, \dots, 199.
Here, \( a = 101, d = 1, a_n = 199 \Rightarrow n = 99 \)
So, \( S_{99} = \frac{99}{2} (101 + 199) = \frac{99}{2} \times 300 = 14850 \)
Sum of the numbers which are not divisible by 9 \( = 14850 - 1683 = 13167 \)

Question. Find the sum: \( \frac{a - b}{a + b} + \frac{3a - 2b}{a + b} + \frac{5a - 3b}{a + b} + \dots \) to 11 terms.
Answer: The first term, \( a_1 = \frac{a - b}{a + b} \). Common difference \( d = \frac{3a - 2b}{a + b} - \frac{a - b}{a + b} = \frac{2a - b}{a + b} \)
\( S_{11} = \frac{11}{2}\left[\frac{2(a - b)}{a + b} + 10\left(\frac{2a - b}{a + b}\right)\right] = \frac{11}{2(a + b)} [2a - 2b + 20a - 10b] = \frac{11}{a + b} [11a - 6b] \)

Question. The sum of the first \( n \) terms of an AP is \( 3n^2 + 6n \). Find the \( n^{th} \) term of this AP.
Answer: We have, \( S_n = 3n^2 + 6n \). \( S_{n - 1} = 3(n - 1)^2 + 6(n - 1) = 3n^2 - 3 \)
The \( n^{th} \) term will be \( a_n = S_n - S_{n - 1} = 3n^2 + 6n - 3n^2 + 3 = 6n + 3 \)

Question. In an AP, the sum of first ten terms is \( - 150 \) and the sum of next ten terms is \( - 550 \). Find the AP.
Answer: Let \( a \) be the first term and \( d \) the common difference of the AP.
We have, \( S_{10} = - 150 \Rightarrow \frac{10}{2} [2a + 9d] = - 150 \Rightarrow 2a + 9d = - 30 \dots(i) \)
And \( S_{20} - S_{10} = - 550 \Rightarrow S_{20} = - 550 - 150 = - 700 \)
\( \Rightarrow \frac{20}{2} [2a + 19d] = - 700 \Rightarrow 2a + 19d = - 70 \dots(ii) \)
From (i) and (ii), \( d = - 4 \) and \( a = 3 \). So, the AP is: 3, -1, -5, \dots

Question. If \( a_n = 3 - 4n \), show that \( a_1, a_2, a_3, \dots \) form an AP. Also find \( S_{20} \).
Answer: We have, \( a_n = 3 - 4n \). \( a_1 = - 1, a_2 = - 5, a_3 = - 9, \dots \)
Since \( a_2 - a_1 = - 4 = a_3 - a_2 \). So, -1, -5, -9, \dots form an AP.
\( S_{20} = \frac{20}{2} [- 2 + 19 \times (- 4)] = 10 [- 2 - 76] = 10 \times (- 78) = - 780 \)

Question. If the sum of the first \( p \) terms of an AP is \( ap^2 + bp \), find its common difference.
Answer: \( a_p = S_p - S_{p - 1} = (ap^2 + bp) - [a(p - 1)^2 + b(p - 1)] = 2ap + b - a \)
\( \therefore a_1 = 2a + b - a = a + b, a_2 = 4a + b - a = 3a + b \)
\( \Rightarrow d = a_2 - a_1 = (3a + b) - (a + b) = 2a \)

Question. The first and the last terms of an AP are 5 and 45 respectively. If the sum of all its terms is 400, find its common difference ‘\( d \)’.
Answer: We have, \( a = 5, T_n = 45, S_n = 400 \)
\( \therefore T_n = a + (n - 1)d \Rightarrow 45 = 5 + (n - 1)d \Rightarrow (n - 1)d = 40 \dots(i) \)
\( S_n = \frac{n}{2} (a + T_n) \Rightarrow 400 = \frac{n}{2} (5 + 45) \Rightarrow n = 2 \times 8 = 16 \)
Substituting the value of \( n \) in (i), we get \( (16 - 1)d = 40 \Rightarrow 15d = 40 \Rightarrow d = \frac{40}{15} = \frac{8}{3} \)

Question. The sum of the first \( n \) terms of an AP is given by \( S_n = 3n^2 - 4n \). Determine the AP and the \( 12^{th} \) term.
Answer: We have, \( S_n = 3n^2 - 4n \). \( S_{n - 1} = 3 (n - 1)^2 - 4 (n - 1) \).
Since, \( a_n = S_n - S_{n - 1} = 6n - 7 \). Substituting \( n = 1, 2, 3, \dots \) we get \( a_1 = -1, a_2 = 5, a_3 = 11 \).
Hence, AP is \( - 1, 5, 11, \dots \). \( a_{12} = 6 \times 12 - 7 = 72 - 7 = 65 \).

Question. If the \( m^{th} \) term of an AP is \( \frac{1}{n} \) and \( n^{th} \) term is \( \frac{1}{m} \), then show that its \( (mn)^{th} \) term is 1.
Answer: Let \( a \) and \( d \) be the first term and the common difference of the AP respectively.
Then, \( a_m = \frac{1}{n} \) and \( a_n = \frac{1}{m} \Rightarrow a + (m - 1)d = \frac{1}{n} \dots(i) \text{ and } a + (n - 1)d = \frac{1}{m} \dots(ii) \)
Subtracting (ii) from (i), we get \( a + (m - 1)d - [a + (n - 1)d] = \frac{1}{n} - \frac{1}{m} \Rightarrow (m - n) d = \frac{m - n}{mn} \Rightarrow d = \frac{1}{mn} \).
Putting the value of \( d \) in (i), we get \( a + (m - 1) \frac{1}{mn} = \frac{1}{n} \Rightarrow a - \frac{1}{mn} = 0 \Rightarrow a = \frac{1}{mn} \).
\( \therefore a_{mn} = a + (mn - 1) d = \frac{1}{mn} + (mn - 1) \frac{1}{mn} = 1 \). Hence proved.

Question. If \( S_n \) denotes the sum of the first \( n \) terms of an AP, prove that \( S_{30} = 3 (S_{20} - S_{10}) \).
Answer: We have, \( S_n = \frac{n}{2} [2a + (n - 1)d] \). \( S_{30} = \frac{30}{2} [2a + (30 - 1)d] = 30a + 435d \dots(i) \)
And \( S_{20} = \frac{20}{2} [2a + (20 - 1)d] = 20a + 190d \). \( S_{10} = \frac{10}{2} [2a + (10 - 1)d] = 10a + 45d \)
\( 3 (S_{20} - S_{10}) = 3 [20a + 190d - 10a - 45d] = 3 [10a + 145d] = 30a + 435d = S_{30} \). Hence proved.

Question. The sum of \( n, 2n, 3n \) terms of an AP are \( S_1, S_2 \) and \( S_3 \) respectively. Prove that \( S_3 = 3(S_2 - S_1) \).
Answer: Let \( a \) be the first term and \( d \) the common difference of the AP.
\( S_1 = \frac{n}{2} [2a + (n - 1) d] \), \( S_2 = \frac{2n}{2} [2a + (2n - 1) d] \), \( S_3 = \frac{3n}{2} [2a + (3n - 1) d] \).
Now, \( S_2 - S_1 = \frac{2n}{2} [2a + (2n - 1) d] - \frac{n}{2} [2a + (n - 1) d] = \frac{n}{2} [4a + 4nd - 2d - 2a - nd + d] = \frac{n}{2} [2a + 3nd - d] \).
\( \therefore 3(S_2 - S_1) = \frac{3n}{2} [2a + (3n - 1) d] = S_3 \). Hence proved.

Question. If the sum of \( m \) terms of an AP is the same as the sum of its \( n \) terms, show that the sum of its \( (m + n) \) terms is zero.
Answer: Let \( a \) and \( d \) be the first term and the common difference of the given AP respectively.
Then, \( S_m = S_n \Rightarrow \frac{m}{2} \{2a + (m - 1) d\} = \frac{n}{2} \{2a + (n - 1) d\} \)
\( \Rightarrow 2a(m - n) + \{m(m - 1) - n (n - 1)] d = 0 \Rightarrow 2a(m - n) + \{(m^2 - n^2) - (m - n)\}d = 0 \)
\( \Rightarrow (m - n)\{2a + (m + n - 1)d\} = 0 \Rightarrow 2a + (m + n - 1) d = 0 \dots(i) \).
Now, \( S_{m+n} = \frac{m + n}{2} \{2a + (m + n - 1) d\} = \frac{m + n}{2} \times 0 = 0 \). Hence proved.

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 the first term of the AP, \( a = 5 \). Let \( d \) be the common difference.
Then, as per the question, \( \sum_{n=1}^{4} a_n = \frac{1}{2} \sum_{n=5}^{8} a_n \Rightarrow a_1 + a_2 + a_3 + a_4 = \frac{1}{2} [a_5 + a_6 + a_7 + a_8] \)
\( \Rightarrow [a + (a + d) + (a + 2d) + (a + 3d)] = \frac{1}{2} [(a + 4d) + (a + 5d) + (a + 6d) + (a + 7d)] \)
\( \Rightarrow 4a + 6d = \frac{1}{2} (4a + 22d) \Rightarrow 4a + 6d = 2a + 11d \Rightarrow 2a = 5d \)
Since \( a = 5, 2(5) = 5d \Rightarrow 10 = 5d \Rightarrow d = 2 \).