CBSE Class 10 Arithmetic Progressions Sure Shot Questions Set 07

I. Very Short Answer Type Questions 

Question. The sum of first five terms of the AP: 3, 7, 11, 15, ... is:
(a) 44
(b) 55
(c) 22
(d) 11
Answer: (b)

Question. If the first term of an AP is 1 and the common difference is 2, then the sum of first 26 terms is
(a) 484
(b) 576
(c) 676
(d) 625
Answer: (c)

Question. If the sum to \( n \) terms of an AP is \( 3n^2 + 4n \), then the common difference of the AP is
(a) 7
(b) 5
(c) 8
(d) 6
Answer: (d)

Question. If \( a, b, c \) are in AP then \( ab + bc = \)
(a) \( b \)
(b) \( b^2 \)
(c) \( 2b^2 \)
(d) \( \frac{1}{b} \)
Answer: (c)

Question. The sum of all natural numbers which are less than 100 and divisible by 6 is
(a) 412
(b) 510
(c) 672
(d) 816
Answer: (d)

Question. Assertion-Reason Type Questions
Assertion (A): Sum of the first 10 terms of the arithmetic progression \( -0.5, -1.0, -1.5, \dots \) is 27.5.
Reason (R): Sum of first \( n \) terms of an AP is given as \( S_n = \frac{n}{2} [2a + (n - 1)d] \) where \( a = \) first term, \( d = \) common difference.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer: (d)

Question. Assertion (A): The sum of the first \( n \) terms of an AP is given by \( S_n = 3n^2 - 4n \). Then its \( n^{th} \) term, \( a_n = 6n - 7 \).
Reason (R): \( n^{th} \) term of an AP, whose sum of \( n \) terms is \( S_n \), is given by \( a_n = S_n - S_{n-1} \).
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer: (a)

Question. Assertion (A): Sum of first hundred even natural numbers divisible by 5 is 500.
Reason (R): Sum of the first \( n \) terms of an AP is given by \( S_n = \frac{n}{2} [a + l] \) where \( l = \) last term.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer: (d)

Question. Find the sum of first 10 terms of the AP: 2, 7, 12, ...
Answer: \( a = 2, d = 5, n = 10 \)
\( S_{10} = \frac{10}{2} [2(2) + (10 - 1)5] = 5(4 + 45) = 5(49) = 245 \)

Question. If the sum of first \( m \) terms of an AP is \( 2m^2 + 3m \), then what is its second term?
Answer: \( S_1 = a_1 = 2(1)^2 + 3(1) = 5 \)
\( S_2 = a_1 + a_2 = 2(2)^2 + 3(2) = 14 \)
\( \therefore a_2 = S_2 - S_1 = 14 - 5 = 9 \)

Question. Find the sum of first 10 multiples of 6.
Answer: AP is 6, 12, 18, ..., 60
\( a = 6, n = 10, d = 6 \)
\( S_{10} = \frac{10}{2} [2(6) + (10 - 1)6] = 5(12 + 54) = 5(66) = 330 \)

Question. What is the sum of five positive integers divisible by 6?
Answer: \( S_5 = 6 + 12 + 18 + 24 + 30 = 90 \)

Question. If the sum of the first \( q \) terms of an AP is \( 2q + 3q^2 \), what is its common difference?
Answer: \( S_1 = T_1 = 5 \)
\( S_2 = T_1 + T_2 = 4 + 12 = 16 \)
\( T_2 = 16 - 5 = 11 \)
\( d = T_2 - T_1 = 11 - 5 = 6 \)

Question. If \( n^{th} \) term of an AP is \( (2n + 1) \), what is the sum of its first three terms?
Answer: \( a_1 = 3, a_3 = 7, S_3 = \frac{3}{2}(3 + 7) = 15 \)

Question. Find the sum of first 100 natural numbers.
Answer: \( S_{100} = \frac{n(n + 1)}{2} = \frac{100 \times 101}{2} = 50 \times 101 = 5050 \)

II. Short Answer Type Questions-I 

Question. Find the sum of first 8 multiples of 3.
Answer: \( S_8 = 3 + 6 + 9 + 12 + \dots + 24 \)
\( = 3(1 + 2 + 3 + \dots + 8) = 3 \times \frac{8 \times 9}{2} = 3 \times 36 = 108 \)

Question. Find the number of terms of the AP: 54, 51, 48, ... so that their sum is 513.
Answer: 18 or 19

Question. In an AP, the first term is -4, the last term is 29 and the sum of all its terms is 150. Find its common difference.
Answer: \( 150 = \frac{n}{2}(-4 + 29) \Rightarrow 300 = 25n \Rightarrow n = 12 \)
Then, \( l = a_{12} = 29 = -4 + 11d \Rightarrow 11d = 33 \Rightarrow d = 3 \)

Question. Find the sum of all three digit natural numbers, which are multiples of 11.
Answer: AP is 110, 121, ..., 990
\( 990 = 110 + (n - 1)11 \Rightarrow n = 81 \)
\( S_{81} = \frac{81}{2}(110 + 990) = \frac{81}{2}(1100) = 44550 \)

Question. The first and the last terms of an AP are 8 and 65 respectively. If sum of all its terms is 730, find its common difference.
Answer: \( S_n = \frac{n}{2}(a + a_n) \Rightarrow 730 = \frac{n}{2}(8 + 65) \Rightarrow \frac{73n}{2} = 730 \Rightarrow n = 20 \)
\( a_{20} = a + 19d \Rightarrow 65 = 8 + 19d \Rightarrow 19d = 57 \Rightarrow d = 3 \)

Question. The sum of the first \( n \) terms of an AP is \( 4n^2 + 2n \). Find the \( n^{th} \) term of this AP.
Answer: \( S_n = 4n^2 + 2n \). \( S_{n-1} = 4(n-1)^2 + 2(n-1) = 4n^2 - 6n + 2 \).
\( a_n = S_n - S_{n-1} = (4n^2 + 2n) - (4n^2 - 6n + 2) = 8n - 2 \)

Question. How many terms of the AP: 18, 16, 14, ... be taken so that their sum is zero?
Answer: \( S_n = 0 \Rightarrow \frac{n}{2}[2(18) + (n-1)(-2)] = 0 \Rightarrow 36 - 2n + 2 = 0 \Rightarrow n = 19 \)

Question. In an AP, if \( S_5 + S_7 = 167 \) and \( S_{10} = 235 \), then find the AP, where \( S_n \) denotes the sum of its first \( n \) terms.
Answer: \( S_5 + S_7 = 167 \Rightarrow \frac{5}{2}(2a + 4d) + \frac{7}{2}(2a + 6d) = 167 \Rightarrow 5a + 10d + 7a + 21d = 167 \Rightarrow 12a + 31d = 167 \dots(i) \)
\( S_{10} = \frac{10}{2}(2a + 9d) = 235 \Rightarrow 2a + 9d = 47 \dots(ii) \)
Multiplying \( (ii) \) by 6: \( 12a + 54d = 282 \dots(iii) \).
Subtracting \( (i) \) from \( (iii) \): \( 23d = 115 \Rightarrow d = 5 \).
From \( (ii) \): \( 2a + 45 = 47 \Rightarrow a = 1 \). AP is 1, 6, 11, ...

Question. The sum of first \( n \) terms of an AP is given by \( S_n = 2n^2 + 3n \). Find the sixteenth term of the AP.
Answer: \( a_1 = S_1 = 5 \). \( S_2 = 14 = a_1 + a_2 \Rightarrow a_2 = 9 \). \( d = 4 \).
\( a_{16} = a_1 + 15d = 5 + 15(4) = 65 \)

III. Short Answer Type Questions

Question. How many multiples of 4 lie between 10 and 250? Also find their sum.
Answer: 60, 7800

Question. Find the sum of first \( n \) terms of an AP whose \( n^{th} \) term is \( 5n - 1 \). Hence find the sum of first 20 terms.
Answer: \( a_1 = 4, a_2 = 9 \Rightarrow d = 5 \). \( S_n = \frac{n}{2}[2(4) + (n-1)5] = \frac{n(5n + 3)}{2} \).
\( S_{20} = \frac{20(103)}{2} = 1030 \)

Question. Find the sum of first \( n \) terms of three A.Ps' are \( S_1, S_2 \) and \( S_3 \). The first term of each AP is 5 and their common differences are 2, 4 and 6 respectively. Prove that \( S_1 + S_3 = 2S_2 \).
Answer: \( S_1 = \frac{n}{2}[10 + (n-1)2] = 4n + n^2 \).
\( S_2 = \frac{n}{2}[10 + (n-1)4] = 3n + 2n^2 \).
\( S_3 = \frac{n}{2}[10 + (n-1)6] = 2n + 3n^2 \).
\( S_1 + S_3 = (4n + n^2) + (2n + 3n^2) = 6n + 4n^2 = 2(3n + 2n^2) = 2S_2 \). Hence Proved.

Question. Find the sum of \( n \) terms of the series \( (4 - \frac{1}{n}) + (4 - \frac{2}{n}) + (4 - \frac{3}{n}) + \dots \)
Answer: \( S_n = (4+4+\dots+4) - \frac{1}{n}(1+2+3+\dots+n) \)
\( = 4n - \frac{1}{n}\left[\frac{n(n+1)}{2}\right] = 4n - \frac{n+1}{2} = \frac{8n - n - 1}{2} = \frac{7n - 1}{2} \)

Question. Solve the equation: \( 1 + 4 + 7 + 10 + \dots + x = 287 \).
Answer: \( a=1, d=3 \). \( 287 = \frac{n}{2}[2(1) + (n-1)3] \Rightarrow 574 = n(3n - 1) \Rightarrow 3n^2 - n - 574 = 0 \).
Solving for \( n \), we get \( n = 14 \). \( x = a_{14} = 1 + 13(3) = 40 \).

IV. Long Answer Type Questions 

Question. The sum of the first three numbers in an arithmetic progression is 18. If the product of the first and the third terms is 5 times the common difference, find the three numbers.
Answer: Let the numbers be \( a-d, a, a+d \). \( 3a = 18 \Rightarrow a = 6 \).
\( (a-d)(a+d) = 5d \Rightarrow 36 - d^2 = 5d \Rightarrow d^2 + 5d - 36 = 0 \).
\( (d+9)(d-4) = 0 \Rightarrow d = 4 \) (Rejecting \( -9 \)).
Numbers are 2, 6, 10.

Question. List of 3-digit number leaving remainder 3 when divided by 4, are 103, 107, 111, ..., 999. Find the middle term and sum of terms before and after it.
Answer: \( a=103, d=4, a_n=999 \Rightarrow n=225 \).
Middle term \( = \frac{225+1}{2} = 113^{th} \) term.
\( a_{113} = 103 + 112(4) = 551 \).
Sum of 112 terms before middle term: \( S_{112} = \frac{112}{2}[2(103) + 111(4)] = 36400 \).
Sum of 112 terms after middle term: \( S_{225} - (S_{112} + a_{113}) = 123975 - (36400 + 551) = 87024 \).

Question. If the ratio of the sum of the first \( n \) terms of two APs is \( (7n + 1) : (4n + 27) \), then find the ratio of their \( 9^{th} \) terms.
Answer: \( \frac{S_n}{S'_n} = \frac{\frac{n}{2}[2a+(n-1)d]}{\frac{n}{2}[2a'+(n-1)d']} = \frac{7n+1}{4n+27} \Rightarrow \frac{a+(\frac{n-1}{2})d}{a'+(\frac{n-1}{2})d'} = \frac{7n+1}{4n+27} \).
To get ratio of \( 9^{th} \) terms, replace \( \frac{n-1}{2} = 8 \Rightarrow n = 17 \).
Ratio \( = \frac{7(17)+1}{4(17)+27} = \frac{120}{95} = \frac{24}{19} \).

Question. In an AP, sum of 14 terms is 1050 and first term is 10. Find the \( 20^{th} \) term.
Answer: \( 1050 = \frac{14}{2}[2(10) + 13d] \Rightarrow d = 10 \).
\( a_{20} = 10 + 19(10) = 200 \).

Question. How many terms of the AP: 24, 21, 18, ... must be taken so that their sum is 78?
Answer: \( a=24, d=-3 \). \( 78 = \frac{n}{2}[48 + (n-1)(-3)] \Rightarrow 156 = n(51 - 3n) \Rightarrow 3n^2 - 51n + 156 = 0 \).
\( n^2 - 17n + 52 = 0 \Rightarrow (n-4)(n-13) = 0 \Rightarrow n = 4 \text{ or } 13 \).

Case Study Based Questions

Pollution—A Major Problem: student thoughts of planting trees in and around the school. Condition I: Sectionplants same as class (Class I = 1, XII = 12). Condition II: Section plants double the class (Class I = 2, XII = 24).

Question. Refer to Condition I: The AP formed by sequence i.e. number of plants by students is
(a) 0, 1, 2, 3, ..., 12
(b) 1, 2, 3, 4, ..., 12
(c) 0, 1, 2, 3, ..., 15
(d) 1, 2, 3, 4, ..., 15
Answer: (b)

Question. If there are two sections of each class, how many trees will be planted by the students?
(a) 126
(b) 152
(c) 156
(d) 184
Answer: (c)

Question. If there are three sections of each class, how many trees will be planted by the students?
(a) 234
(b) 260
(c) 310
(d) 326
Answer: (a)

Question. Refer to Condition II: If there are two sections of each class, how many trees will be planted by the students?
(a) 422
(b) 312
(c) 360
(d) 540
Answer: (b)

Question. If there are three sections of each class, how many trees will be planted by the students?
(a) 468
(b) 590
(c) 710
(d) 620
Answer: (a)

Case Study II: Loan of ₹ 1,18,000. 1st instalment ₹ 1000. Increase ₹ 100 every month.

Question. The amount paid by him in \( 30^{th} \) installment is
(a) ₹ 3900
(b) ₹ 3500
(c) ₹ 3700
(d) ₹ 3600
Answer: (a)

Question. The total amount paid by him upto 30 installments is
(a) ₹ 37000
(b) ₹ 73500
(c) ₹ 75300
(d) ₹ 75000
Answer: (b)

Question. What amount does he still have to pay after \( 30^{th} \) installment?
(a) ₹ 45500
(b) ₹ 49000
(c) ₹ 44500
(d) ₹ 54000
Answer: (c)

Question. If total installments are 40, then amount paid in the last installment is
(a) ₹ 4900
(b) ₹ 3900
(c) ₹ 5900
(d) ₹ 9400
Answer: (a)

Question. The ratio of the \( 1^{st} \) installment to the last installment is
(a) 1 : 49
(b) 10 : 49
(c) 10 : 39
(d) 39 : 10
Answer: (b)

IMPORTANT FORMULAE

• The \( n^{th} \) term of an AP, \( a_n = a + (n - 1) d \)
• The \( n^{th} \) term of an AP from end, \( a_n = l - (n - 1) d \)
• Sum of finite terms of an AP: \( S_n = \frac{n}{2} [2a + (n - 1) d] \) or \( S_n = \frac{n}{2} (a + a_n) \)
• If there are only \( n \) terms in an AP, then \( a_n = l \), the last term: \( S_n = \frac{n}{2} (a + l) \)
• Note: \( a_n = S_n - S_{n-1} \)