CBSE Class 10 Mathematics Arithmetic Progressions VBQs Set E

OBJECTIVE TYPE QUESTIONS

Question. \( 30^{th} \) term of the A.P., : 10, 7, 4,………, is:
(a) 97
(b) 77
(c) –77
(d) –87
Answer: (c)
Explanation: In the given A.P., \( a = 10 \) and \( d = 7 – 10 = –3 \)
Thus, the \( 30^{th} \) term, \( t_{30} = 10 + (30 – 1) (–3) = –77 \)

Question. \( 11^{th} \) term of the A.P., : \( -3, -\frac{1}{2}, 2 \dots \) is:
(a) 28
(b) 22
(c) –38
(d) \( -48\frac{1}{2} \)
Answer: (b)
Explanation: In the given A.P., \( a = –3 \) and \( d = -\frac{1}{2} + 3 = \frac{5}{2} \)
Thus, the \( 11^{th} \) term, \( t_{11} = -3 + (11 – 1)\left(\frac{5}{2}\right) = -3 + 25 = 22 \)

Question. In an A.P., if \( d = –4, n = 7, a_n = 4 \), then \( a \) is;
(a) 6
(b) 7
(c) 20
(d) 28
Answer: (d)
Explanation: In the given A.P., \( d = –4, n = 7, a_n = 4 \)
\( a_n = a + (n – 1)d \Rightarrow 4 = a + (7 – 1)(–4) \Rightarrow a = 28 \)

Question. In an A.P., if \( a = 3.5, d = 0, n = 101 \), then \( a_n \) will be
(a) 0
(b) 3.5
(c) 103.5
(d) 104.5
Answer: (b)
Explanation: In the given A.P., \( a = 3.5, d = 0, n = 101 \)
\( a_n = a + (n - 1)d \Rightarrow a_n = 3.5 + (101 - 1)0 \Rightarrow a_n = 3.5 \)

Question. The list of numbers – 10, – 6, – 2, 2,…… is:
(a) an A.P., with \( d = – 16 \)
(b) an A.P., with \( d = 4 \)
(c) an A.P., with \( d = – 4 \)
(d) not an A.P.
Answer: (b)
Explanation: In the given numbers – 10, – 6, – 2, 2,…
\( (-6) - (-10) = 4 \)
\( (-2) - (-6) = 4 \)
\( 2 - (-2) = 4 \)
Since, the common difference is same, thus, the given numbers are in A.P. with \( d = 4 \).

Question. The \( 11^{th} \) term of the A.P.,: \( -5, -\frac{5}{2}, 0, \frac{5}{2}, \dots \) is:
(a) –20
(b) 20
(c) –30
(d) 30
Answer: (b)
Explanation: In the given A.P., \( a = –5, d = -\frac{5}{2} - (-5) = \frac{5}{2}, n = 11 \)
\( t_n = a + (n - 1)d \Rightarrow t_{11} = -5 + (11 - 1)\left(\frac{5}{2}\right) \Rightarrow t_{11} = -5 + 25 = 20 \)

Question. The first four terms of an A.P., whose first term is –2 and the common difference is –2, are:
(a) – 2, 0, 2, 4
(b) – 2, 4, – 8, 16
(c) – 2, – 4, – 6, – 8
(d) – 2, – 4, – 8, –16
Answer: (c)
Explanation: In the given A.P., \( a = –2, d = –2 \),
\( t_1 = -2 \)
\( t_2 = (-2) + (-2) = -4 \)
\( t_3 = (-4) + (-2) = -6 \)
\( t_4 = (-6) + (-2) = -8 \)

Question. The \( 21^{st} \) term of the A.P., whose first two terms are –3 and 4 is :
(a) 17
(b) 137
(c) 143
(d) –143
Answer: (b)
Explanation: In the given A.P., \( t_1 = –3 \) and \( t_2 = 4 \)
\( d = t_2 - t_1 = 4 - (-3) = 7 \)
\( t_{21} = a + (n - 1)d = (-3) + (21 - 1)(7) = -3 + 140 = 137 \)

Question. The famous mathematician associated with finding the sum of the first 100 natural numbers is:
(a) Pythagoras
(b) Newton
(c) Gauss
(d) Euclid
Answer: (c)
Explanation: The famous mathematician associated with finding the sum of the first 100 natural numbers is Gauss.

Question. If the first term of an A.P. is –5 and the common difference is 2, then the sum of the first 6 terms is:
(a) 0
(b) 5
(c) 6
(d) 15
Answer: (a)
Explanation: In the given A.P., \( a = –5 \) and \( d = 2 \)
\( S_n = \frac{n}{2} [2a + (n - 1)d] \)
\( S_6 = \frac{6}{2} [2(-5) + (6 - 1) \times 2] = 3[-10 + 10] = 0 \)

Question. The sum of first 16 terms of the A.P.,: 10, 6, 2,... is:
(a) –320
(b) 320
(c) –352
(d) –400
Answer: (a)
Explanation: In the given A.P., \( a = 10, d = 6 – 10 = –4 \)
Thus, \( S_{16} = \frac{16}{2} [2 \times 10 + (16 – 1) \times (–4)] = 8[20 - 60] = -320 \)

Question. In an A.P., if \( a = 1, a_n = 20 \) and \( S_n = 399 \), then \( n \) is:
(a) 19
(b) 21
(c) 38
(d) 42
Answer: (c)
Explanation: In the given A.P., \( a = 1, a_n = 20 \) and \( S_n = 399 \)
\( S_n = \frac{n}{2}(a + a_n) \Rightarrow 399 = \frac{n}{2}(1 + 20) \Rightarrow 399 = \frac{21n}{2} \Rightarrow n = 38 \)

Question. The sum of first five multiples of 3 is:
(a) 45
(b) 55
(c) 65
(d) 75
Answer: (a)
Explanation: In the given A.P., \( a = 3, d = 3 \) and \( n = 5 \)
\( S_5 = \frac{5}{2} [2 \times 3 + (5 - 1) \times 3] = \frac{5}{2} [6 + 12] = \frac{5 \times 18}{2} = 45 \)

Question. The sum of first five positive integers divisible by 6 is:
(a) 180
(b) 90
(c) 45
(d) 30
Answer: (b)
Explanation: Positive integers divisible by 6 are 6, 12, 18, 24, 30. This is an A.P. with \( a=6, d=6, n=5 \).
\( S_5 = \frac{5}{2} [2 \times 6 + (5 - 1) \times 6] = \frac{5}{2} [12 + 24] = 90 \)

Case-based MCQs

Your friend Veer wants to participate in a 200 m race. He can currently run that distance in 51 seconds and with each day of practice it takes him 2 seconds less. He wants to do in 31 seconds.

Question. Which of the following terms are in A.P. for the given situation
(a) 51, 53, 55….
(b) 51, 49, 47….
(c) – 51, – 53, – 55….
(d) 51, 55, 59…
Answer: (b)
Explanation: \( a = 51, d = – 2 \). A.P. = 51, 49, 47 ...... .

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
Answer: (c)
Explanation: Goal = 31 second. Let \( n = \) number of days.
\( a_n = 31 \Rightarrow a + (n – 1)d = 31 \Rightarrow 51 + (n – 1)(– 2) = 31 \Rightarrow 51 – 2n + 2 = 31 \Rightarrow – 2n = – 22 \Rightarrow n = 11 \)

Question. Which of the following term is not in the A.P. for the above given situation
(a) 41
(b) 30
(c) 37
(d) 39
Answer: (b)

Question. If \( n^{th} \) term of an A.P. is given by \( a_n = 2n + 3 \) then common difference of an A.P. is:
(a) 2
(b) 3
(c) 5
(d) 1
Answer: (a)
Explanation: Here, \( a_1 = 2(1) + 3 = 5 \) and \( a_2 = 2(2) + 3 = 7 \).
\( d = a_2 – a_1 = 7 – 5 = 2 \)

Question. The value of \( x \), for which \( 2x, x + 10, 3x + 2 \) are three consecutive terms of an A.P.
(a) 6
(b) – 6
(c) 18
(d) – 18
Answer: (a)
Explanation: Since, \( 2x, x + 10, 3x + 2 \) are in A.P., then common difference will remain same.
\( (x + 10) – 2x = (3x + 2) – (x + 10) \Rightarrow 10 – x = 2x – 8 \Rightarrow 3x = 18 \Rightarrow x = 6 \)

Your elder brother wants to buy a car and plans to take loan from a bank for his car. He repays his total loan of ₹ 1,18,000 by paying every month starting with the first instalment of ₹ 1000. If he increases the instalment by ₹ 100 every month.

Question. The amount paid by him in \( 30^{th} \) instalment is:
(a) ₹ 3900
(b) ₹ 3500
(c) ₹ 3700
(d) ₹ 3600
Answer: (a)
Explanation: \( a = 1000, d = 100 \). \( a_{30} = a + (30 – 1)d = 1000 + 29 \times 100 = 3900 \).

Question. The amount paid by him in the 30 instalments is:
(a) ₹ 37000
(b) ₹ 73500
(c) ₹ 75300
(d) ₹ 75000
Answer: (b)
Explanation: Sum of 30 instalments \( = \frac{n}{2} [2a + (n – 1)d] = \frac{30}{2} [2 \times 1000 + (30 – 1)100] = 15[2000 + 2900] = 15 \times 4900 = 73500 \).
Total amount paid in 30 instalments = ₹ 73500.

Question. What amount does he still have to pay after \( 30^{th} \) instalment ?
(a) ₹ 45500
(b) ₹ 49000
(c) ₹ 44500
(d) ₹ 54000
Answer: (c)
Explanation: Remaining amount \( = 1,18,000 – 73,500 = 44,500 \).

Question. If total instalments are 40, then amount paid in the last instalment ?
(a) ₹ 4900
(b) ₹ 3900
(c) ₹ 5900
(d) ₹ 9400
Answer: (a)
Explanation: Amount paid in \( 40^{th} \) instalment, \( a_{40} = 1000 + (40 – 1)100 = 1000 + 3900 = Rs 4900 \).

Question. The ratio of the \( 1^{st} \) instalment to the last instalment is:
(a) 1 : 49
(b) 10 : 49
(c) 10 : 39
(d) 39 : 10
Answer: (b)
Explanation: \( 1^{st} \) instalment : last instalment \( = 1000 : 4900 = 10 : 49 \).

Very Short Answer Type Questions

Question. Which term of the following A.P. 27, 24, 21, ........ is zero ?
Answer: We know that \( a_n = a + (n – 1)d \). Here, \( a_n = 0, a = 27, d = 24 - 27 = -3 \).
\( 0 = 27 + (n – 1)(– 3) \Rightarrow 3n = 30 \Rightarrow n = 10 \).
\( 10^{th} \) term of the given A.P. is zero.

Question. In an Arithmetic Progression, if \( d = – 4, n = 7, a_n = 4 \), then find \( a \).
Answer: We know that \( a_n = a + (n – 1)d \).
\( 4 = a + 6 \times (– 4) \Rightarrow a = 28 \).

Question. If the first term of an A.P. is \( p \) and the common difference is \( q \), then find its \( 10^{th} \) term.
Answer: We have, first term \( (a) = p \), common difference \( (d) = q \) and \( n = 10 \).
Then, \( a_{10} = p + (10 – 1)q = p + 9q \).

Question. Find the common difference of the A.P. \( \frac{1}{p}, \frac{1-p}{p}, \frac{1-2p}{p}, \dots \)
Answer: Given A.P. \( = \frac{1}{p}, \frac{1-p}{p}, \frac{1-2p}{p} \dots \)
Common difference \( = a_2 – a_1 = \frac{1-p}{p} - \frac{1}{p} = \frac{1-p-1}{p} = \frac{-p}{p} = -1 \).

Question. Find the \( n^{th} \) term of the A.P. \( a, 3a, 5a, \dots \)
Answer: Given A.P. \( = a, 3a, 5a, \dots \)
Here first term \( = a \) and \( d = 3a – a = 2a \).
\( n^{th} \text{ term} = a + (n – 1)d = a + (n – 1)2a = a + 2na – 2a = (2n – 1)a \).

Question. How many two digit numbers are divisible by 3 ?
Answer: Numbers are 12, 15, 18, ..., 99. Here \( a = 12, d = 3, l = 99 \).
\( 99 = 12 + (n – 1) \times 3 \Rightarrow n = 30 \).
Therefore, there are 30 two digit numbers divisible by 3.

Question. Find the sum of the first 10 multiples of 6.
Answer: First 10 multiples of 6 are 6, 12, 18, ..., 60. This is an A.P. with \( a=6, l=60, n=10 \).
\( S_{10} = \frac{10}{2} [6 + 60] = 5 \times 66 = 330 \).

Question. If \( n^{th} \) term of an A.P. is \( (2n + 1) \), what is the sum of its first three terms ?
Answer: Since, \( a_1 = 3, a_2 = 5 \) and \( a_3 = 7 \).
\( S_3 = \frac{3}{2}(3 + 7) = 15 \).

Question. Which term of the A.P. 8, 14, 20, 26, ....... will be 72 more than its \( 41^{st} \) term.
Answer: Given \( a = 8 \) and \( d = 6 \). Let \( n^{th} \) term be 72 more than its \( 41^{st} \) term.
\( t_n – t_{41} = 72 \Rightarrow [8 + (n – 1)6] – [8 + 40 \times 6] = 72 \Rightarrow (n – 1)6 = 312 \Rightarrow n = 53 \).

Question. Write the \( n^{th} \) term of the A.P. \( \frac{1}{m}, \frac{1+m}{m}, \frac{1+2m}{m}, \dots \)
Answer: We have, \( a = \frac{1}{m} \) and \( d = \frac{1+m}{m} - \frac{1}{m} = 1 \).
\( a_n = \frac{1}{m} + (n - 1)1 = \frac{1 + (n - 1)m}{m} \).

Question. If the \( n^{th} \) term of the A.P. – 1, 4, 9, 14, .... is 129. Find the value of \( n \).
Answer: Given, \( a = – 1 \) and \( d = 4 – (– 1) = 5 \).
\( a_n = – 1 + (n – 1) \times 5 = 129 \Rightarrow (n – 1)5 = 130 \Rightarrow n = 27 \).

Question. What is the common difference of an A.P. in which \( a_{21} – a_7 = 84 \) ?
Answer: Let \( a \) be the \( 1^{st} \) term and \( d \) be the common difference.
\( a_{21} - a_7 = 84 \Rightarrow [a + (21 - 1)d] - [a + (7 - 1)d] = 84 \Rightarrow 14d = 84 \Rightarrow d = 6 \).
Common difference is 6.

Question. For what value of \( k \) will \( k + 9, 2k – 1 \) and \( 2k + 7 \) be the consecutive terms of an A.P. ?
Answer: Since, \( k + 9, 2k – 1, 2k + 7 \) are in A.P., then \( (k + 9) + (2k + 7) = 2(2k - 1) \).
\( 3k + 16 = 4k - 2 \Rightarrow k = 18 \).

Question. Find the tenth term of the sequence: \( \sqrt{2}, \sqrt{8}, \sqrt{18}, \dots \)
Answer: Given sequence is \( \sqrt{2}, 2\sqrt{2}, 3\sqrt{2} \dots \).
Here, \( a = \sqrt{2}, d = \sqrt{2} \) and \( n = 10 \).
\( a_{10} = \sqrt{2} + (10 - 1)\sqrt{2} = 10\sqrt{2} = \sqrt{200} \).

Question. Is series \( \sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots \) an A.P. ? Give reason.
Answer: Common difference, \( d_1 = \sqrt{6} - \sqrt{3} = \sqrt{3}(\sqrt{2} - 1) \).
\( d_2 = \sqrt{9} - \sqrt{6} = 3 - \sqrt{6} \).
As common differences are not equal, the given series is not an A.P.

Short Answer Type Questions-I

Question. Find the number of natural numbers between 102 and 998 which are divisible by 2 and 5 both.
Answer: The numbers divisible by 2 and 5 both must end with 0. Such numbers between 102 and 998 are 110, 120, ..., 990.
\( a_n = 990 \Rightarrow 110 + (n – 1) \times 10 = 990 \Rightarrow n = 89 \).

Question. Show that \( (a – b)^2, (a^2 + b^2) \) and \( (a + b)^2 \) are in A.P.
Answer: Common difference \( d_1 = (a^2 + b^2) – (a – b)^2 = a^2 + b^2 – (a^2 + b^2 – 2ab) = 2ab \).
\( d_2 = (a + b)^2 – (a^2 + b^2) = a^2 + b^2 + 2ab – (a^2 + b^2) = 2ab \).
Since, \( d_1 = d_2 \), the terms are in A.P.

Question. Find the sum of first 20 terms of the following A.P.: 1, 4, 7, 10, ........
Answer: Here, \( a = 1, d = 3 \) and \( n = 20 \).
\( S_{20} = \frac{20}{2} [2 \times 1 + (20 – 1)3] = 10(2 + 57) = 590 \).

Question. The sum of the first 7 terms of an A.P. is 63 and that of its next 7 terms is 161. Find the A.P.
Answer: \( S_7 = 63 \Rightarrow \frac{7}{2}(2a + 6d) = 63 \Rightarrow 2a + 6d = 18 \) ...(i).
Sum of 14 terms \( S_{14} = 63 + 161 = 224 \).
\( \frac{14}{2}(2a + 13d) = 224 \Rightarrow 2a + 13d = 32 \) ...(ii).
Subtracting (i) from (ii), we get \( 7d = 14 \Rightarrow d = 2 \).
Putting \( d = 2 \) in (i), \( 2a + 12 = 18 \Rightarrow a = 3 \).
A.P. is 3, 5, 7, 9, ... .

Question. If \( S_n \), the sum of first \( n \) terms of an A.P. is given by \( S_n = 3n^2 – 4n \). Find the \( n^{th} \) term.
Answer: \( a_1 = S_1 = 3 - 4 = -1 \). \( a_2 = S_2 - S_1 = [3(2)^2 - 4(2)] - (-1) = 5 \).
\( d = a_2 - a_1 = 6 \).
\( a_n = a_1 + (n - 1)d = -1 + (n - 1)6 = 6n - 7 \).

Question. Which term of the A.P. 3, 15, 27, 39,... will be 120 more than its \( 21^{st} \) term ?
Answer: \( a_{21} = 3 + (20 \times 12) = 243 \).
\( a_n = 243 + 120 = 363 \).
\( 363 = 3 + (n - 1) \times 12 \Rightarrow n = 31 \).
Thus, \( 31^{st} \) term is 120 more than \( a_{21} \).

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

Question. Find the \( 20^{th} \) term from the last term of the A.P.: 3, 8, 13, ...... 253.
Answer: \( 20^{th} \) term from end \( = l - (n - 1)d = 253 - (20 - 1)5 = 253 - 95 = 158 \).

Question. Find how many integers between 200 and 500 are divisible by 8.
Answer: Numbers are 208, 216, ..., 496. This is an A.P. with \( a = 208, d = 8, l = 496 \).
\( 496 = 208 + (n - 1)8 \Rightarrow (n - 1)8 = 288 \Rightarrow n - 1 = 36 \Rightarrow n = 37 \).

Question. The fifth term of an A.P. is 26 and its \( 10^{th} \) term is 51. Find the A.P.
Answer: \( a_5 = a + 4d = 26 \) and \( a_{10} = a + 9d = 51 \).
Subtracting equations, \( 5d = 25 \Rightarrow d = 5 \).
Then, \( a + 20 = 26 \Rightarrow a = 6 \).
The A.P. is 6, 11, 16, ... .

Question. How many terms of the A.P. \( -6, -\frac{11}{2}, -5 \dots \) are needed to give their sum zero.
Answer: \( a = -6, d = \frac{1}{2} \). Let \( S_n = 0 \).
\( \frac{n}{2} [2(-6) + (n - 1)\frac{1}{2}] = 0 \Rightarrow -12 + \frac{n-1}{2} = 0 \Rightarrow \frac{n-25}{2} = 0 \Rightarrow n = 25 \).

Question. In an A.P. of 50 terms, the sum of the first 10 terms is 210 and the sum of its last 15 terms is 2565. Find the A.P.
Answer: \( S_{10} = \frac{10}{2} [2a + 9d] = 210 \Rightarrow 2a + 9d = 42 \) ...(i).
Sum of last 15 terms is between \( a_{36} \) and \( a_{50} \).
Sum \( = \frac{15}{2} [a_{36} + a_{50}] = \frac{15}{2} [(a + 35d) + (a + 49d)] = 2565 \).
\( \frac{15}{2} [2a + 84d] = 2565 \Rightarrow a + 42d = 171 \) ...(ii).
Solving (i) and (ii), we get \( a = 3, d = 4 \).
The A.P. is 3, 7, 11, ... .