CBSE Class 10 Mathematics Arithmetic Progression MCQs Set N

Question. Which of the following is not an A.P.?
(a) \( - 1.2, 0.8, 2.8, .... \)
(b) \( 3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, ... \)
(c) \( \frac{4}{3}, \frac{7}{3}, \frac{9}{3}, \frac{12}{3}, ... \)
(d) \( \frac{-1}{5}, \frac{-2}{5}, \frac{-3}{5}, ... \)
Answer: (c) \( \frac{4}{3}, \frac{7}{3}, \frac{9}{3}, \frac{12}{3}, .... \)
Explanation:
Here, \( a_2 - a_1 = \frac{7}{3} - \frac{4}{3} = \frac{3}{3} = 1 \)
\( a_3 - a_2 = \frac{9}{3} - \frac{7}{3} = \frac{2}{3} \)
Since \( a_2 - a_1 \neq a_3 - a_2 \), it does not form an A.P.

Question. In an AP, if \( a = 3.5, d = 0 \) and \( n = 101 \), then \( a_n \) will be:
(a) 0
(b) 3.5
(c) 103.5
(d) 104.5
Answer: (b) 3.5
Explanation:
It is given that \( a = 3.5, d = 0, n = 101 \)
We know that in an AP,
\( a_n = a + (n - 1)d \)
\( = 3.5 + (101 - 1) \times 0 \)
\( a_n = 3.5 + 0 \)
\( \therefore a_n = 3.5 \)

Question. The list of numbers –10, –6, –2, 2, ... is:
(a) an AP with \( d = –16 \)
(b) an AP with \( d = 4 \)
(c) an AP with \( d = –4 \)
(d) not an AP
Answer: (b) an AP with \( d = 4 \)
Explanation:
The given list of number is –10, –6, –2, 2, ...
Here, \( a_1 = -10, a_2 = -6, a_3 = -2 \) and \( a_4 = 2 ... \)
Since \( a_2 - a_1 = -6 - (-10) = -6 + 10 = 4 \)
\( a_3 - a_2 = -2 - (-6) = -2 + 6 = 4 \)
\( a_4 - a_3 = 2 - (-2) = 2 + 2 = 4 \)
From the above, we can see that each successive term has the same difference i.e. 4
Hence, the given list forms an AP with common difference \( d = 4 \).

Question. The first term of an A.P. is 5 and the last term is 45. If the sum of all the terms is 400, the number of terms is
(a) 20
(b) 8
(c) 10
(d) 16
Answer: (d) 16
Explanation:
Let there be ‘n’ terms in AP.
Here, \( a = 5 \) and \( a_n = 45 \)
Also, \( S_n = \frac{n}{2}[a + a_n] = 400 \)
\( \Rightarrow \frac{n}{2} (5 + 45) = 400 \)
\( \Rightarrow n = 16 \)
Thus, AP has 16 terms.

Question. The common difference of the A.P. \( \frac{1}{p}, \frac{1-p}{p}, \frac{1-2p}{p}, .......... \) is
(a) 1
(b) \( \frac{1}{p} \)
(c) – 1
(d) \( -\frac{1}{p} \)
Answer: (c) – 1
Explanation:
The common difference = \( \frac{1-p}{p} - \frac{1}{p} = \frac{1-p-1}{p} = \frac{-p}{p} = -1 \)

Question. The \( n^{th} \) term of the A.P. \( a, 3a, 5a, ....... \) is
(a) \( na \)
(b) \( (2n - 1)a \)
(c) \( (2n + 1)a \)
(d) \( 2na \)
Answer: (b) \( (2n - 1)a \)
Explanation: Here, first term = a and common difference, \( d = 2a \)
So, \( a_n = a + (n - 1) d = a + 2a (n - 1) \)
\( = 2an - a = a (2n - 1) \)

Question. The \( 11^{th} \) term of the AP: \( -5, -\frac{5}{2}, 0, \frac{5}{2}, ... \) is
(a) –20
(b) 20
(c) –30
(d) 30
Answer: (b) 20
Explanation:
The given list of numbers is \( -5, -\frac{5}{2}, 0, \frac{5}{2}, ... \)
Here, \( a = a_1 = -5, a_2 = -\frac{5}{2}, a_3 = 0, a_4 = \frac{5}{2} \)
\( d = -\frac{5}{2} - (-5) = -\frac{5}{2} + 5 = \frac{5}{2} \) [\( \because d = a_2 - a_1 \)]
\( \therefore a_{11} = ? \)
We know that \( a_n = a + (n - 1)d \)
\( \Rightarrow a_{11} = a + (11 - 1)d \)
\( = (-5) + (10)\left(\frac{5}{2}\right) \)
\( = -5 + (5)(5) = -5 + 25 = 20 \)
\( \Rightarrow a_{11} = 20 \)

Question. The first four terms of an AP, 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) –2, –4, –6, –8
Explanation: It is given that the first term, \( a = -2 \)
and common difference, \( d = -2 \)
We know that \( a_n = a + (n - 1)d \)
\( a_1 = -2 + (1 - 1)(-2) = -2 \)
\( a_2 = -2 + (2 - 1)(-2) = -2 - 2 = -4 \)
\( a_3 = -2 + (3 - 1)(-2) \)
\( = -2 + (2)(-2) = -2 - 4 = -6 \)
\( a_4 = -2 + (4 - 1)(-2) \)
\( = -2 + (3 )(-2) = -2 - 6 = -8 \)
Hence, the first four terms of the AP are –2, –4, –6, –8

Question. The \( 21^{st} \) term of the AP whose first two terms are –3 and 4 is:
(a) 17
(b) 137
(c) 143
(d) –143
Answer: (b) 137
Explanation: It is given that
\( a_1 = -3 \), \( a_2 = 4 \)
We know that \( a_n = a + (n - 1)d \)
\( \Rightarrow a_1 = a + (1 - 1)d = a \)
\( \Rightarrow a_2 = a + (2 - 1)d = a + d \)
\( \Rightarrow a_1 = -3 \) and \( a_2 = a + d = 4 \)
\( \Rightarrow -3 + d = 4 \)
\( \Rightarrow d = 4 + 3 = 7 \)
\( \therefore a_{21} = a + (21 - 1)d = -3 + (20)7 \)
\( = -3 + 140 = 137 \)
\( \Rightarrow a_{21} = 137 \)

Question. Which term of the AP: 21, 42, 63, 84, ... is 210?
(a) \( 9^{th} \)
(b) \( 10^{th} \)
(c) \( 11^{th} \)
(d) \( 12^{th} \)
Answer: (b) \( 10^{th} \)
Explanation: The given series is 21, 42, 63, 84, ...
Here, the first term, \( a = 21 \)
and common difference, \( d = 42 - 21 = 21 \)
Let the \( n^{th} \) term of the given AP be 210
We know that \( a_n = a + (n - 1)d \)
\( \Rightarrow 210 = 21 + (n - 1)21 \)
\( \Rightarrow 210 = 21 + 21n - 21 \)
\( \Rightarrow 210 = 21n \)
\( \Rightarrow n = 10 \)
Hence, 210 is the \( 10^{th} \) term of the AP.

Question. The value of x for which 2x, (x + 10) and (3x + 2) are the three consecutive terms of an AP, is
(a) 6
(b) –6
(c) 18
(d) –18
Answer: (a) 6
Explanation:
Since 2x, (x + 10) and (3x + 2) are the three consecutive terms of an AP,
\( 2(x + 10) = 2x + (3x + 2) \) [\( \because 2b = a + c \)]
\( \Rightarrow 2x + 20 = 5x +2 \)
\( \Rightarrow 3x = 18 \)
\( \Rightarrow x = 6 \)

Question. The first term of an AP is p and the common difference is q, then its 10th term is
(a) q + 9p
(b) p – 9q
(c) p + 9q
(d) 2p + 9q
Answer: (c) p + 9q
Explanation:
Here, \( a = p \) and \( d = q \). Then
\( a_{10} = a + (10 - 1)d = p + 9q \)

Question. If the common difference of an AP is 5, then what is \( a_{18} - a_{13} \)?
(a) 5
(b) 20
(c) 25
(d) 30
Answer: (c) 25
Explanation: It is given that common difference, \( d = 5 \)
\( a_{18} - a_{13} = ? \)
We know \( a_n = a + (n - 1)d \)
\( a_{18} = a + (18 - 1)d = a + 17d \)
\( a_{13} = a + (13 - 1)d = a + 12d \)
Now, \( a_{18} - a_{13} = (a + 17d) - (a + 12d) \)
\( = a + 17d - a - 12d \)
\( = 5d = 5 \times 5 = 25 \) [As \( d = 5 \)]
\( \Rightarrow a_{18} - a_{13} = 25 \)

Question. Two APs have the same common difference. The first term of one of these is –1 and that of the other is –8. Then the difference between their \( 4^{th} \) terms is:
(a) –1
(b) –8
(c) 7
(d) –9
Answer: (c) 7
Explanation: Let d be the common difference of the two AP’s and a be the first term of the first AP and \( A_1 \) be the first term of the second AP. We have to find out \( |a_4 - A_4| = ? \)
Also \( a_1 = -1 \), \( A_1 = -8 \)
We know that \( a_n = a + (n - 1)d \)
\( a_4 = a_1 + (4 - 1)d = a + 3d = -1 + 3d \)
\( A_4 = A_1 + (4 - 1)d = -8 + 3d \)
Now, the difference between their \( 4^{th} \) terms will be
\( |a_4 - A_4| = (-1 + 3d) - (-8 + 3d) \)
\( = -1 + 3d + 8 - 3d \)
\( = 7 \)
Hence, the required difference is 7.

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) Gauss
Explanation: Newton is famous for his laws of physics.
Pythagoras is famous for the pythagorean theorem of a right angled triangle.
Gauss is the famous mathematician associated with finding the sum of the first 100 natural numbers.
Euclid is most famous for his work in geometry.

Question. If k, 2k – 1 and 2k + 1 are three consecutive terms of an AP, then the value of k is:
(a) 2
(b) 3
(c) –3
(d) 5
Answer: (b) 3
Explanation:
Here, \( k + (2k + 1) = 2 (2k - 1) \)
i.e. \( 3k + 1 = 4k - 2 \)
\( \Rightarrow k = 3 \)

Question. If the first term of an AP 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) 0
Explanation: It is given that the first term, \( a = -5 \) and common difference, \( d = 2 \)
\( S_6 = ? \)
We know that the sum of n terms of an AP is
\( S_n = \frac{n}{2} \{2a + (n - 1)d\} \)
\( \Rightarrow S_6 = \frac{6}{2} \{2(-5) + (6 - 1)(2)\} \)
\( \Rightarrow S_6 = 3\{-10 + 5(2)\} = 3\{-10 + 10\} = 3(0) = 0 \)
\( \Rightarrow S_6 = 0 \)

Question. The \( 11^{th} \) term of the AP: \( \sqrt{2}, 3\sqrt{2}, 5\sqrt{2}, .... \) is:
(a) \( 17\sqrt{2} \)
(b) \( 19\sqrt{2} \)
(c) \( 21\sqrt{2} \)
(d) \( 23\sqrt{2} \)
Answer: (c) \( 21\sqrt{2} \)
Explanation:
Here, \( 11^{th} \) term \( = a + 10d = \sqrt{2} + 10(2\sqrt{2}) \)
\( = 21\sqrt{2} \)

Question. The sum of the first 16 terms of the AP: 10, 6, 2, ... is:
(a) –320
(b) 320
(c) –352
(d) –400
Answer: (a) –320
Explanation: The given series of AP is 10, 6, 2 ...
Here, the first term, \( a = 10 \)
and common difference, \( d = a_2 - a_1 = 6 - 10 = -4 \)
Sum of 16 terms, \( S_{16} = ? \)
We know that \( S_n = \frac{n}{2} \{2a + (n - 1)d\} \)
\( \Rightarrow S_{16} = \frac{16}{2} \{2(10) + (16 - 1)(-4)\} \)
\( = 8\{20 + 15(-4)\} = 8\{20 - 60\} \)
\( = 8(-40) = -320 \)
\( \Rightarrow S_{16} = -320 \)

Question. In an AP if \( a = 1, a_n = 20 \) and \( S_n = 399 \), then n is:
(a) 19
(b) 21
(c) 38
(d) 42
Answer: (c) 38
Explanation: It is given that the first term, \( a = 1 \)
and \( n^{th} \) term, \( a_n = 20 \)
Sum of n terms, \( S_n = 399 \)
We know that
\( a_n = a + (n - 1)d \)
\( \Rightarrow 20 = 1 + (n - 1)d \)
\( \Rightarrow (n - 1)d = 19 \) ... (i)
Also we know that
\( S_n = \frac{n}{2} \{2a + (n - 1)d\} \)
\( \Rightarrow 399 = \frac{n}{2} \{2(1) + (n - 1)d\} \)
\( \Rightarrow 798 = n [2 + (n - 1)d] \)
\( \Rightarrow 798 = 2n + n(n - 1)d \) ... (ii)
Using equation (i) & (ii) we get
\( 798 = 2n + 19n \)
\( \Rightarrow 798 = 21n \)
\( \Rightarrow n = \frac{798}{21} = 38 \)
i.e., \( n = 38 \)

Fill in the Blanks

Fill in the blanks/tables with suitable information:

Question. Fill the two blanks in the sequence 2, ....., 26, ..... so that the sequence forms an A.P.
Answer: 14, 38

Question. The sum of first 16 terms of the AP 5, 8, 11, 14, ...... is ............................... .
Answer: 440
Explanation: Here first term = 5
Common difference = 8 - 5 = 3
Number of terms = 16
\( S_n = \frac{n}{2} [2a + (n - 1)d] \)
\( S_{16} = \frac{16}{2} [2 \times 5 + (16 - 1) \times 3] \)
\( = 8 [10 + 45] \)
\( = 8 \times 55 = 440 \)

Question. The common difference of an A.P. is 6, then \( a_{15} - a_{11} \) .................................. .
Answer: 24
Explanation: Let a be the first term and d be the common difference.
\( n^{th} \) term = \( a_n = a + (n - 1) d \)
Now, \( a_{15} = a + (15 - 1) d = a + 14d \)
\( a_{11} = a + (11 - 1) d = a + 10d \)
\( a_{15} - a_{11} = (a + 14d) - (a + 10d) = 4d \)
As \( d = 6 \)
\( a_{15} - a_{11} = 4 \times 6 = 24 \)

Question. If 4/5, a, 2 are three consecutive terms of an AP then the value of a is ............................... .
Answer: \( \frac{7}{5} \)
Explanation: Given \( \frac{4}{5}, a, 2 \) are in AP
Then, \( a - \frac{4}{5} = 2 - a \)
\( \Rightarrow 2a = 2 + \frac{4}{5} \Rightarrow 2a = \frac{14}{5} \)
\( \Rightarrow a = \frac{7}{5} \)

Question. If \( 4, x_1, x_2, x_3, 28 \) are in AP then \( x_3 \) = ...................... .
Answer: 22
Explanation: Given, \( 4, x_1, x_2, x_3, 28 \) are in AP
Let d be the common difference
Now, first term, a = 4
and fifth term, \( a_5 = 28 \)
\( a_5 = a + (5 - 1) d = a + 4d \)
\( \Rightarrow 28 = 4 + 4d \Rightarrow 4d = 24 \Rightarrow d = 6 \)
\( x_3 = a + 3d = 4 + 3 \times 6 = 22 \)

Question. If \( S_n = 5n^2 + 3n \), then \( n^{th} \) term is ........................... .
Answer: \( 10n - 2 \)
Explanation: \( a_n = S_n - S_{n-1} \)
\( = 5n^2 + 3n - [5(n - 1)^2 + 3(n - 1)] \)
\( = 5n^2 + 3n - [5n^2 + 5 - 10n + 3n - 3] \)
\( = 10n - 2 \)

Short Answer Questions

Question. Find the \( 16^{th} \) term of the AP: 2, 7, 12, 17, ....... .
Answer: 77
Explanation:
Here, \( a = 2 \)
\( d = 7 - 2 = 12 - 7 = 5 \)
\( a_{16} = a + (16 - 1)d \)
\( = 2 + 15 \times 5 \)
\( = 2 + 75 = 77 \)

Question. The number of terms of AP: 18, 16, 14, ... that make the sum zero, is ................
Answer: 19
Explanation:
Let n terms of the given AP make the sum zero.
Then,
\( \frac{n}{2} [18 \times 2 + (n - 1)(-2)] = 0 \)
\( \Rightarrow 36 - 2 ( n - 1) = 0 \Rightarrow 36 - 2n + 2 = 0 \)
\( \Rightarrow 2n = 38 \)
\( \Rightarrow n = 19 \)

Question. Second term of the AP if its \( S_n = n^2 + 2n \) is .....................
Answer: 5
Explanation:
Here, \( a_2 = S_2 - S_1 = (2^2 + 2 \times 2) - (1^2 + 2 \times 1) \)
\( = (4 + 4) - (1 + 2) \)
\( = 5 \)

Question. \( 10^{th} \) term from end of AP: 4, 9, 14, ...., 254 is ............................... .
Answer: 209
Explanation:
The given AP in reverse form is:
254, 249, 244, ...., 14, 9, 4.
Here, \( a = 254, d = -5 \)
So, \( a_{10} = 254 + 9(-5) \)
\( = 254 - 45 = 209 \)