CBSE Class 10 Mathematics Arithmetic Progressions VBQs Set 02

Multiple Choice Questions

Question. Which of the following is not an A.P.?
(a) – 1.2, 0.8, 2.8, ....
(b) \( 3, 3 + \sqrt{2}, 3 + 2\sqrt{2}, 3 + 3\sqrt{2}, \dots \)
(c) \( \frac{4}{3}, \frac{7}{3}, \frac{9}{3}, \frac{12}{3}, \dots \)
(d) \( -\frac{1}{5}, -\frac{2}{5}, -\frac{3}{5}, \dots \)
Answer: (c)
Explanation: Here, \( \frac{7}{3} - \frac{4}{3} \neq \frac{9}{3} - \frac{7}{3} \) as \( 1 \neq \frac{2}{3} \).
So, 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)
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)
Explanation: The given list of number is –10, –6, –2, 2, ...
Here, \( a_1 = -10, a_2 = -6, a_3 = -2 \) and \( a_4 = 2 \dots \)
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)
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. \( \frac{1}{p}, \frac{1-p}{p}, \frac{1-2p}{p}, \dots \) is:
(a) 1
(b) \( \frac{1}{p} \)
(c) – 1
(d) \( -\frac{1}{p} \)
Answer: (c)
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. \( a, 3a, 5a, \dots \) is:
(a) na
(b) (2n – 1)a
(c) (2n + 1)a
(d) 2na
Answer: (b)
Explanation: Here, first term = \( a \) and common difference, \( d = 2a \).
So, \( a_n = a + (n - 1)d = a + 2a(n - 1) \)
\( = a + 2an - 2a = 2an - a = a(2n - 1) \)

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)
Explanation: The given list of numbers is \( -5, -\frac{5}{2}, 0, \frac{5}{2}, \dots \)
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 \( a_n = a + (n - 1)d \)
\( a_{11} = a + (11 - 1)d \)
\( = (-5) + (10)\left(\frac{5}{2}\right) \)
\( = -5 + (5)(5) = -5 + 25 = 20 \)
\( a_{11} = 20 \)

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)
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)
Explanation: It is given that \( a_1 = -3, a_2 = 4 \).
We know that \( a_n = a + (n - 1)d \)
\( a_1 = a + (1 - 1)d = a \)
\( a_2 = a + (2 - 1)d = a + d \)
\( \therefore a_1 = -3 \) and \( a_2 = a + d = 4 \)
\( \Rightarrow -3 + d = 4 \Rightarrow d = 4 + 3 = 7 \)
\( \therefore a_{21} = a + (21 - 1)d = -3 + (20)7 = -3 + 140 = 137 \)
\( a_{21} = 137 \)

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)
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 \( a_n = a + (n - 1)21 \)
\( 210 = 21 + (n - 1)21 \)
\( 210 = 21 + 21n - 21 \)
\( 210 = 21n \Rightarrow n = 10 \)
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)
Explanation: Since \( 2x, (x + 10) \) and \( (3x + 2) \) are the three consecutive terms of an AP,
\( 2(x + 10) = 2x + (3x + 2) \) [\( \because 2b = a + c \)]
\( 2x + 20 = 5x + 2 \)
\( 3x = 18 \Rightarrow x = 6 \)

Question. The first term of an AP is \( p \) and the common difference is \( q \), then its \( 10^{th} \) term is:
(a) \( q + 9p \)
(b) \( p - 9q \)
(c) \( p + 9q \)
(d) \( 2p + 9q \)
Answer: (c)
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)
Explanation: It is given that common difference, \( d = 5 \).
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, \( a_{18} - a_{13} = (a + 17d) - (a + 12d) = 5d \)
\( = 5 \times 5 = 25 \) [As \( d = 5 \)]
\( a_{18} - a_{13} = 25 \)

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)
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.
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)
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)
Explanation: Here, \( k + (2k + 1) = 2 (2k - 1) \)
i.e. \( 3k + 1 = 4k - 2 \)
\( k = 3 \)

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

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

Question. The sum of the first 16 terms of the AP 10, 6, 2, ... is:
(a) –320
(b) 320
(c) –352
(d) –400 [2
Answer: (a)
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 \( S_n = \frac{n}{2} \{2a + (n - 1)d\} \)
\( S_{16} = \frac{16}{2} \{2(10) + (16 - 1)(-4)\} \)
\( = 8[20 + 15(-4)] = 8\{20 - 60\} \)
\( = 8(-40) = -320 \)
\( S_{16} = -320 \)

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)
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 \( a_n = a + (n - 1)d \)
\( 20 = 1 + (n - 1)d \Rightarrow (n - 1)d = 19 \) ... (i)
Also we know that \( S_n = \frac{n}{2} \{2a + (n - 1)d\} \)
\( 399 = \frac{n}{2} \{2(1) + (n - 1)d\} \)
\( 798 = n [2 + 19] \) [Using equation (i)]
\( 798 = 21n \Rightarrow n = \frac{798}{21} = 38 \)
i.e., \( n = 38 \)

Fill in the Blanks

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 \( a = 5 \), common difference \( d = 8 - 5 = 3 \).
Number of terms \( n = 16 \).
\( S_n = \frac{n}{2} [2a + (n - 1)d] \)
\( S_{16} = \frac{16}{2} [2 \times 5 + (16 - 1) \times 3] \)
\( = 8 [10 + 45] = 8 \times 55 = 440 \)

Question. The common difference of an A.P. 6, then \( a_{15} - a_{11} \) is .................................. .
Answer: 24
Explanation: Let \( a \) be the first term and \( d \) be the common difference.
\( n^{th} \text{ term} = a_n = a + (n - 1)d \)
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 \( \frac{4}{5}, a, 2 \) 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, \( a_5 = 28 \).
\( a_5 = a + (5 - 1)d = 28 \)
\( \Rightarrow 4 + 4d = 28 \Rightarrow 4d = 24 \Rightarrow d = 6 \).
\( x_3 = a + 3d = 4 + 3 \times 6 = 22 \).

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

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

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, \( \frac{n}{2} [2 \times 18 + (n - 1)(-2)] = 0 \)
\( 36 - 2(n - 1) = 0 \Rightarrow 36 - 2n + 2 = 0 \)
\( 2n = 38 \Rightarrow n = 19 \).

Question. Second term of the AP if its \( S_n = n^2 + 2n \) is .....................
Answer: 5
Explanation: Here, \( a_2 = S_2 - S_1 = (2^2 + 2 \times 2) - (1^2 + 2 \times 1) = 8 - 3 = 5 \).

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, \( a_{10} = 254 + 9(-5) = 254 - 45 = 209 \).

Very Short Questions

Question. Find the sum of the first 100 natural numbers.
Answer: The list of first 100 natural numbers is 1, 2, 3, ......, 100, which forms an AP with \( a = 1, d = 1 \).
So, \( S_{100} = \frac{100}{2} [2(1) + (100 - 1)(1)] = 50 [101] = 5050 \).

Question. If the mean of the first \( n \) natural number is 15, then find \( n \).
Answer: The natural number are 1, 2, 3 ..... \( n \).
Their mean \( = \frac{S_n}{n} \) and \( S_n = \frac{n}{2} [2a + (n - 1)d] \).
The mean of first \( n \) natural numbers is \( \frac{n(n + 1)}{2n} = \frac{n + 1}{2} \).
\( \therefore \frac{n + 1}{2} = 15 \) (given)
\( \Rightarrow n = 29 \).

Question. If in an A.P., \( a = 15, d = -3 \) and \( a_n = 0 \), then find the value of \( n \).
Answer: Given, for an A.P., \( a = 15, d = -3 \) and \( a_n = 0 \).
Now, \( a_n = a + (n - 1)d \)
\( \Rightarrow 0 = 15 + (n - 1) \times (-3) \)
\( \Rightarrow (n - 1) = 5 \)
\( \Rightarrow n = 6 \).
Hence, the value of \( n \) is 6.

Question. Find the number of terms in the A.P. : 18, \( 15\frac{1}{2} \), 13, ..., – 47. 
Answer: Given A.P. is 18, \( 15\frac{1}{2} \), 13, ....., – 47.
Here, first term, \( a = 18 \).
Common difference, \( d = \frac{31}{2} - 18 = \frac{31 - 36}{2} = -\frac{5}{2} \).
Last term, \( a_n = -47 \).
Now, \( a_n = a + (n - 1)d \), where, ‘n’ is the number of terms.
\( -47 = 18 + (n - 1) \times \left(-\frac{5}{2}\right) \)
\( (n - 1) \times \left(-\frac{5}{2}\right) = -65 \)
\( (n - 1) = 26 \Rightarrow n = 27 \).
Hence, the number of terms in the given A.P. is 27.

Question. Find the common difference of the Arithmetic Progression (A.P.) \( \frac{1}{a}, \frac{3-a}{3a}, \frac{3-2a}{3a}, \dots (a \neq 0) \). 
Answer: Given Arithmetic progression (AP) is \( \frac{1}{a}, \frac{3-a}{3a}, \frac{3-2a}{3a}, \dots (a \neq 0) \).
In the given progression,
\( a_1 = \frac{1}{a}, a_2 = \frac{3-a}{3a}, a_3 = \frac{3-2a}{3a} \).
Common difference,
\( d = a_2 - a_1 = \frac{3-a}{3a} - \frac{1}{a} = \frac{3-a-3}{3a} = \frac{-a}{3a} = -\frac{1}{3} \).
Hence, the common difference of the A.P. is \( -\frac{1}{3} \).

Question. Justify whether it is true to say that –1, \( \frac{3}{2} \), –2, \( \frac{5}{2} \), ... form an AP as \( a_2 - a_1 = a_3 - a_2 \).
Answer: False
Explanation: The given series of numbers is –1, \( \frac{3}{2}, -2, \frac{5}{2}, \dots \)
\( a_2 - a_1 = \frac{3}{2} - (-1) = \frac{3}{2} + 1 = \frac{5}{2} \).
\( a_3 - a_2 = -2 - \frac{3}{2} = -\frac{7}{2} \).
As \( a_2 - a_1 \neq a_3 - a_2 \), it does not form an AP.