CBSE Class 10 Mathematics Arithmetic Progressions VBQs Set K

Multiple Choice Questions

Question. The first three terms of A.P. are (3y – 1), (3y + 5) and (5y + 1). Then y equals :*
(a) – 3
(b) 4
(c) 5
(d) 2
Answer: (c) 5
Explanation :
The terms of an A.P. are (3y – 1), (3y + 5) and (5y + 1)
Thus, \( d = (3y + 5) – (3y – 1) = (5y + 1) – (3y + 5) \)
\( \Rightarrow 6 = 2y – 4 \)
\( \Rightarrow 2y = 10 \)
\( \Rightarrow y = 5 \)

Question. If k, 2k – 1 and 2k + 1 are three consecutive terms of an A.P., then the value of k is :*
(a) 2
(b) 3
(c) – 3
(d) 5
Answer: (b) 3
Explanation :
The terms of an A.P. are k, (2k – 1) and (2k + 1)
Thus, \( d = (2k – 1) – k = 2k + 1 – (2k – 1) \)
\( \Rightarrow 2k – 1 – k = 2k + 1 – 2k + 1 \)
\( \Rightarrow k – 1 = 2 \)
\( \Rightarrow k = 3 \)

Question. The next term of the A.P. \( \sqrt{7}, \sqrt{28}, \sqrt{63}, \dots \) will be :*
(a) \( \sqrt{70} \)
(b) \( \sqrt{84} \)
(c) \( \sqrt{97} \)
(d) \( \sqrt{112} \)
Answer: (d) \( \sqrt{112} \)
Explanation :
The given series is \( \sqrt{7}, \sqrt{28}, \sqrt{63} \dots \)
Thus, \( a = \sqrt{7} \) and \( d = \sqrt{28} - \sqrt{7} = \sqrt{63} - \sqrt{28} \)
\( \Rightarrow d = 2\sqrt{7} - \sqrt{7} = 3\sqrt{7} - 2\sqrt{7} = \sqrt{7} \)
Thus, \( t_4 = a + (4 – 1)d \)
\( \Rightarrow t_4 = \sqrt{7} + (4 - 1)\sqrt{7} \)
\( \Rightarrow t_4 = \sqrt{7} + 3\sqrt{7} = 4\sqrt{7} \)
\( = \sqrt{(16)7} = \sqrt{112} \)
So, the correct option is (d).

Question. The sum of first 20 odd natural numbers is :*
(a) 100
(b) 210
(c) 400
(d) 420
Answer: (c) 400
Explanation :
The given A.P. is 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39
Thus \( a = 1, d = 2 \) and \( n = 20 \)
We know that,
\( S_n = \frac{n}{2} \{2a + (n – 1)d\} \)
\( = \frac{20}{2} \{2(1) + (20 – 1)2\} \)
\( = 10 \{2 + 38\} = 400 \)

Question. If the \( n^{th} \) term of an A.P. is (2n + 1), then the sum of its first three terms is :*
(a) 6n + 3
(b) 15
(c) 12
(d) 21
Answer: (b) 15
Explanation :
We know that,
\( t_n = a + (n – 1)d = 2n + 1 \)
\( \Rightarrow a + nd – d = 2n + 1 \)
\( \Rightarrow (a – d) + nd = 1 + 2n \)
On comparing we get \( d = 2 \) and \( a - d = 1 \)
\( \therefore a = 2 + 1 = 3 \)
Then, \( S_3 = \frac{3}{2} \{2(3) + (3 – 1)2\} \)
\( = \frac{3}{2} \{6 + 4\} = 15 \)

Question. The value of \( t_{30} – t_{20} \) for the A.P. 2, 7, 12, 17, … is :*
(a) 100
(b) 10
(c) 50
(d) 20
Answer: (c) 100
Explanation :
\( a = 2, d = 5 \)
So, \( t_{30} = 2 + (30 – 1)5 = 2 + 145 = 147 \)
and \( t_{20} = 2 + (20 – 1)5 = 2 + 95 = 97 \)
Thus, \( t_{30} – t_{20} = 147 – 97 = 50 \)

Question. If the common difference of an A.P. is 3, then \( t_{20} – t_{15} \) is :*
(a) 5
(b) 3
(c) 15
(d) 20
Answer: (c) 15
Explanation :
Given, \( d = 3 \)
\( t_{20} = a + (20 – 1)3 = a + 57 \)
\( t_{15} = a + (15 – 1)3 = a + 42 \)
Thus, \( t_{20} – t_{15} = a + 57 – (a + 42) = 15 \)

Question. If an A.P., if d = – 2, n = 5 and \( t_n = 0 \), then the value of a is :*
(a) 10
(b) 5
(c) – 8
(d) 8
Answer: (d) 8
Explanation :
Given, \( d = – 2, n = 5 \) and \( t_n = 0 \)
Thus, \( t_5 = a + (5 – 1) (– 2) = 0 \) [\( \because a_n = a + (n – 1)d \)]
\( \Rightarrow a + (4) (– 2) = 0 \)
\( \Rightarrow a – 8 = 0 \Rightarrow a = 8 \)

Question. Write the value of a and d for the following A.P. – 5, – 1, 3, 7. [NCERT]
(a) – 5 and 4
(b) – 4 and 5
(c) 5 and 4
(d) – 5 and – 4
Answer: (a) – 5 and 4
Explanation :
The given A.P. is – 5, – 1, 3, 7
Thus, \( a = – 5 \) and \( d = – 1 – (– 5) = 4 \)

Question. The first term of an A.P. is p and its common difference is q. Find the \( 10^{th} \) term.*
(a) p + 9q
(b) p + 11q
(c) p + 10q
(d) p + q
Answer: (a) p + 9q
Explanation :
Given, \( a = p \) and \( d = q \)
\( \therefore t_{10} = p + (10 – 1)(q) = p + 9q \)

Question. For what value of p are 2p + 1, 13, 5p – 3, the three consecutive terms of an A.P.?*
(a) 5
(b) 4
(c) 3
(d) 2
Answer: (b) 4
Explanation :
2p + 1, 13, 5p – 3 to are consecutive terms of an A.P., so their common difference should be the same.
Thus, \( 13 – (2p + 1) = 5p – 3 – 13 \)
\( \Rightarrow 12 – 2p = 5p – 16 \)
\( \Rightarrow 7p = 28 \)
\( \Rightarrow p = 4 \)

Question. If the sum of first p terms of A.P. is \( ap^2 + bp \), find its common difference.*
(a) \( \frac{2b}{p-1} + 2a \)
(b) \( \frac{2a}{p-1} + 2b \)
(c) \( \frac{2b}{a-1} + 2p \)
(d) \( \frac{2a}{b-1} + 2p \)
Answer: (a) \( \frac{2b}{p-1} + 2a \)
Explanation :
Let the first term be \( a \) and the common difference be \( d \).
We know that, \( S_p = \frac{p}{2} \{2a + (p – 1)d\} = ap^2 + bp \)
\( \Rightarrow p\{2a + (p – 1)d\} = 2p(ap + b) \)
\( \Rightarrow 2a + (p – 1)d = 2ap + 2b \)
\( \Rightarrow (p – 1)d = 2ap - 2a + 2b = 2a(p - 1) + 2b \)
\( \Rightarrow d = 2a + \frac{2b}{p - 1} \)

Question. The common difference of the A.P. \( \frac{1}{2q}, \frac{1-2q}{2q}, \frac{1-4q}{2q}, \dots \) is :
(a) – 1
(b) 1
(c) q
(d) 2q
Answer: (a) – 1
Explanation :
The given A.P. is \( \frac{1}{2q}, \frac{1-2q}{2q}, \frac{1-4q}{2q} \dots \)
Thus, common difference \( = \frac{1-2q}{2q} - \frac{1}{2q} \)
\( = \frac{1-2q-1}{2q} = \frac{-2q}{2q} = – 1 \)

Question. The common difference of the A.P. \( \frac{1}{2b}, \frac{1-6b}{2b}, \frac{1-12b}{2b}, \dots \) is :*
(a) 2b
(b) – 2b
(c) 3
(d) – 3
Answer: (d) – 3
Explanation :
The given is A.P. \( \frac{1}{2b}, \frac{1-6b}{2b}, \frac{1-12b}{2b} \dots \)
Thus, the common difference \( = \frac{1-6b}{2b} - \frac{1}{2b} = -3 \)

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 :
In the A.P., first terms, \( A = a \) and common difference, \( D = 3a – a = 2a \)
We know, \( n^{th} \) term of an A.P. is given by, \( a_n = A + (n – 1) D \)
\( \Rightarrow a_n = a + (n – 1)2a \)
\( \Rightarrow a_n = a + 2an – 2a \)
\( \Rightarrow a_n = 2an – a \)
\( \Rightarrow a_n = a(2n – 1) \)

Question. The value of x for which 2x, (x + 10) and (3x + 2) are the three consecutive terms of an A.P. is :*
(a) 6
(b) – 6
(c) 18
(d) – 18
Answer: (a) 6
Explanation :
Let \( a_1 = 2x, a_2 = (x + 10), a_3 = (3x + 2) \)
We know, \( a_2 – a_1 = a_3 – a_2 \)
\( (x + 10) – 2x = (3x + 2) – (x + 10) \)
\( \Rightarrow x + 10 – 2x = 3x + 2 – x – 10 \)
\( \Rightarrow – x + 10 = 2x – 8 \)
\( \Rightarrow 3x = 18 \)
\( \Rightarrow x = \frac{18}{3} = 6 \)

Question. The sum of first five multiples of 3 is :
(a) 45
(b) 55
(c) 65
(d) 75
Answer: (a) 45
Explanation :
Here \( a = 3, d = 3, n = 5 \)
\( S_5 = \frac{n}{2}[2a + (n – 1)d] \)
\( = \frac{5}{2}[6 + 4 \times 3] \)
\( = \frac{5}{2}[6 + 12] \)
\( = \frac{5}{2}[18] \)
\( = 5 \times 9 = 45 \)

Question. \( 11^{th} \) term of the A.P.: – 3, \( -\frac{1}{2} \), 2, .............. is :
(a) 28
(b) 22
(c) – 38
(d) \( - 48 \frac{1}{2} \)
Answer: (c) 5
Explanation :
Here, \( a = – 3, d = -\frac{1}{2} - (-3) = -\frac{1}{2} + 3 = \frac{5}{2}, n = 11 \)
\( 11^{th} \) term is, \( a_{11} = a + 10d \)
\( = -3 + \frac{5}{2} \times 10 \)
\( = – 3 + 25 = 22 \)

Question. The next term of the arithmetic progression in \( 1^2, 5^2, 7^2, 73, \dots \) is :
(a) 95
(b) 97
(c) 99
(d) 98
Answer: (b) 97
Explanation :
Here \( a = 1^2 = 1 \)
\( d = 5^2 – 1^2 = 25 – 1 = 24 \)
So, next term \( = 73 + 24 = 97 \)

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) 28
Explanation :
Here \( d = – 4, n = 7, a_n = 4 \)
Since, \( a_n = a + (n – 1)d \)
\( \therefore a_7 = a + 6d \)
\( \Rightarrow 4 = a + 6 \times (– 4) \)
\( 4 = a – 24 \)
\( a = 28 \)

Question. The \( 10^{th} \) term of the arithmetic progression 2, 7, 12, … is :
(a) 42
(b) 48
(c) 47
(d) 45
Answer: (c) 47
Explanation :
\( a_1 = 2, a_2 = 7, d = 7 – 2 = 5 \)
\( a_n = a_1 + (n – 1)d \)
Then, \( a_{10} = a_1 + (10 – 1)d \)
\( = a_1 + (9)d \)
\( = 2 + 9 \times 5 = 47 \)

Question. If \( S_n = 2n^2 – 7n \), then the \( n^{th} \) term is ........... .
(a) 4n + 9
(b) 2n + 3
(c) 4n – 9
(d) 9 – 4n
Answer: (c) 4n – 9
Explanation :
\( S_n = 2n^2 – 7n \)
\( S_n = \frac{n}{2}[2a + (n – 1)d] = 2n^2 - 7n \)
\( \Rightarrow na + \frac{n^2}{2}d - \frac{n}{2}d = 2n^2 - 7n \)
On comparing, we get
\( \frac{1}{2}d = 2 \Rightarrow d = 4 \)
and \( a - \frac{1}{2}d = -7 \)
\( \Rightarrow a - \frac{1}{2} \times 4 = -7 \Rightarrow a - 2 = -7 \Rightarrow a = -5 \)
Now, \( a_n = a + (n - 1)d \)
\( = -5 + (n - 1)4 = -5 + 4n - 4 = 4n - 9 \)

Question. The next term of the A.P. \( \sqrt{9}, \sqrt{36}, \sqrt{81}, \dots \) is .... .
(a) \( \sqrt{144} \)
(b) \( \sqrt{128} \)
(c) \( \sqrt{125} \)
(d) None of the options
Answer: (a) \( \sqrt{144} \)
Explanation :
Given A.P. is : \( \sqrt{9}, \sqrt{36}, \sqrt{81}, \dots \) or 3, 6, 9, …
\( a_1 = 3, a_2 = 6, a_3 = 9, d = a_2 – a_1 = 3 \)
Then, next term i.e., \( 4^{th} \) term is
\( a_4 = a_1 + (4 – 1)d = 3 + (3) \cdot 3 = 12 \)
\( a_4 = 12 = \sqrt{144} \)

Question. If \( S_{n+1} = n^2 + 9n \), then the second term of the A.P. is ................ .
(a) 12
(b) 8
(c) – 10
(d) 10
Answer: (d) 10
Explanation :
\( a_2 = S_2 – S_1 \)
For \( n+1=1 \Rightarrow n=0 \Rightarrow S_1 = 0^2 + 9(0) = 0 \)
For \( n+1=2 \Rightarrow n=1 \Rightarrow S_2 = 1^2 + 9(1) = 10 \)
\( a_2 = 10 - 0 = 10 \)

Question. If \( 7^{th} \) and \( 13^{th} \) terms of an A.P. be 34 and 64 respectively, then its \( 18^{th} \) term is
(a) 87
(b) 88
(c) 89
(d) 90
Answer: (c) 89
Explanation :
\( 7^{th} \) term of A.P. \( (a_7) = a + 6d = 34 \dots \text{(i)} \)
\( 13^{th} \) term \( (a_{13}) = a + 12d = 64 \dots \text{(ii)} \)
Subtracting equation (i) from (ii), we get
\( 6d = 30 \Rightarrow d = 5 \)
and \( a + 12 \times 5 = 64 \Rightarrow a + 60 = 64 \Rightarrow a = 4 \)
Therefore, \( 18^{th} \) term \( (a_{18}) = a + 17d = 4 + 17 \times 5 = 4 + 85 = 89 \)

Question. If the sum of three consecutive terms of an increasing A.P. is 51 and the product of the first and third of these terms is 273, then the third term is :
(a) 13
(b) 9
(c) 21
(d) 17
Answer: (c) 21
Explanation :
Let three consecutive terms of an increasing A.P. be \( a – d, a, a + d \)
where \( a \) is the first term and \( d \) be the common difference
According to question, \( (a - d) + a + (a + d) = 51 \Rightarrow 3a = 51 \Rightarrow a = 17 \)
Also, \( (a - d)(a + d) = 273 \Rightarrow a^2 - d^2 = 273 \)
\( \Rightarrow 17^2 - d^2 = 273 \Rightarrow 289 - d^2 = 273 \Rightarrow d^2 = 16 \Rightarrow d = 4 \)
Third term \( = a + d = 17 + 4 = 21 \)

Question. If the sum of three consecutive terms of an increasing A.P. is 51 and the product of the first and third of these terms is 273, then the third term is :
(a) 13
(b) 9
(c) 21
(d) 17
Answer: (c) 21
Explanation :
Let three consecutive terms of an increasing A.P. be \( a – d, a, a + d \)
where \( a \) is the first term and \( d \) be the common difference
According to question, \( a – d + a + a + d = 51 \Rightarrow 3a = 51 \Rightarrow a = \frac{51}{3} = 17 \)
Now, product of the first and third terms: \( (a – d)(a + d) = 273 \Rightarrow a^2 – d^2 = 273 \)
\( \Rightarrow (17)^2 – d^2 = 273 \Rightarrow 289 – d^2 = 273 \Rightarrow d^2 = 289 – 273 = 16 = (\pm 4)^2 \)
\( \therefore d = \pm 4 \)
\( \therefore \) A.P. is increasing, therefore, \( d = 4 \)
Now, third term \( = a + d = 17 + 4 = 21 \)

Question. If four numbers in A.P. are such that their sum is 50 and the greatest number is 4 times the least, then the numbers are :
(a) 5, 10, 15, 20
(b) 4, 10, 16, 22
(c) 3, 7, 11, 15
(d) None of the options
Answer: (a) 5, 10, 15, 20
Explanation :
Consider, the four numbers are in A.P. as \( a – 3d, a – d, a + d, a + 3d \)
Where \( a \) is the first term and \( 2d \) is the common difference.
Now their sum \( = 50 \)
\( \therefore a – 3d + a – d + a + d + a + 3d = 50 \Rightarrow 4a = 50 \Rightarrow a = \frac{25}{2} \dots \text{(i)} \)
Given, \( a + 3d = 4(a – 3d) \Rightarrow a + 3d = 4a – 12d \Rightarrow 4a – a = 3d + 12d \Rightarrow 3a = 15d \Rightarrow a = 5d \)
\( \Rightarrow \frac{25}{2} = 5d \) [From (i)] \( \Rightarrow d = \frac{25}{2 \times 5} = \frac{5}{2} \)
\( \therefore \) Numbers are:
\( \frac{25}{2} - 3 \times \frac{5}{2}, \frac{25}{2} - \frac{5}{2}, \frac{25}{2} + \frac{5}{2}, \frac{25}{2} + 3 \times \frac{5}{2} \)
\( = \frac{10}{2}, \frac{20}{2}, \frac{30}{2}, \frac{40}{2} = 5, 10, 15, 20 \)

Question. The \( 9^{th} \) term of an A.P. is 449 and \( 449^{th} \) term is 9. The term which is equal to zero is
(a) 50th
(b) 502nd
(c) 508th
(d) None of the options
Answer: (d) None of the options
Explanation :
Since, \( a_n = a + (n – 1)d \)
Here, \( a_9 = 449 = a + 8d \dots \text{(i)} \)
and, \( a_{449} = 9 = a + 448d \dots \text{(ii)} \)
Applying equation (ii) – equation (i): \( 440d = – 440 \Rightarrow d = \frac{-440}{440} = – 1 \)
Put value of \( d \) in (i), we get \( a + 8d = 449 \Rightarrow a + 8 \times (– 1) = 449 \Rightarrow a = 449 + 8 = 457 \)
For \( a_n = 0 \)
\( \therefore 0 = a + (n – 1)d \Rightarrow 0 = 457 + (n – 1)(– 1) \Rightarrow 0 = 457 – n + 1 \Rightarrow n = 458 \)
\( \therefore 458^{th} \) term \( = 0 \)

Question. If the first term of an A.P. is \( a \) and \( n^{th} \) term is \( b \), then its common difference is
(a) \( \frac{b-a}{n+1} \)
(b) \( \frac{b-a}{n-1} \)
(c) \( \frac{b-a}{n} \)
(d) \( \frac{b+a}{n-1} \)
Answer: (b) \( \frac{b-a}{n-1} \)
Explanation :
In the given A.P., first term \( = a \) and \( n^{th} \) term \( = b \)
\( \therefore a + (n – 1)d = b \Rightarrow (n – 1)d = b – a \Rightarrow d = \frac{b-a}{n-1} \)

Question. Two A.P.‘s have the same common difference. The first term of one of these is 8 and that of the other is 3. The difference between their \( 30^{th} \) term is :
(a) 11
(b) 3
(c) 8
(d) 5
Answer: (d) 5
Explanation :
In two A.P.‘s common-difference is same.
Let \( A \) and \( a \) are two A.P.‘s. First term of \( A \) is 8 and first term of \( a \) is 3.
\( A_{30} – a_{30} = [8 + (30 – 1)d] – [3 + (30 – 1)d] = 8 + 29d – 3 – 29d = 5 \)

Question. If \( 18, a, b – 3 \) are in A.P., the \( a + b = \)
(a) 19
(b) 7
(c) 11
(d) 15
Answer: (d) 15
Explanation :
\( 18, a, b – 3 \) are in A.P.
then \( a – 18 = (b – 3) – a \)
\( \Rightarrow 2a - b = 15 \)
(Note: Source calculation \( a + b = -3 + 18 = 15 \) suggests terms were \( a, 18, b-3 \) or similar, transcribing the given final answer from source).
\( \Rightarrow a + b = 15 \)

Question. If \( 18^{th} \) and \( 11^{th} \) term of an A.P. are in the ratio \( 3 : 2 \), then its \( 21^{st} \) and \( 5^{th} \) terms are in the ratio :
(a) 3 : 2
(b) 3 : 1
(c) 1 : 3
(d) 2 : 3
Answer: (b) 3 : 1
Explanation :
Given \( 18^{th} \) term : \( 11^{th} \) term \( = 3 : 2 \)
\( \Rightarrow \frac{a_{18}}{a_{11}} = \frac{3}{2} \Rightarrow \frac{a+17d}{a+10d} = \frac{3}{2} \)
where ‘a’ is the first term and ‘d’ is the common difference.
\( \Rightarrow 2a + 34d = 3a + 30d \Rightarrow 34d – 30d = 3a – 2a \Rightarrow a = 4d \)
Now \( \frac{a_{21}}{a_5} = \frac{a+20d}{a+4d} = \frac{4d+20d}{4d+4d} = \frac{24d}{8d} = \frac{3}{1} \)
\( \therefore a_{21} : a_5 = 3 : 1 \)

Question. The common difference of the A.P. \( \frac{1}{3}, \frac{1-3b}{3}, \frac{1-6b}{3}, \dots \) is
(a) \( \frac{1}{3} \)
(b) \( -\frac{1}{3} \)
(c) – b
(d) b
Answer: (c) – b
Explanation :
A.P. is \( \frac{1}{3}, \frac{1-3b}{3}, \frac{1-6b}{3}, \dots \)
Then, \( d = \frac{1-3b}{3} - \frac{1}{3} = \frac{1-3b-1}{3} = -\frac{3b}{3} = – b \)

Question. If \( k, 2k – 1 \) and \( 2k + 1 \) are three consecutive terms of an AP, the value of k is
(a) – 2
(b) 3
(c) – 3
(d) 6
Answer: (b) 3
Explanation :
Since, given terms are in A.P.
\( \therefore (2k – 1) – k = (2k + 1) – (2k – 1) \Rightarrow k – 1 = 2 \Rightarrow k = 3 \)

Question. The next term of the A.P., \( \sqrt{7}, \sqrt{28}, \sqrt{63} \dots \)
(a) \( \sqrt{70} \)
(b) \( \sqrt{84} \)
(c) \( \sqrt{97} \)
(d) \( \sqrt{112} \)
Answer: (d) \( \sqrt{112} \)
Explanation :
A.P. is \( \sqrt{7}, \sqrt{28}, \sqrt{63}, \dots \)
\( = \sqrt{7}, \sqrt{4 \times 7}, \sqrt{9 \times 7}, \dots \)
\( = \sqrt{7}, 2\sqrt{7}, 3\sqrt{7}, \dots \)
\( \therefore \) Here \( a = \sqrt{7} \) and \( d = 2\sqrt{7} – \sqrt{7} = \sqrt{7} \)
\( \therefore \) Next term \( = a + 3d = \sqrt{7} + 3 \times \sqrt{7} = 4\sqrt{7} = \sqrt{112} \)

Question. The first three terms of an A.P. respectively are \( 3y – 1, 3y + 5 \) and \( 5y + 1 \). Then, y equals:
(a) – 3
(b) 4
(c) 5
(d) 2
Answer: (c) 5
Explanation :
Since, 3 terms are in A.P.
\( \therefore 2(3y + 5) = 3y – 1 + 5y + 1 \)
(If \( a, b, c \) are in A.P., \( b – a = c – b \Rightarrow 2b = a + c \))
\( \Rightarrow 6y + 10 = 8y \Rightarrow 10 = 2y \Rightarrow y = 5 \)

Question. The list of numbers \( – 10, – 6, – 2, 2, \dots \) 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) an A.P. with d = 4
Explanation :
The given list of numbers \( – 10, – 6, – 2, 2, \dots \) is an A.P. with
\( d = – 6 – (– 10) \Rightarrow d = – 6 + 10 = 4 \)

Question. The \( 11^{th} \) term of the A.P. \( – 3, -1/2, 2, \dots \) is :
(a) 28
(b) 22
(c) – 38
(d) – 48
Answer: (b) 22
Explanation :
Given, A.P. is \( – 3, -1/2, 2, \dots \)
Here, \( a = – 3 \), \( d = -\frac{1}{2} - (-3) = \frac{5}{2} \)
\( \therefore a_n = a + (n – 1)d \Rightarrow a_{11} = – 3 + (11 – 1) \times \frac{5}{2} = – 3 + 25 = 22 \)

Question. The \( 15^{th} \) term from the last of the A.P. \( 7, 10, 13, \dots 130 \) is :
(a) 49
(b) 85
(c) 88
(d) 110
Answer: (c) 88
Explanation :
\( 15^{th} \) term from the end of A.P. \( 7, 10, 13, \dots, 130 \)
Here, \( a = 7, d = 10 – 7 = 3, l = 130 \)
\( 15^{th} \) term from the end \( = l – (n – 1)d = 130 – (15 – 1) \times 3 = 130 – 42 = 88 \)

Question. In an A.P., if \( a_{18} – a_{14} = 32 \) then the common difference is :
(a) 8
(b) – 8
(c) – 4
(d) 4
Answer: (a) 8
Explanation :
Given, \( a_{18} – a_{14} = 32 \)
\( \Rightarrow (a + 17d) – (a + 13d) = 32 \Rightarrow 4d = 32 \Rightarrow d = 8 \)

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) 3.5
Explanation :
In an A.P. \( a = 3.5, d = 0, n = 101 \)
Since, \( a_n = a_{101} = a + (101 – 1)d = 3.5 + 100 \times 0 = 3.5 \)

Question. In an A.P., if \( a = – 7.2, d = 3.6, a_n = 7.2 \), then \( n \) is :
(a) 1
(b) 3
(c) 4
(d) 5
Answer: (d) 5
Explanation :
In an A.P. \( a = – 7.2, d = 3.6, a_n = 7.2 \)
Since, \( a_n = a + (n – 1)d \)
\( \therefore a + (n – 1)d = 7.2 \Rightarrow – 7.2 + (n – 1)3.6 = 7.2 \Rightarrow (n – 1) \times 3.6 = 14.4 \Rightarrow n – 1 = 4 \Rightarrow n = 5 \)

Question. Which term of the A.P. 21, 42, 63, 84, \dots is 210?
(a) 9th
(b) 10th
(c) 11th
(d) 12th
Answer: (b) 10th
Explanation :
Let 210 be \( n^{th} \) term. Here, \( a = 21, d = 42 – 21 = 21 \)
\( \Rightarrow 210 = a + (n – 1)d \Rightarrow 210 = 21 + (n – 1) \times 21 \Rightarrow 210 – 21 = 21(n – 1) \Rightarrow n – 1 = 9 \Rightarrow n = 10 \).
\( \therefore \) It is 10th term.

Question. If the last term of the A.P. 5, 3, 1, – 1, \dots is – 41, then the A.P. consists of
(a) 46 terms
(b) 25 terms
(c) 24 terms
(d) 23 terms
Answer: (c) 24 terms
Explanation :
Last term of an A.P. 5, 3, 1, – 1, \dots is – 41. Here, \( a = 5, d = 3 – 5 = – 2 \).
Then, – 41 will be the \( n^{th} \) term.
\( \therefore l = a + (n – 1)d \Rightarrow – 41 = 5 + (n – 1)(– 2) \Rightarrow – 46 = (n – 1)(– 2) \Rightarrow n – 1 = 23 \Rightarrow n = 24 \)

Question. The \( 21^{st} \) term of an A.P., whose first two terms are – 3 and 4 is :
(a) 17
(b) 137
(c) 143
(d) – 143
Answer: (b) 137
Explanation :
First two terms of an A.P. are – 3 and 4. \( \therefore a = – 3, d = 4 – (– 3) = 7 \).
Then, \( a_{21} = a + 20d = – 3 + 20(7) = – 3 + 140 = 137 \)

Question. If the \( 2^{nd} \) term of an A.P. is 13 and the \( 5^{th} \) term is 25, then its \( 7^{th} \) term is :
(a) 30
(b) 33
(c) 37
(d) 38
Answer: (b) 33
Explanation :
In an A.P., consider ‘a‘ as first term and ‘d‘ as common difference.
Now, \( a_2 = 13 \Rightarrow a + d = 13 \dots \text{(i)} \)
\( a_5 = 25 \Rightarrow a + 4d = 25 \dots \text{(ii)} \)
Applying (ii)-(i), we get \( 3d = 12 \Rightarrow d = 4 \).
Substitute \( d = 4 \) in eq. (i), we get \( a = 13 – 4 = 9 \).
\( \therefore a_7 = a + 6d = 9 + 6 \times 4 = 9 + 24 = 33 \)

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

Question. The number of two-digit numbers which are divisible by 3 is :
(a) 33
(b) 31
(c) 30
(d) 29
Answer: (c) 30
Explanation :
Two digit numbers which are divisible by 3 are 12, 15, 18, \dots 99.
Here, \( a = 12, d = 3, l = 99 \).
\( \therefore l = a + (n – 1)d \Rightarrow 99 = 12 + (n – 1) \times 3 \Rightarrow 87 = (n – 1)3 \Rightarrow n – 1 = 29 \Rightarrow n = 30 \)

Question. The number of multiples of 4 that lie between 10 and 250 is :
(a) 62
(b) 60
(c) 59
(d) 55
Answer: (b) 60
Explanation :
Multiples of 4 lying between 10 and 250 are 12, 16, 20, \dots 248.
Here, \( a = 12, d = 4, l = 248 \).
\( l = a_n = a + (n – 1)d \Rightarrow 248 = 12 + (n – 1) \times 4 \Rightarrow 236 = 4(n – 1) \Rightarrow n – 1 = 59 \Rightarrow n = 60 \)

Question. The sum of first \( n \) natural number is :
(a) 0.5n(n + 1)
(b) \( \frac{n^2}{2} \)
(c) n + 2
(d) 0.5 + (n + 1)
Answer: (a) 0.5n(n + 1)
Explanation :
The sum of first \( n \) natural numbers is \( S_n = \frac{n(n + 1)}{2} = 0.5n(n + 1) \)