CBSE Class 10 Maths HOTs Arithmetic Progressions Set B

Multiple Choice Questions

Question. Which of the following form of an AP? 
(a) \( -1, -1, -1, -1, \dots \)
(b) \( 0, 2, 0, 2, \dots \)
(c) \( 1, 1, 2, 2, 3, 3, \dots \)
(d) \( \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots \)
Answer: (a)

Question. Which of the following is not an AP? 
(a) \( -1.2, 0.8, 2.8, \dots \)
(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)

Question. If \( -\frac{5}{7}, a, 2 \) are consecutive terms in an Arithmetic Progression, then the value of ‘a’ is 
(a) \( \frac{9}{7} \)
(b) \( \frac{9}{14} \)
(c) \( \frac{19}{7} \)
(d) \( \frac{19}{14} \)
Answer: (b)

Question. The common difference of an AP, whose nth term is \( a_n = (3n + 7) \), is
(a) 3
(b) 7
(c) 10
(d) 6
Answer: (a)

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)

Question. The value of p for which \( (2p + 1), 10 \) and \( (5p + 5) \) are three consecutive terms of an AP is
(a) \( -1 \)
(b) \( -2 \)
(c) 1
(d) 2
Answer: (c)

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)

Question. Let a be a sequence defined by \( a_1 = 1, a_2 = 1 \) and \( a_n = a_{n-1} + a_{n-2} \) for all \( n > 2 \), then the value of \( \frac{a_4}{a_3} \) is
(a) \( \frac{2}{3} \)
(b) \( \frac{5}{4} \)
(c) \( \frac{4}{5} \)
(d) \( \frac{3}{2} \)
Answer: (d)

Question. If an AP have 8 as the first term and \( -5 \) as the common difference and its first three terms are 8, A, B, then \( (A + B) \) is equal to
(a) 0
(b) \( -1 \)
(c) 1
(d) 2
Answer: (c)

Question. In an AP, if \( d = -4, n = 7 \) and \( a_n = 4 \), then a is equal to
(a) 6
(b) 7
(c) 20
(d) 28
Answer: (d)

Question. The 11th term of an AP \( -5, -\frac{5}{2}, 0, \frac{5}{2}, \dots \) 
(a) \( -20 \)
(b) 20
(c) \( -30 \)
(d) 30
Answer: (b)

Question. The 21st term of an AP whose first two terms are \( -3 \) and 4, is 
(a) 17
(b) 137
(c) 143
(d) \( -143 \)
Answer: (b)

Question. If the 2nd term of an AP is 13 and 5th term is 25, what is its 7th term?
(a) 30
(b) 33
(c) 37
(d) 38
Answer: (b)

Question. Which term of an AP : 21, 42, 63, 84, ... is 210? [NCERT Exemplar]
(a) 9th
(b) 10th
(c) 11th
(d) 12th
Answer: (b)

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)

Question. What is the common difference of an AP in which \( a_{18} - a_{14} = 32 \)? 
(a) 8
(b) \( -8 \)
(c) \( -4 \)
(d) 4
Answer: (a)

Question. Two APs have the same common difference. The first term of one of these is \( -1 \) and that of the other is \( -8 \). The difference between their 4th terms is 
(a) \( -1 \)
(b) \( -8 \)
(c) 7
(d) \( -9 \)
Answer: (c)

Question. If 7 times the 7th term of an AP is equal to 11 times its 11th term, then its 18th term will be
(a) 7
(b) 11
(c) 18
(d) 0
Answer: (d)

Question. The 4th term from the end of an AP \( -11, -8, -5, \dots, 49 \) is
(a) 37
(b) 40
(c) 43
(d) 58
Answer: (b)

Question. Which term of the AP 5, 15, 25, ... will be 130 more than its 31st term?
(a) 42
(b) 44
(c) 46
(d) 48
Answer: (b)

Question. The number of terms of an AP 5, 9, 13, ..., 185 is [NCERT Exemplar, 
(a) 31
(b) 51
(c) 41
(d) 46
Answer: (d)

Question. The sum of AP, sequence \( -37, -33, -29, \dots \) upto 12 term is
(a) 180
(b) \( -180 \)
(c) 170
(d) \( -170 \)
Answer: (b)

Question. The sum of first 16 terms of the AP 10, 6, 2, ... is 
(a) \( -320 \)
(b) 320
(c) \( -352 \)
(d) \( -400 \)
Answer: (a)

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)

Question. In an AP, if \( a = 1, a_n = 20 \), and \( S_n = 399 \), then n is equal to
(a) 19
(b) 21
(c) 38
(d) 42
Answer: (c)

Question. The sum of first five multiples of 3 is 
(a) 45
(b) 55
(c) 65
(d) 75
Answer: (a)

Question. Examine that the sequence 13, 10, 7, 4,... is an AP.
Answer: Given, AP is 13, 10, 7, 4, ......
Here, \( a_1 = 13, a_2 = 10, a_3 = 7, a_4 = 4, ......... \)
Here, we have \( a_2 - a_1 = 10 - 13 = -3 \),
\( a_3 - a_2 = 7 - 10 = -3 \),
\( a_4 - a_3 = 4 - 7 = -3 \) and so on.
Since, difference of any two consecutive terms is same.
So, the given sequence is an AP.

Question. Find the common difference of the following AP’s.
(i) 3, -2, -7, -12, ...
(ii) 11, 11, 11, 11, ...
(iii) \( 5\frac{1}{2}, 9\frac{1}{2}, 13\frac{1}{2}, 17\frac{1}{2}, ... \)
Answer: (i) Given, AP is 3, -2, -7, -12, ...
Here, \( a_1 = 3, a_2 = -2, a_3 = -7, a_4 = -12 \) and so on.
\( \therefore \) Common difference \( (d) = a_2 - a_1 = -2 - 3 = -5 \)
(ii) Given, AP is 11, 11, 11, 11, ...
Here, \( a_1 = a_2 = a_3 = a_4 = 11 \) and so on.
\( \therefore \) Common difference \( (d) = a_2 - a_1 = 11 - 11 = 0 \)
(iii) Given, AP is \( 5\frac{1}{2}, 9\frac{1}{2}, 13\frac{1}{2}, 17\frac{1}{2}, ... \)
Here, \( a_1 = 5\frac{1}{2}, a_2 = 9\frac{1}{2}, a_3 = 13\frac{1}{2}, a_4 = 17\frac{1}{2} \) and so on.
\( \therefore \) Common difference \( (d) = a_2 - a_1 = 9\frac{1}{2} - 5\frac{1}{2} = \frac{19}{2} - \frac{11}{2} = \frac{8}{2} = 4 \)

Question. Write an AP having 4 as the first term and -3 as the common difference.
Answer: Given, first term \( (a) = 4 \) and common difference \( (d) = -3 \)
On putting the values of \( a \) and \( d \) in general form
\( a, a + d, a + 2d, a + 3d, ... \), we get
\( 4, 4 + (-3), 4 + 2(-3), 4 + 3(-3), ... \)
\( 4, 1, 4 - 6, 4 - 9, \dots \) or \( 4, 1, -2, -5, ... \)
Which is the required AP.

Question. Find the 20th term of the sequence 7, 3, -1, -5 ...
Answer: Given, sequence is 7, 3, -1, -5, ...
Here, \( 3 - 7 = -4, -1 - 3 = -4, -5 - (-1) = -4 \) and so on.
So, given sequence is an AP, in which \( a = 7 \) and \( d = -4 \).
Since, nth term, \( a_n = a + (n - 1)d \)
On putting \( n = 20 \), we get
\( a_{20} = a + (20 - 1)d = 7 + 19(-4) \) [\( \because a = 7, d = -4 \)]
\( = 7 - 19 \times 4 = 7 - 76 = -69 \)
Hence, 20th term of given sequence is -69.

Question. How many terms are there in the sequence 3, 6, 9, 12, ..., 111?
Answer: Given, sequence is 3, 6, 9, 12, ..., 111.
Here, \( 6 - 3 = 9 - 6 = 12 - 9 = \dots = 3 \)
So, it is an AP with first term, \( a = 3 \) and common difference, \( d = 3 \). Let there be \( n \) terms in the given sequence.
Then, \( \text{nth term} = 111 \)
\( \Rightarrow a + (n - 1)d = 111 \) [\( \because a_n = a + (n - 1)d \)]
\( \Rightarrow 3 + (n - 1) \times 3 = 111 \)
\( \Rightarrow 3(1 + n - 1) = 111 \)
\( \Rightarrow n = \frac{111}{3} \Rightarrow n = 37 \)
Hence, the given sequence contains 37 terms.

Question. Which term of the AP: 21, 18, 15, ... is -81?
Answer: Given, AP is 21, 18, 15, ... .
Here, \( a = 21 \) and \( d = 18 - 21 = -3 \)
Let nth term of given AP be -81
Then, \( a_n = -81 \)
\( \Rightarrow a + (n - 1)d = -81 \) [\( \because a_n = a + (n - 1)d \)]
On putting the values of \( a \) and \( d \), we get
\( 21 + (n - 1)(-3) = -81 \Rightarrow 21 - 3n + 3 = -81 \)
\( \Rightarrow 24 - 3n = -81 \Rightarrow -3n = -81 - 24 = -105 \)
\( \Rightarrow n = \frac{-105}{-3} = 35 \)
Hence, 35th term of given AP is -81.

Question. How many numbers of two digits are divisible by 7?
Answer: Two-digits numbers are 10, 11, 12, 13, 14, 15, ..., 97, 98, 99, in which only 14, 21, 28, ..., 98 are divisible by 7.
Here, \( 21 - 14 = 28 - 21 = \dots = 7 \).
So, this list of numbers forms an AP, whose first term \( (a) = 14 \), common difference \( (d) = 7 \).
Let there are \( n \) terms in the above sequence, then \( a_n = 98 \)
\( \Rightarrow a + (n - 1)d = 98 \) [\( \because a_n = a + (n - 1)d \)]
\( \Rightarrow 14 + (n - 1)7 = 98 \Rightarrow 14 + 7n - 7 = 98 \)
\( \Rightarrow 7n = 91 \Rightarrow n = \frac{91}{7} = 13 \)
Hence, 13 numbers of two digits are divisible by 7.

Question. Determine the 10th term from the end of the AP : 4, 9, 14, ..., 254.
Answer: Given, AP is 4, 9, 14, ..., 254.
Here, \( l = \text{last term} = 254 \)
\( d = \text{common difference} = 9 - 4 = 5 \)
\( \therefore \) 10th term from the end \( = l - (10 - 1)d = l - 9d \)
\( = 254 - 9 \times 5 = 254 - 45 = 209 \)

Question. Determine the general term of an AP whose 7th term is -1 and 16th term is 17.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the AP, whose 7th term is -1 and 16th term is 17.
Since, \( a_7 = -1 \) and \( a_{16} = 17 \)
\( \therefore \) We have, \( a + (7 - 1)d = -1 \Rightarrow a + 6d = -1 \) ...(i)
and \( a + (16 - 1)d = 17 \Rightarrow a + 15d = 17 \) ...(ii)
[\( \because a_n = a + (n - 1)d \)]
On subtracting Eq. (i) from Eq. (ii), we get
\( (a + 15d) - (a + 6d) = 17 - (-1) \)
\( \Rightarrow 9d = 18 \Rightarrow d = 2 \)
On substituting \( d = 2 \) in Eq. (i), we get
\( a + 6 \times 2 = -1 \)
\( \Rightarrow a + 12 = -1 \)
\( \Rightarrow a = -13 \)
Hence, general term,
\( a_n = a + (n - 1)d \)
\( = -13 + (n - 1)2 \) [\( \because a = -13 \) and \( d = 2 \)]
\( = -13 + 2n - 2 = 2n - 15 \)

Question. Find four numbers in AP whose sum is 20 and the sum of whose squares is 120.
Answer: Let the numbers be \( a - 3d, a - d, a + d \) and \( a + 3d \).
Then, according to the given condition, we have
\( (a - 3d) + (a - d) + (a + d) + (a + 3d) = 20 \) …(i)
and \( (a - 3d)^2 + (a - d)^2 + (a + d)^2 + (a + 3d)^2 = 120 \) …(ii)
From Eq. (i), we get
\( 4a = 20 \Rightarrow a = 5 \)
From Eq. (ii), we get
\( a^2 + 9d^2 - 6ad + a^2 + d^2 - 2ad + a^2 + d^2 + 2ad + a^2 + 9d^2 + 6ad = 120 \)
\( \Rightarrow 4a^2 + 20d^2 = 120 \)
\( \Rightarrow a^2 + 5d^2 = 30 \)
\( \Rightarrow 25 + 5d^2 = 30 \) [\( \because a = 5 \)]
\( \Rightarrow 5d^2 = 5 \Rightarrow d^2 = 1 \Rightarrow d = \pm 1 \)
If \( d = 1 \), then the numbers are 2, 4, 6, 8 and if \( d = -1 \), then the numbers are 8, 6, 4, 2.
Hence, the numbers are 2, 4, 6, 8 or 8, 6, 4, 2.

Question. A sum of ₹ 2000 is invested at 7% simple interest per year. Calculate the interest at the end of each year. Do these interest form an AP? If so, then find the interest at the end of 20th year making use of this fact.
Answer: Given, initial money \( P = Rs 2000 \)
Rate of interest, \( R = 7\% \) per year; Time, \( T = 1, 2, 3, 4, \dots \)
We know that, simple interest is given by the following formula
\( SI = \frac{PRT}{100} \)
\( \therefore SI \) at the end of 1st year \( = \frac{2000 \times 7 \times 1}{100} = Rs 140 \)
\( SI \) at the end of 2nd year \( = \frac{2000 \times 7 \times 2}{100} = Rs 280 \)
\( SI \) at the end of 3rd year \( = \frac{2000 \times 7 \times 3}{100} = Rs 420 \)
Thus, required list of numbers is 140, 280, 420, ... .
Here, \( 280 - 140 = 420 - 280 = \dots = 140 \)
So, above list of numbers forms an AP, whose first term \( (a) = 140 \) and common difference \( (d) = 140 \).
Now, SI at the end of 20th year will be equal to 20th term of the above AP.
\( \because a_{20} = a + (20 - 1)d = 140 + 19 \times 140 = 140 + 2660 = 2800 \)
Hence, the interest at the end of 20th year will be Rs 2800.

Question. Each year, a tree grow 5 cm less than the preceding year. If it grew by 1m in the first year, then in how many years will it have ceased growing?
Answer: Given that, tree grow 5 cm or 0.05 m less than preceding year.
\( \therefore \) The following sequence can be formed.
\( 1, (1 - 0.05), (1 - 2 \times 0.05), \dots, 0 \)
i.e. 1, 0.95, 0.90, ... , 0 which is an AP.
Here, \( a = 1, d = 0.95 - 1 = -0.05 \) and \( l = 0 \)
Let \( l = a_n = a + (n - 1)d \)
Then, \( 0 = 1 + (n - 1)(-0.05) \)
\( \Rightarrow (n - 1)(0.05) = 1 \)
\( \Rightarrow n - 1 = \frac{1}{0.05} \)
\( \Rightarrow n - 1 = \frac{1 \times 100}{5} \)
\( \Rightarrow n - 1 = 20 \)
\( \Rightarrow n = 21 \)
Hence, in 21 yr, tree will have ceased growing.

Question. The eighth term of an AP is half its second term and the eleventh term exceeds one-third of its fourth term by 1. Find the 15th term. 
Answer: Let \( a \) and \( d \) be the first term and common difference of an AP. Then,
\( a_8 = \frac{1}{2}a_2 \) and \( a_{11} = \frac{1}{3}a_4 + 1 \)
\( \Rightarrow a + (8 - 1)d = \frac{1}{2}[a + (2 - 1)d] \)
and \( a + (11 - 1)d = \frac{1}{3}[a + (4 - 1)d] + 1 \)
\( \Rightarrow a + 7d = \frac{1}{2}(a + d) \)
\( \Rightarrow 2a + 14d = a + d \)
\( \Rightarrow a + 13d = 0 \) …(i)
and \( a + 10d = \frac{1}{3}(a + 3d) + 1 \)
\( \Rightarrow 3a + 30d = a + 3d + 3 \)
\( \Rightarrow 2a + 27d = 3 \) …(ii)
On solving Eqs. (i) and (ii), we get
\( a = -13, d = 1 \)
\( \therefore a_{15} = a + (15 - 1)d \)
\( = -13 + 14(1) = 1 \)

Question. The fourth term of an AP is 11. The sum of the fifth and seventh terms of the AP is 24. Find its common difference. 
Answer: Let \( a \) be the first term and \( d \) be the common difference. Then,
\( a_4 = 11 \Rightarrow a + (4 - 1)d = 11 \)
\( \Rightarrow a + 3d = 11 \) …(i)
Also, given \( a_5 + a_7 = 24 \)
\( \Rightarrow [a + (5 - 1)d] + [a + (7 - 1)d] = 24 \)
\( \Rightarrow a + 4d + a + 6d = 24 \)
\( \Rightarrow 2a + 10d = 24 \)
\( \Rightarrow a + 5d = 12 \) [divide by 2] …(ii)
On subtracting Eq. (i) from Eq. (ii), we get
\( 2d = 1 \Rightarrow d = \frac{1}{2} \)
Hence, common difference is \( \frac{1}{2} \).

Question. Find the sum of the first 22 terms of the AP : 8, 3, -2, ...
Answer: Given, AP is 8, 3, -2, ... .
Here, first term, \( (a) = 8 \)
Common difference, \( (d) = 3 - 8 = -5 \) and \( n = 22 \)
\( \because \) Sum of first n terms, \( S_n = \frac{n}{2}[2a + (n - 1)d] \)
\( \therefore S_{22} = \frac{22}{2}[2 \times 8 + (22 - 1)(-5)] \)
\( = 11[16 + 21(-5)] \)
\( = 11[16 - 105] \)
\( = 11(-89) = -979 \)
Hence, sum of first 22 terms of an AP is -979.

Question. Find the sum of first 24 terms of an AP, whose nth term is given by \( a_n = 3 + 2n \).
Answer: Given, nth term of an AP, \( a_n = 3 + 2n \)
Clearly, sum of first 24 terms, \( (S_{24}) \)
\( = \frac{24}{2}(a_1 + a_{24}) = 12(5 + 51) \)
[\( \because a_1 = 3 + 2 = 5 \) and \( a_{24} = 3 + 2 \times 24 = 3 + 48 = 51 \)]
\( = 12 \times 56 = 672 \)

Question. If the sum of first 10 terms of an AP is 140 and the sum of first 16 terms is 320, then find the sum of first m terms.
Answer: Let the first term of this AP be \( a \) and common difference be \( d \).
Given, sum of first 10 terms, \( (S_{10}) = 140 \)
\( \Rightarrow \frac{10}{2}[2a + (10 - 1)d] = 140 \Rightarrow 2a + 9d = \frac{140}{5} \)
[\( \because S_n = \frac{n}{2}[2a + (n - 1)d] \)]
\( \Rightarrow 2a + 9d = 28 \) ...(i)
Also, given sum of first 16 terms, \( (S_{16}) = 320 \)
\( \Rightarrow \frac{16}{2}[2a + (16 - 1)d] = 320 \)
\( \Rightarrow 2a + 15d = \frac{320}{8} \Rightarrow 2a + 15d = 40 \) ...(ii)
On subtracting Eq. (i) from Eq. (ii), we get
\( 6d = 12 \Rightarrow d = 2 \)
On putting \( d = 2 \) in Eq. (i), we get
\( 2a + 9(2) = 28 \Rightarrow 2a = 28 - 18 \)
\( \Rightarrow a = \frac{10}{2} = 5 \)
Thus, \( a = 5 \) and \( d = 2 \).
Hence, sum of first m terms, \( (S_m) = \frac{m}{2}[2a + (m - 1)d] \)
\( = \frac{m}{2}[2(5) + (m - 1)2] = \frac{m}{2}[10 + 2m - 2] \)
\( = \frac{m}{2}(2m + 8) = m(m + 4) = m^2 + 4m \)

Question. Find the sum of all three-digit natural numbers, which are multiples of 11. 
Answer: All three-digit natural numbers, multiples of 11 are 110, 121, 132, …, 990.
Here, common difference, \( 121 - 110 = 132 - 121 = \dots = 11 \).
So, it is an AP with first term, \( a = 110 \), common difference, \( d = 11 \) and last term, \( l = 990 \).
Let \( l = a_n = a + (n - 1)d \)
\( \therefore 990 = 110 + (n - 1) \times 11 \)
\( \Rightarrow 990 - 110 = 11n - 11 \)
\( \Rightarrow 11n = 891 \Rightarrow n = 81 \)
\( \because S_n = \frac{n}{2}[a + l] \)
\( \therefore S_{81} = \frac{81}{2}[110 + 990] \)
\( = \frac{81}{2} \times 1100 = 81 \times 550 = 44550 \)

Question. If \( S_n \), the sum of first \( n \) terms of an AP is given by \( S_n = 3n^2 - 4n \), find the nth term. 
Answer: Given, \( S_n = 3n^2 - 4n \) …(i)
On replacing \( n \) by \( (n - 1) \) in Eq. (i), we get
\( S_{n-1} = 3(n - 1)^2 - 4(n - 1) \)
nth term of the AP \( a_n = S_n - S_{n-1} \)
\( \therefore a_n = (3n^2 - 4n) - [3(n - 1)^2 - 4(n - 1)] \)
\( a_n = 3n^2 - 4n - [3(n^2 - 2n + 1) - 4n + 4] \)
\( a_n = 3n^2 - 4n - [3n^2 - 6n + 3 - 4n + 4] \)
\( a_n = 3n^2 - 4n - [3n^2 - 10n + 7] \)
\( a_n = 6n - 7 \)
Thus, the nth term of the AP \( = 6n - 7 \).