CBSE Class 10 Mathematics Arithmetic Progression Worksheet Set H

Arithmetic Progressions

Points to Remember

• 1. \( n^{th} \) term of an A.P. (\( a_n \)) with the first term \( a \) and common difference \( d \) is given by, \( a_n = a + (n - 1)d \)
• 2. \( n^{th} \) term from the end = Last term + \( (n - 1) (- d) = l - (n - 1)d \), where \( l \) = Last term of A.P.
• 3. The sum of \( n \) terms of an A.P. with the first term \( a \) and common difference \( d \) is given by, \( S_n = \frac{n}{2} [2a + (n - 1)d] \). Also, \( S_n = \frac{n}{2} (a + l) \), where \( l \) = Last term of A.P.

Multiple Choice Questions

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) Sol. We know, Common difference, \( d = a_2 - a_1 = \frac{1 - p}{p} - \frac{1}{p} = \frac{1 - p - 1}{p} = \frac{- p}{p} = - 1 \).

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) Sol. In the A.P., \( a = a \) and \( d = 3a - a = 2a \). 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 \( a \) and \( d \) for the following A.P. – 5, – 1, 3, 7 is :
(a) – 5 and 4
(b) – 4 and 5
(c) 5 and 4
(d) – 5 and – 4
Answer: (a) Sol. The given A.P. is – 5, – 1, 3, 7. Thus, \( a = - 5 \) and \( d = - 1 - (- 5) = 4 \).

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) Sol. Here, \( a = 2 \) and \( d = 5 \). 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 9, \( p \), 17 are three consecutive terms of an A.P., then the value of \( p \) is :
(a) 8
(b) 4
(c) 26
(d) 13
Answer: (d) Sol. Since 9, \( p \), 17 are in A.P. \( \therefore d = p - 9 = 17 - p \Rightarrow 2p = 26 \Rightarrow p = \frac{26}{2} = 13 \).

Question. If 2, \( a - 2 \) and \( 3a \) are in A.P., then the value of \( a \) is :
(a) 3
(b) 2
(c) – 3
(d) – 2
Answer: (d) Sol. 2, \( a - 2, 3a \) are in A.P. \( \therefore (a - 2) - 2 = 3a - (a - 2) \Rightarrow a - 4 = 2a + 2 \Rightarrow - 6 = a \). [Correction based on step calculation: \( a - 4 = 3a - a + 2 \Rightarrow a - 4 = 2a + 2 \Rightarrow - 6 = a \). Manual check: If \( a = -6 \), terms are \( 2, -8, -18 \). CD is -10. Wait, Sol text says -2=a. Let's re-read Sol text verbatim: \( (a-2)-2 = 3a-(a-2) \Rightarrow a-4=2a-2 \Rightarrow -2=a \).]

Fill in the Blanks

Question. The sum of first \( n \) positive integers is .............
Answer: \( \frac{n(n + 1)}{2} \)

Question. The \( n^{th} \) term of the A.P. : 7, 3, – 1... is given by, \( a_n = \) ..............
Answer: \( 11 - 4n \)

Question. The sum of first and third terms of an A.P. is equal to .................. the second term.
Answer: Twice

Question. \( n^{th} \) term from the end = \( l - (\dots\dots\dots) d \)
Answer: \( n - 1 \)

True/False

Question. If a constant term is added to each term of an A.P., then the result pattern of numbers are also in A.P.
Answer: True

Question. List of numbers 2, 4, 8, 16, ...... form an A.P.
Answer: False. In the series 2, 4, 8, 16, ...... \( 4 - 2 = 2 \); \( 8 - 4 = 4 \); \( 16 - 8 = 8 \). i.e., Difference between two consecutive terms is not same.

Question. The sum of first 10 natural numbers is 55.
Answer: True.

Very Short Answer Type Questions

Question. The first three terms of A.P. are \( (3y - 1), (3y + 5) \) and \( (5y + 1) \). Then find \( y \).
Answer: 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 \( \frac{4}{5}, a, 2 \) are three consecutive terms of an A.P., then find the value of \( a \).
Answer: Since \( \frac{4}{5}, a, 2 \) are the three consecutive terms of an A.P., then
\( d = a - \frac{4}{5} = 2 - a \)
\( \Rightarrow 2a = 2 + \frac{4}{5} \)
\( \Rightarrow 10a = 10 + 4 \)
\( \Rightarrow 10a = 14 \)
\( \Rightarrow a = \frac{7}{5} \).

Question. What is the common difference of an A.P. in which \( a_{21} - a_7 = 84 \)?
Answer: Given : \( a_{21} - a_7 = 84 \)
\( \Rightarrow a + (21 - 1)d - \{ a + (7 - 1)d \} = 84 \)
\( \Rightarrow a + 20d - a - 6d = 84 \)
\( \Rightarrow 14d = 84 \)
\( \Rightarrow d = 6 \).

Question. Find the common difference of the A.P. \( \frac{1}{2q}, \frac{1 - 2q}{2q}, \frac{1 - 4q}{2q} \dots \)
Answer: The given A.P. is \( \frac{1}{2q}, \frac{1 - 2q}{2q}, \frac{1 - 4q}{2q} \dots \)
Thus, common difference \( = \frac{1 - 2q}{2q} - \frac{1}{2q} \)
\( = \frac{- 2q}{2q} = - 1 \).
So, the common difference of A.P. is \( - 1 \).

Question. Write the next term of the A.P. \( \sqrt{2}, \sqrt{8}, \sqrt{18}, \dots \)
Answer: The given A.P. is \( \sqrt{2}, \sqrt{8}, \sqrt{18}, \dots \)
Thus, \( a = \sqrt{2} \)
and \( d = \sqrt{8} - \sqrt{2} = 2\sqrt{2} - \sqrt{2} = \sqrt{2} \)
Hence, \( t_4 = \sqrt{2} + (4 - 1)\sqrt{2} \)
\( = \sqrt{2} + 3\sqrt{2} = 4\sqrt{2} = \sqrt{32} \).

Question. For the A.P. \( \frac{3}{2}, \frac{1}{2}, -\frac{1}{2}, -\frac{3}{2}, \dots \) write the first term \( a \) and the common difference \( d \).
Answer: Given A.P. is,
\( \frac{3}{2}, \frac{1}{2}, -\frac{1}{2}, -\frac{3}{2}, \dots \)
So, \( a = \frac{3}{2} \) and \( d = \frac{1}{2} - \frac{3}{2} = \frac{- 2}{2} = - 1 \).

Question. Check whether the following series form an A.P. \( 1, - 1, - 3, - 5, \dots \)
Answer: Given series is, \( 1, - 1, - 3, - 5, \dots \)
Here, \( - 1 - 1 = - 2; - 3 - ( - 1) = - 2; - 5 - ( - 3) = - 2 \)
Since, the common difference between two consecutive series is same.
\(\therefore\) The given series forms an A.P.

Question. In an AP, if the common difference \( (d) = - 4 \) and the seventh term \( (a_7) \) is 4, then find the first term.
Answer: Given \( d = -4, a_7 = 4 \).
\( a_n = a + (n - 1)d \)
\( a_7 = a + (7 - 1)d \)
\( 4 = a + 6(-4) \)
\( 4 = a - 24 \)
\( a = 28 \).

Question. Which term of the A.P. \( 3, 8, 13, 18, \dots \) is 78?
Answer: Given A.P. is \( 3, 8, 13, 18, \dots \)
Here, \( a = 3, d = 8 - 3 = 5 \)
Let \( a_n = 78 \)
Then \( a + (n - 1)d = 78 \)
\( \Rightarrow 3 + (n - 1) 5 = 78 \)
\( \Rightarrow 5n - 2 = 78 \)
\( \Rightarrow 5n = 80 \)
\( \Rightarrow n = 16 \).
Thus, 16th term of given A.P. is 78.

Question. If the \( n^{th} \) term of the A.P. \( - 1, 4, 9, 14, \dots \) is 129, find the value of \( n \).
Answer: Given A.P. is,
\( - 1, 4, 9, 14, \dots 129 \).
Here, \( a = - 1 \) and \( d = 4 - ( - 1) = 5 \)
Now, \( a_n = 129 \)
\( \Rightarrow a + (n - 1)d = 129 \)
\( \Rightarrow - 1 + (n - 1) 5 = 129 \)
\( \Rightarrow 5n - 6 = 129 \)
\( \Rightarrow 5n = 135 \)
\( \Rightarrow n = 27 \).

Question. Find the eleventh term from the last term of the A.P. : \( 27, 23, 19, \dots, - 65 \).
Answer: Given A.P. is,
\( 27, 23, 19, \dots, - 65 \).
Here, \( a = 27, d = 23 - 27 = - 4 \) and \( l = - 65 \)
We know, \( n^{th} \) term from the last \( = l - (n - 1)d \)
\(\therefore\) 11th term from the last \( = - 65 - (11 - 1) ( - 4) \)
\( = - 65 + 40 \)
\( = - 25 \).

Question. If \( n^{th} \) term of an A.P. is \( (2n + 1) \), what is the sum of its first three terms?
Answer: Given : \( a_n = 2n + 1 \)
So, \( a_1 = 2(1) + 1 = 3 \) and \( a_3 = 2(3) + 1 = 7 \)
We know, \( S_n = \frac{n}{2} (a + a_n) \)
\(\therefore S_3 = \frac{3}{2} (a_1 + a_3) \)
\( = \frac{3}{2} (3 + 7) = \frac{3}{2} \times 10 = 15 \).

Question. Find the sum of the first 22 terms of the A.P. : \( 8, 3, - 2, \dots \)
Answer: Given A.P. is,
\( 8, 3, - 2, \dots \)
Here, \( a = 8, d = 3 - 8 = - 5, n = 22 \)
We know, \( S_n = \frac{n}{2} [2a + (n - 1)d] \)
\(\therefore S_{22} = \frac{22}{2} [2(8) + (22 - 1) ( - 5)] \)
\( = 11 [16 - 105] \)
\( = 11 \times ( - 89) \)
\( = - 979 \).

Question. If 7 times the 7th term of an A.P. is equal to 11 times its 11th term, then find its 18th term.
Answer: Given : \( 7 \times a_7 = 11 \times a_{11} \)
\( \Rightarrow 7 (a + 6d) = 11 (a + 10d) \)
\( \Rightarrow 7a + 42d = 11a + 110d \)
\( \Rightarrow 4a = - 68d \)
\( \Rightarrow a = - 17d \)
Now, \( a_{18} = a + 17d = - 17d + 17d = 0 \).

Question. How many terms of the A.P. \( 27, 24, 21, \dots \) should be taken so that their sum is zero?
Answer: Here, \( a = 27, d = 24 - 27 = - 3 \) and \( S_n = 0 \).
Now, \( S_n = 0 \)
\( \Rightarrow \frac{n}{2} [2a + (n - 1)d] = 0 \)
\( \Rightarrow n [2 \times 27 + (n - 1) ( - 3)] = 0 \)
\( \Rightarrow n [57 - 3n] = 0 \)
\( \Rightarrow n = 0 \) or \( n = \frac{57}{3} = 19 \)
\(\therefore n = 19 \) [\( \because n \neq 0 \)].

Short Answer Type Questions-I

Question. The first and the last terms of an A.P. are 5 and 45 respectively. If the sum of all the terms is 400, find the common difference.
Answer: Given, \( S_n = 400, a = 5 \) and \( l = 45 \)
We know, \( S_n = \frac{n}{2} [a + l] \)
\( \Rightarrow 400 = \frac{n}{2} [5 + 45] \)
\( \Rightarrow n = \frac{400 \times 2}{50} = 16 \)
Thus, applying \( t_n = a + (n - 1)d \), we have
\( 45 = 5 + (16 - 1)d \)
\( \Rightarrow 15d = 40 \)
\( \Rightarrow 3d = 8 \)
\( \Rightarrow d = \frac{8}{3} \).
Thus, the common difference is \( \frac{8}{3} \).

Question. The 10th term of an A.P. is \( ( - 4) \) and its 22nd term is \( ( - 16) \). Find its 38th term.
Answer: Given : \( a_{10} = - 4 \) and \( a_{22} = - 16 \)
\( \Rightarrow a + 9d = - 4 \) and \( a + 21d = - 16 \)
Subtracting the equations, we get
\( a + 21d = - 16 \)
\( a + 9d = - 4 \)
——————
\( 12d = - 12 \)
\( \Rightarrow d = - 1 \)
Now, \( a + 9d = - 4 \)
\( \Rightarrow a + 9( - 1) = - 4 \)
\( \Rightarrow a = - 4 + 9 = 5 \)
So, \( a_{38} = a + 37d = 5 + 37 ( - 1) = - 32 \).

Question. Find whether \( - 150 \) is a term of the A.P. \( 11, 8, 5, 2, \dots \)
Answer: Given A.P. is,
\( 11, 8, 5, 2, \dots \)
Here, \( a = 11, d = 8 - 11 = - 3 \)
Let \( a_n = - 150 \)
\( \Rightarrow a + (n - 1)d = - 150 \)
\( \Rightarrow 11 + (n - 1) ( - 3) = - 150 \)
\( \Rightarrow - 3n + 14 = - 150 \)
\( \Rightarrow - 3n = - 164 \)
\( \Rightarrow n = \frac{164}{3} \).
\(\because n \) is not a whole number. \(\therefore - 150 \) is not a term of given A.P.

Question. Find a, b and c such that the numbers a, 7, b, 23 and c are in A.P.
Answer: Given A.P. is,
\( a, 7, b, 23, c \)
\( \therefore 7 - a = b - 7 = 23 - b = c - 23 \)
\( \Rightarrow b - 7 = 23 - b \)
\( \Rightarrow 2b = 30 \)
\( \Rightarrow b = 15 \)
Also, \( 7 - a = b - 7 \) and \( 23 - b = c - 23 \)
\( \Rightarrow 7 - a = 15 - 7 \Rightarrow 23 - 15 = c - 23 \)
\( \Rightarrow a = 7 - 8 = - 1 \Rightarrow c = 8 + 23 = 31 \)
So, \( a = - 1, b = 15 \) and \( c = 31 \).

Question. Find the sum of first fifteen multiples of 8.
Answer: First fifteen multiples of 8 are, 8, 16, 24, ........... 120.
The above series is an A.P. with \( a = 8, d = 8, a_n = 120 \) and \( n = 15 \).
\( \therefore S_{15} = \frac{15}{2} [2 \times 8 + (15 - 1)8] \)
\( = \frac{15}{2} [16 + 112] \)
\( = \frac{15}{2} \times 128 \)
\( = 960 \).

Question. In an A.P., a = 1, \( a_n = 20 \) and \( S_n = 399 \), find n.
Answer: Given : \( a_n = 20 \)
\( \Rightarrow a + (n - 1)d = 20 \)
\( \Rightarrow 1 + (n - 1)d = 20 \)
\( \Rightarrow (n - 1)d = 19 \)
Also, \( S_n = 399 \)
\( \Rightarrow \frac{n}{2} [2a + (n - 1)d] = 399 \)
\( \Rightarrow \frac{n}{2} [2 \times 1 + 19] = 399 \)
\( \Rightarrow \frac{n}{2} \times 21 = 399 \)
\( \Rightarrow n = \frac{399 \times 2}{21} = 38 \)
\( \therefore n = 38 \)

Question. The 8th term of an A.P. is zero. Prove that the 38th term is triple of its 18th term.
Answer: Let the first term of the A.P. be \( a \) and the common difference be \( d \).
Thus, \( t_8 = a + (8 - 1)d = 0 \) [Given]
or \( a + 7d = 0 \dots(i) \)
Also, \( t_{18} = a + (18 - 1)d = a + 17d = (a + 7d) + 10d = 10d \) [From (i)]
and \( t_{38} = a + (38 - 1)d = a + 37d = (a + 7d) + 30d = 30d \) [From (i)]
\( = 3(10d) = 3t_{18} \).
Hence Proved.

Question. Which term of the A.P. 3, 15, 27, 39, ...... will be 120 more than its 21st term?
Answer: The given A.P. is 3, 15, 27, 39, ....
Here \( a = 3, d = 12 \)
\( \therefore a_{21} = a + 20d = 3 + 20 \times 12 = 3 + 240 = 243 \)
Now, \( a_n = a_{21} + 120 = 243 + 120 = 363 \)
\( \because a_n = a + (n - 1)d \)
\( \therefore 363 = 3 + (n - 1) 12 \)
or \( 360 = (n - 1) 12 \)
or \( n - 1 = 30 \Rightarrow n = 31 \)
Hence, the term which is 120 more than its 21st term will be its 31st term.

Question. How many two-digit numbers are divisible by 7 ?
Answer: Two-digit numbers divisible by 7 are, 14, 21, 28, ...... 98.
The above series is an A.P. with \( a = 14, d = 7 \) and \( a_n = 98 \).
Now, \( a_n = 98 \)
\( \Rightarrow a + (n - 1)d = 98 \)
\( \Rightarrow 14 + (n - 1)7 = 98 \)
\( \Rightarrow 7n + 7 = 98 \)
\( \Rightarrow 7n = 91 \)
\( \Rightarrow n = \frac{91}{7} = 13 \)
Thus, there are 13 two-digit numbers that are divisible by 7.

Question. The 19th term of an A.P. is equal to three times its 6th term. If its 9th term is 19, find the A.P.
Answer: Let the first term be \( a \) and the common difference be \( d \).
Given, \( T_{19} = 3T_6 \)
\( \Rightarrow \{a + (19 - 1)d\} = 3\{a + (6 - 1)d\} \)
\( \Rightarrow a + 18d = 3a + 15d \)
\( \Rightarrow 2a = 3d \)
and \( T_9 = \{a + (9 - 1)d\} = 19 \)
\( \Rightarrow a + 8d = 19 \)
Multiplying the above equation by 2, we get
\( \Rightarrow 2a + 16d = 38 \dots(i) \)
Substituting \( 2a = 3d \) in equation (i), we get
\( 3d + 16d = 38 \)
\( \Rightarrow 19d = 38 \)
\( \Rightarrow d = 2 \)
Thus \( a = \frac{3}{2} (2) = 3 \)
Hence, the required A.P. is 3, 5, 7, 9, …

Question. Find the sum of all natural numbers that are less than 100 and divisible by 4.
Answer: Natural numbers less than 100 that are divisible by 4 are,
4, 8, 12, ...... 96
The above series is an A.P. with \( a = 4, d = 4 \) and \( a_n = 96 \).
We know, \( a_n = a + (n - 1)d \)
\( \Rightarrow 96 = 4 + (n - 1)4 \)
\( \Rightarrow 96 = 4n \)
\( \Rightarrow n = \frac{96}{4} = 24 \)
Now, \( S_n = \frac{n}{2} (a + a_n) \)
\( \therefore S_{24} = \frac{24}{2} (4 + 96) = 12 \times 100 = 1200 \).

Question. The sum of first n terms of an A.P. is given by \( S_n = 2n^2 + 3n \). Find the sixteenth term of the A.P.
Answer: Given : \( S_n = 2n^2 + 3n \)
We know, \( a_n = S_n - S_{n - 1} \)
\( \therefore a_{16} = S_{17} - S_{16} \)
\( = \{2(17)^2 + 3(17)\} - [2(16)^2 + 3(16)] \)
\( = \{2(289) + 51\} - [2(256) + 48] \)
\( = \{578 + 51\} - (512 + 48) \)
\( = 629 - 560 \)
\( = 69 \).

Question. How many terms of an A.P. – 6, \( -\frac{11}{2} \), – 5, \( -\frac{9}{2} \) are needed to give their sum zero ?
Answer: Given A.P. is,
\( - 6, -\frac{11}{2}, - 5, -\frac{9}{2} \)
Here, \( a = - 6 \) and \( d = -\frac{11}{2} - (- 6) = \frac{- 11 + 12}{2} = \frac{1}{2} \)
Now, \( S_n = 0 \)
\( \Rightarrow \frac{n}{2} \{2a + (n - 1)d\} = 0 \)
\( \Rightarrow \frac{n}{2} \{2(- 6) + (n - 1)\frac{1}{2}\} = 0 \)
\( \Rightarrow \frac{n}{2} \{- 12 + \frac{1}{2}n - \frac{1}{2}\} = 0 \)
\( \Rightarrow \frac{n}{2} \{\frac{1}{2}n - \frac{25}{2}\} = 0 \)
\( \Rightarrow n(n - 25) = 0 \)
\( \Rightarrow n = 25 \) (\( \because n \neq 0 \))
Thus, 25 terms are needed to give the sum of given A.P. zero.

Question. For what value of n, are the nth terms of two A.Ps. 63, 65, 67,....... and 3, 10, 17,...... equal ?
Answer: 1st A.P. is 63, 65, 67, .....
Here, \( a = 63, d = 65 - 63 = 2 \)
\( \therefore a_n = a + (n - 1)d = 63 + (n - 1) (2) = 63 + 2n - 2 = 61 + 2n \)
2nd A.P. is 3, 10, 17....
Here, \( a = 3, d = 10 - 3 = 7 \)
\( \therefore a_n = a + (n - 1)d = 3 + (n - 1) (7) = 3 + 7n - 7 = 7n - 4 \)
According to question,
\( 61 + 2n = 7n - 4 \)
\( 61 + 4 = 7n - 2n \)
\( 65 = 5n \)
\( n = \frac{65}{5} = 13 \)
Hence, 13th term of both A.Ps is equal.

Question. If m times the mth term of an A.P. is equal to n times the nth term and m ≠ n, then show that its (m + n)th term is zero.
Answer: Let the first term be \( a \) and the common difference be \( d \).
Hence \( t_m = a + (m - 1)d \) and \( t_n = a + (n - 1)d \)
Now \( m\{a + (m - 1)d\} = n\{a + (n - 1)d\} \)
\( \Rightarrow ma + dm^2 - dm = na + dn^2 - dn \)
\( \Rightarrow (m - n)a + (m^2 - n^2)d - (m - n)d = 0 \)
\( \Rightarrow (m - n)a + (m - n)(m + n)d - (m - n)d = 0 \)
Dividing by \( (m - n) \):
\( \Rightarrow a + (m + n)d - d = 0 \)
\( \Rightarrow a + \{(m + n) - 1\}d = 0 \)
\( \Rightarrow t_{m + n} = 0 \).
Hence Proved.