CBSE Class 10 Mathematics Arithmetic Progression Worksheet Set G

Question. Write the n^th term of odd numbers.
Answer. 1, 3, 5, ......
a_n = 1 + (n – 1)2 = 2n – 1.

Question. Find the nth term of the A.P. – 10, – 15, – 20, – 25, ...........
Answer.  a_n = a + (n – 1)d = – 5(n + 1)

Question. Find the common difference of A.P. 4(1/9), 4(2/9), 4(1/3), .............
Answer.  d = a₂ – a₁ = 1/9

Question. For what value of p, the following terms are three consecutive terms of an A.P. 4/5, p, 2.
Answer. p = 7/5

Question. What is value of a₁₆ for the A.P. – 10, – 12, – 14, – 16, .......
Answer. a₁₆ = a + 15d = – 40

Question. Write the sum of first n even numbers.
Answer. 2 + 4 + 6 + ... + 2n = n/2 [2 + 2n] = n(n + 1)

Question. The 11^th term from the last term of an A.P. 10, 7, 4, ...., – 62 is
(A) 25
(B) –32
(C) 16
(D) 0
Answer. (B) –32

Question. In an A.P. if d = – 4, n = 7, a_n = 4, then a is
(A) 6
(B) 7
(C) 120
(D) 28
Answer. (D) 28

Question. 30th term of the A.P. 10, 7, 4 .... is
(A) 97
(B) 77
(C) –77
(D) –87
Answer. (C) –77

Question. The common difference of the A.P. 2 , 2 2 , 3 2 , 4 2 ...... is
(A) √2
(B) 1
(C) 2√2
(D) –√2
Answer. (A) √2

Question. The list of numbers – 10, – 6, – 2, 2, ... 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

Question. What is the common difference of an A.P. in which a₁₈ – a₁₄ = 32 ?
(A) 8
(B) – 8
(C) – 4
(D) 40
Answer. (A) 8

Question. 2, 5, 9, 14, .... is an A.P. True or False.
Answer. False, ∵ a₂ – a₁ = 5 – 2 = 3
∵ a₃ – a₂ = 9 – 5 = 4

Question. 301 is a term of an A.P. 5, 11, 17, 23 .... True or False.
Answer. False, 301 = 5 + (n – 1) 6
Solving we get n = 151/3 which is not a natural number.
∴ 301 is not a term of this A.P.

SHORT ANSWER TYPE QUESTIONS-I

Question. Find the sum of first 15 multiples of 8.
Answer. S₁₅ = 15/2 [2a + 14d]
where a = 8, d = 8
∴ Answer is 960

Question. Is 144 a term of the A.P. 3, 7, 11, ......... ? Justify your answer.
Answer. 144 = 3 + (n – 1) 4
(141/4) +1 = n which is not possible

Question. How many terms of the A.P. 22, 20, 18, ....... should be taken so that their sum is zero.
Answer. S_n = 0 ⇒ n/2 [44 + (n – 1) (– 2)] = 0.
Solve n = 23

Question. The first term, common difference and last term of an A.P. are 12, 6 and 252 respectively, Find the sum of all terms of this A.P.
Answer. a = 12, d = 6, a_n = 252 ⇒ n = 41
Find S₄₁ = 5412, use S_n = n/2 [2a + (n – 1) d]

Question. Find the middle term of an A.P. – 6, – 2, 2, .... 58.
Answer. a_n = a + (n – 1) d
58 = – 6 + (n – 1) 4
find n = 17
Find Middle term using concept of median
= [(n+1)/2]^th term = 9th term
a₉ = – 6 + 8(4) = 26

Question. Find whether (– 150) is a term of A.P. 11, 8, 5, 2, ..... ?
Answer. Let a_n = –150
11 + (n – 1) (– 3) = – 150
Solve and get n is not a natural number. ( n = 164/3 )
∴ Answer is No.

Question. Is the sequence formed in the following situations an A.P.
(i) Number of students left in the school auditorium from the total strength of 1000 students when they leave the auditorium in batches of 25.
(ii) The amount of money in the account every year when Rs. 100 are deposit annually to accumulate at compound interest at 4% per annum.
Answer. (i) Yes → (1000, 975, 950, 925 .....)
(ii) No → (104, 108.16, 112.48 .....)

SHORT ANSWER TYPE QUESTIONS-II

Question. Find the middle terms of the A.P. 7, 13, 19, ......., 241.
Answer.  n = 40 Middle terms are a₂₀, a₂₁
∴ Answer is 121, 127

Question. Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.
Answer. Numbers divisible by both 2 and 5
⇒ Numbers divisible by 10.
Numbers between 101 and 999 divisible by 2 and 5 both 110, 120, 130, 140, ..., 990.
Use a_n = 990 to get n = 89.

Question. If S_n = 4n – n² in an A.P. find the A.P.
Answer. S_n = 4n – n²
S₁ = a₁ = 4 – 1 = 3
S₂ = a₁ + a₂ ⇒ a₂ = 1 , A.P. 3, 1, – 1, ...
S₃ = a₁ + a₂ + a₃ ⇒ a₃ – 1

Question. How many terms of the A.P. 9, 17, 25, ..... must be taken to give a sum of 636?
Answer. n = 12, n = 53/4

Question. Find the sum of odd numbers between 0 and 50.
Answer. Odd numbers between 0 to 50
1, 3, 5, 7, ..., 49
a_n = 49
a + (n – 1)d = 49
1 + (n – 1)2 = 49
n = 25
S_n = n/2 [a + l]
S₂₅ = 25/2 [1 + 49] = 25 × 25 = 625

Question. The sum of 5^th and 9^th terms of an A.P. is 72 and the sum of 7^th and 12^th term is 97. Find the A.P.
Answer. a₅ + a₉ = 72
a₇ + a₁₂ = 97
Solve these equations to get a and d, a = 6, d = 5
∴ A.P., 6, 11, 16, 21, 26, .......

Question. If S_n, the sum of first n terms of an A.P. is given by Sn = 5n² + 3n, then find its n^th term and common difference.
Answer. S_n = 5n² + 3n
Find a_n = S_n – S_n–1 = 10 n – 2
Use it to get d = 10

Question. Which term of the A.P. 3, 15, 27, 39 .... will be 120 more than its 21^st term?
Answer. Let a_n = 120 + a₂₁
3 + (n – 1)d = 120 + [3 + 20d]
3 + (n – 1)12 = 120 + [3 + 20 × 12]
= 120 + 243
(n – 1)12 = 363 – 3 = 360
n = 31

LONG ANSWER TYPE QUESTIONS

Question. The sum of the first 9 terms of an A.P. is 171 and the sum of its first 24 terms is 996. Find the first term and common difference of the A.P.
Answer. S₉ = 171, S₂₄ = 996
a + 4d = 19, 2a + 23d = 83
Solve to get,
d = 3, a = 7

Question. The sum of first 7 terms of an A.P. is 63 and the sum of its next 7 term is 161. Find the 28^th term of this A.P.
Answer. ATQ S₇ = 63, ...(1)
Sum of next 7 terms = S₁₄ – S₇ = 161 ...(2)
Use S_n = n/2 [2a + (n – 1) d]
Solve (1) and (2) to get a and d then find a₂₈ using a_n = a + (n – 1) d.
a = 3, d = 2
∴ Answer is a₂₈ = 57

Question. If the 4^th term of an A.P. is zero, prove that the 25^th term of the A.P. is three times its 11^th term.
Answer. a₄ = 0 ⇒ a + 3d = 0 ⇒ a = – 3d
a₂₅ = a + 24d = – 3d + 24d = 21d
a₁₁ = a + 10d = – 3d + 10d = 7d
∴ a₂₅ = 3a₁₁

Question. In an A.P. if S₅ + S₇ = 167 and S₁₀ = 235. Find the A.P., where S_n denotes the sum of its first n terms.
Answer. Use S_n = n/2 [2a + (n – 1) d]
S₅ + S₇ = 167 S₁₀ = 235
Solve to get a = 1, d = 5
A.P. = 1, 6, 11, 16, 21, .........

Question. The sum of first n terms of an A.P. is 5n² + 3n. If the m^th term is 168, find the value of m. Also find the 20^th term of the A.P.
Answer. S_n = 5n² + 3n
S₁ = a₁ = 8
S₂ = a₁ + a₂
26 = 8 + a₂ ⇒ a₂ = 18
d = 18 – 8 = 10
a_m = 168 ⇒ a + (m – 1)d = 168
8 + (m – 1)10 = 168 ⇒ m = 17

Question.  Your friend Veer wants to participate in a 200 m race. Presently, he can run 200 m in 51 seconds and during each day practice it takes him 2 seconds less. He wants to do in 31 seconds.

Question. Which of the following term is not in the AP of the above given situation?   
(a) 41
(b) 30
(c) 37
(d) 39

Answer: B

Question. The value of x, for which 2x, x + 10, 3x + 2 are three consecutive terms of an AP is   
(a) 6
(b) – 6
(c) 18
(d) –18

Answer: A

Question. Which of the following terms are in AP for the given situation?   
(a) 51, 53, 55, ...
(b) 51, 49, 47, ...
(c) –51, –53, –55, ...
(d) 51, 55, 59, …

Answer: B

Question. If nth term of an AP is given by an = 2n + 3 then common difference of an AP is   
(a) 2
(b) 3
(c) 5
(d) 1

Answer: A

Question. What is the minimum number of days he needs to practice till his goal is achieved?   
(a) 10
(b) 12
(c) 11
(d) 9

Answer: C

India is competitive manufacturing location due to the low cost of manpower and strong technical and engineering capabilities contributing to higher quality production runs. The production of TV sets in a factory increases uniformly by a fixed number every year. It produced 16000 sets in 6^th year and 22600 in 9^th year.

Question. In which year, the production is 29,200?     
(a) 10th year
(b) 12th year
(c) 15th year
(d) 18th year

Answer: B

Question. The production during first year is     
(a) 3000 TV sets
(b) 5000 TV sets
(c) 7000 TV sets
(d) 10000 TV sets

Answer: B

Question. The production during first 3 years is       
(a) 12800
(b) 19300
(c) 21600
(d) 25200

Answer: C

Question. The difference of the production during 7^th year and 4^th year is       
(a) 6600
(b) 6800
(c) 5400
(d) 7200

Answer: A

Question. The production during 8th year is       
(a) 10500
(b) 11900
(c) 12500
(d) 20400

Answer: D

Pollution—A Major Problem: One of the major serious problems that the world is facing today is the environmental pollution. Common types of pollution include light, noise, water and air pollution.
In a school, students thoughts of planting trees in and around the school to reduce noise pollution and air pollution.
Condition I: It was decided that the number of trees that each section of each class will plant be the same as the class in which they are studying, e.g. a section of class I will plant 1 tree a section of class II will plant 2 trees and so on a section of class XII will plant 12 trees.
Condition II: It was decided that the number of trees that each section of each class will plant be the double of the class in which they are studying, e.g. a section of class I will plant 2 trees, a section of class II will plant 4 trees and so on a section of class XII will plant 24 trees.

Question. If there are two sections of each class, how many trees will be planted by the students?     
(a) 422
(b) 312
(c) 360
(d) 540

Answer: B

Question. If there are three sections of each class, how many trees will be planted by the students?     
(a) 234
(b) 260
(c) 310
(d) 326

Answer: A

Question. The AP formed by sequence i.e. number of plants by students is    
(a) 0, 1, 2, 3, ..., 12
(b) 1, 2, 3, 4, ..., 12
(c) 0, 1, 2, 3, ..., 15
(d) 1, 2, 3, 4, ..., 15

Answer: B

Question. If there are three sections of each class, how many trees will be planted by the students?     
(a) 468
(b) 590
(c) 710
(d) 620

Answer: A

Question. If there are two sections of each class, how many trees will be planted by the students?   
(a) 126
(b) 152
(c) 156
(d) 184

Answer: C

II. Your elder brother wants to buy a car and plans to take loan from a bank for his car. He repays his total loan of ₹ 1,18,000 by paying every month starting with the first instalment of ₹ 1000. If he increases the instalment by ₹ 100 every month , answer the following:

Question. What amount does he still have to pay after 30^th installment?     
(a) ₹ 45500
(b) ₹ 49000
(c) ₹ 44500
(d) ₹ 54000

Answer: C

Question. The total amount paid by him upto 30 installments is     
(a) ₹ 37000
(b) ₹ 73500
(c) ₹ 75300
(d) ₹ 75000

Answer: B

Question. The ratio of the 1st installment to the last installment is     
(a) 1 : 49
(b) 10 : 49
(c) 10 : 39
(d) 39 : 10

Answer: B

Question. If total installments are 40, then amount paid in the last installment is     
(a) ₹ 4900
(b) ₹ 3900
(c) ₹ 5900
(d) ₹ 9400

Answer: A

Question. The amount paid by him in 30th installment is     
(a) ₹ 3900
(b) ₹ 3500
(c) ₹ 3700
(d) ₹ 3600

Answer: A