CBSE Class 10 Mathematics Arithmetic Progression Worksheet Set A

More MCQs for NCERT Class 10 Mathematics Arithmetic Progression........

Question. In an Arithmetic progression, the 4^th term is 11 and the 12^th term is 35, then the first term of the series is
(A) 5
(B) 4
(C) 3
(D) 2
Answer : D

Question. The first term of the A.P whose third term is 16 and the difference of 5^th and 7th term is 12 is
(A) 7
(B) 6
(C) 5 
(D) 4
Answer : D

Question. If a,b,c,d,e are in A.P. find the value of a -4b+6c -4d +e ?
(A) 0 
(B) 1
(C) 2 
(D) 3
Answer : A

Question. In a certain A.P., 5 times the 5th term is equal to 8 times the 8^th term find the 13^th term?
(A) 5 
(B) 2
(C) 0 
(D) 1
Answer : C

Question. Find the smallest positive term of the series 25,22³/₄ ,20¹/₂ ,18¹/₄ ..............?
(A) 9^th
(B) 10^th
(C) 11^th 
(D) 12^th
Answer : D

Question. The sum of first four terms of an A.P. is 56. The sum of last four terms is 112. If its first term is 11, find number of terms?
(A) 8 
(B) 9
(C) 10 
(D) 11
Answer : D

Question. The sum of first 20 even natural number is
(a) 420
(b) 100
(c) 220
(d) 400
Answer : A

Question. How many natural numbers are there between 200 and 500, which are divisible by 7?
(a) 42
(b) 43
(c) 63
(d) 53
Answer : B

Question. The 4th term from the end of an AP , -11 , -8 , -5 , ……..49
(a) 17
(b) 25
(c) 40
(d) - 40
Answer : C

Question. The sum of first five multiple of 3 is
(a) 35
(b) 45
(c) 63
(d) 53
Answer : B

Question. Find 15th terms of an AP 15, 10 ,5, 0 -5 -------------
(a) -55
(b) -60
(c) -65
(d) none of these
Answer : A

Question. Given two A.P.’s 2,5,8,11..................T₆₀ and 3,5,7,9.................T₅₀. Find the number of terms which are identical?
(A) 17 
(B) 18
(C) 19 
(D) 20
Answer : A

Question. If pth, qth, rth terms of an A.P. are a,b,c, respectively. Find the value of a(q-r) + b (r-p) +c(p-q)
(A) 2 
(B) 1
(C) 0 
(D) 5
Answer : C

Question. If the sum of three numbers in A.P. is 24 and their product is 440. Find the numbers?
(A) 5,8,11 
(B) 5,9,11
(C) 2,4,9 
(D) 2,6,9
Answer : A

Question. Divide 32 into four parts which are in A.P. such that the ratio of product of extremes to the product of means is 7:15
(A) 1,5,9,13 
(B) 3,7,11,15
(C) 2,6,10,14 
(D) 4,8,12,16
Answer : C

Question. If the sum of series 2,5,8,11.................. is 60100, find n?
(A) 200 
(B) 210
(C) 220 
(D) 240
Answer : A

Question. The sum of n terms of two A.P’s are in ratio 5n+4:9n+6 find ratio of their 18^th terms?
(A) 179:321 
(B) 180:322
(C) 170:320 
(D) 171:329
Answer : A

Question. If log 5 x 2^x + 1 , log4 (2^1-x +1) and 1 are in A.P. find x?
(A) log⁵₂ 
(B)1-log⁵₂
(C) log²₅ 
(D) 1-log²₅
Answer : B

Question. If 3 positive real nos, a,b,c, are in A.P. such that abc=4, find minimum value of b?
(A) 2^1/3
(B) 2^1/2
(C) 2^2/3
(D) 2^3/2
Answer : C

Question. If the sides of right angled triangle are in A.P., then the sines of acute angles are?
(A) 3/5 , 4/5
(B) 1/√2 , √(2/3)
(C) 1/2 , √3/2
(D) 1/√2 , √3/2
Answer : A

Question. Concentric circles of radii 1,2,3,........100 cm are drawn. The interior of the smallest circle is coloured red and the angular regions are colored alternatively green & red, so that no two adjacent regions are of same coloured find total area of green regions in sq. cm.?
(A) 1000π 
(B) 5050π
(C) 4950π 
(D) 5151π
Answer : B

Question. In an AP, the sum of first n term is (3n² /2 +5n/2). Find its 25^th term.
(A) 66 
(B) 86
(C) 76 
(D) 96
Answer : C

Question. If the 12th term of an A.P. is -13 and the sum of the first four terms is 24 what is the sum of the first 10 terms.
(A) 150 
(B) -1
(C) 180 
(D) zero
Answer : D

Question. If n AMs are inserted between 2 & 38, the sum of the resulting series obtained is 200. The value of n (total number of terms) is
(A) 8 
(B) 10
(C) 9 
(D) 11
Answer : A

Question. Find t₅ and t₆ of the arithmetic progression 0, 1/4, 1/2, 3/4,……. respectively
(A) 1, 5/4 
(B) 5/4, 1
(C) 1, 7/4 
(D) 7/4, 1
Answer : A

Question. If t_n = 6n + 5, then t_{n + 1} =
(A) 6n –1 
(B) 6n+11
(C) 6n + 6 
(D) 6n – 5
Answer : B

Question. Which term of the arithmetic progression 21, 42, 63, 84, ……. is 420?
(A) 19 
(B) 20
(C) 21 
(D) 22
Answer : B

Question. Find the 15 term of the arithmetic progression 10, 4, –2,……
(A) –72 
(B) –74
(C) –76 
(D) –78
Answer : B

Question. If the k^th term of the arithmetic progression 25, 50, 75, 100,…….. is 1000, then k is ________.
(A) 20 
(B) 30
(C) 40 
(D) 50
Answer : C

Question. The sum of the first 20 terms of an arithmetic progression whose first term is 5 and common difference is 4, is
(A) 820 
(B) 830
(C) 850 
(D) 860
Answer : D

Question. Two arithmetic progressions have equal common differences. The first term of one of these is 3 and that of the other is 8, then the difference between their 100^th terms is
(A) 4 
(B) 5
(C) 6 
(D) 3
Answer : B

Question. If a, b and c are in arithmetic progression, then b + c, c + a and a + b are in
(A) arithmetic progression 
(B) geometric progression
(C) harmonic progression 
(D) none of these
Answer : A

Question. The sum of the first 51 terms of the arithmetic progression whose 2nd term is 2 and 4^th tem is 8, is
(A) 3774 
(B) 3477
(C) 7548 
(D) 7458
Answer : A

Question. Three alternate terms of an arithmetic progression are x + y,x - y and 2x +3y, then x =
(A) -y 
(B) -2y
(C) -4y 
(D) -6y
Answer : D

Question. Find the 15^th term of the series 243, 81, 27,……..
(A) 1/3¹⁴ 
(B) 1/ 3⁸
(C) (1/3)⁹
(D) (1/3)¹⁰
Answer : C

Question. In a right triangle, the lengths of the sides are in arithmetic progression. If the lengths of the sides of the triangle are integers, which of the following could be the length of the shortest side?
(A) 1225 
(B) 1700
(C) 1275 
(D) 1150
Answer : C

Question. If S₁ = 3,7,11,15,........ upto 125 terms and S₂ = 4,7,10,13,16........ upto 125 terms, then how many terms are there in S₁ that are there in S₂?
(A) 29 
(B) 30
(C) 31 
(D) 32
Answer : C

Question. Find the sum of all natural numbers and lying between 100 and 200 which leave a remainder of 2 when divided by 5 in each case.
(A) 2990 
(B) 2847
(C) 2936 
(D) none of these
Answer : A

Question. An AP starts which a positive fraction and every alternate term is an integer. If the sum of the first 11 terms is 33, then find the fourth term.
(A) 2 
(B) 3
(C) 5 
(D) 6
Answer : A

Question. If the sum of 16 terms of an AP is 1624 and the first term is 500 times the common difference, then find the common difference.
(A) 5 
(B) 1/2
(C) 1/5 
(D) 2
Answer : C

Question. Let Sn denote the sum of n terms of an A.P. whose first term is a. If the common difference d is given by d = S_n - kS_n-1 + S_n-2 then k =
(A) 1 
(B) 2
(C) 3 
(D) none of these
Answer : B

Question. The first and last term of an A.P. are a and l respectively. If S is the sum of all the terms of the A.P, and the common difference is given by l²-a² /k-(l+a)then k =
(A) S 
(B) 2S
(C) 3S 
(D) none of these
Answer : B

Question. If the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers, then k =
(A) 1/n
(B) n - 1 /n
(C) n + 1/2n
(D) n + 1 /n
Answer : D

Question. If the first, second and last term of an A.P. are a, b and 2a respectively, its sum is
(A) ab /2(b - a)
(B) ab /b - a
(C) 3ab /2(b - a)
(D) none of these
Answer : C

Question. If S₁ is the sum of an arithmetic progression of ‘n’ odd number of terms and S₂ the sum of the terms of the series in odd places, then S₁/S₂ =
(A) 2n /n +1
(B) n /n +1
(C) n + 1 /2n
(D) n + 1/n
Answer : A

CASE STUDY QUESTIONS

1. 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 6th year and 22600 in 9th year.

Question. Find the production during first year.
Answer : Rs 5000

Question. Find the production during 8th year.
Answer : Production during 8th year is (a+7d) = 5000 + 2(2200) = 20400

Question. Find the production during first 3 years.
Answer : Production during first 3 year = 5000 + 7200 + 9400 = 21600

Question. In which year, the production is Rs 29,200.
Answer : N = 12 5. Difference = 18200 - 11600 = 6600

Q2. Students of a school thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will 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 till class XII. There are three sections of each class.

Question. Find the total number of trees planted by the students of the school.?
Answer : 324

Question.. Find total number of trees planted by primary 1 to 5 class students?
Answer : 45

Question.. Find total number of classes ?
Answer : 12

Very Short Answer Type Questions

Question. In a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be double of the class in which they are studying. If there are 1 to 12 classes in the school and each class has two sections, find how many trees were planted by the students.
Answer : 312

Question. Reshma wanted to save at least ? 6,500 for sending her daughter to school next year (after 12 months). She saved Rs 450 in the first month and raised her savings by ? 20 every next month. How much will she be able to save in next 12 months? Will she be able to send her daughter to the school next year?
Answer : Rs 6720

Question. The n^th term of an AP is 6n + 2. Find its common difference.
Answer : 6

Question. The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles of the triangle.
Answer : 40°, 60°, 80°

Question. For what value of n, the nth term of two APs: 63, 65, 67, ... and 3, 10, 17, ... are equal.
Answer : 13

Question. If the sum of the first n terms of an AP is 4n − n², what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly find the 3rd, the 10th and the nth terms.
Answer : The second term is 1, The 3rd, 10th, and nth terms are −1, −15, and 5 − 2n respectively

Question. Determine k so that k² + 4k + 8, 2k² + 3k + 6, 3k² + 4k + 4 are three consecutive terms of an AP. 
Answer : k = 0

Question. Which term of the AP: 115, 110, 105, ..... is its first negative term?
Answer : 25^th term

Question. A sum of Rs 1600 is to be used to give ten cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.
Answer : 250,230,210,190,170,150,130,110,90,70

Question. A thief, after committing a theft, runs at a uniform speed of 50 m/minute. After 2 minutes, a policeman runs to catch him. He goes 60 m in first minute and increases his speed by 5 m/minute every succeeding minute. After how many minutes, the policeman will catch the thief?
Answer : police man catch the thief after 5 minutes

Question. Find the 10^th term from end of the AP: 4, 9, 14, ..., 254.
Answer : 209

Question. The 8^th term of an AP is 37 and its 12^th term is 57. Find the AP.
Answer : 2, 7, 12, 17, 22, ...

Short Answer Type Questions

Question. Find how many integers between 200 and 500 are divisible by 8.
Answer : AP formed is 208, 216, 224, ...., 496
a_n = 496 
⇒ 208 + (n – 1) × 8 = 496
⇒ n = 37

Question. Find the number of two-digit numbers which are divisible by 6.
Answer : Two-digit numbers which are divisible by 6 are 12, 18, 24,..., 96
∵ Last term,
a_n = 96
⇒ 12 + (n – 1)6 = 96 
⇒ (n – 1)6 = 96 – 12 = 84
⇒ n = 15
∴ There are 15 two-digit numbers divisible by 6.

Question. Is –150 a term of the AP: 17, 12, 7, 2, ...?
Answer : Let a_n = –150
a + (n – 1)d = –150
⇒ 17 + (n – 1)(–5) = –150
⇒ (n – 1)(–5) = – 167
⇒ n = 167+5/5 = 172/5 = 34(2/5)
Here, n is not a natural number.
∴ –150 is not a term of the given AP.

Question. Which term of the AP: 3, 15, 27, 39, ... will be 120 more than its 21^st term?
Answer : AP: 3, 15, 27, 39, ...
a = 3, d = 15 – 3 = 12
a₂₁ = a + 20d = 3 + 20 × 12
= 3 + 240 = 243
120 more than a₂₁ = 243 + 120 = 363 
Let 363 be n^th term.
So, 363 = 3 + (n – 1) 12
⇒ 360 = 12(n – 1)
30 = n – 1 ⇒ n = 31
Thus, 31st term of the given AP is 120 more than its 21^st term.

Question. If 1/X+2, 1/x+3. and 1/x+5 , and are in AP, find the value of x. 
Answer : Given term are in AP
So, 2/x+3 = 1/2+1 + 1/x+5
⇒ 2/x + 3 = (x+5) + (x+2)/(x+2)(x+5)
⇒ 2x² + 14x + 20 = 2x² + 13x + 21
∴  x = 1

Question. How many three digit numbers are divisible by 11? 
Answer : Three-digit numbers which are divisible by 11 are 110, 121, 132, ..., 990
Let a_n = 990
⇒ a + (n – 1)d = 990
⇒ 110 + 11(n – 1) = 990
∴  n = 81
Hence, there are 81 three-digit numbers which are divisible by 11.

Question. How many natural numbers are there between 200 and 500, which are divisible by 7?
Answer : Natural numbers between 200 and 500 which are divisible by 7 are as 203, 210, 217, ..., 497
Let above are n numbers and an = 497
a + (n – 1)d = 497 
⇒ 203 + 7(n – 1) = 497
⇒ n = 43
∴ There are 43 natural numbers between 200 and 500 divisible by 7

Question. How many two-digit numbers are divisible by 7? 
Answer : Two-digit numbers which are divisible by 7 are 14, 21, 28,..., 98.
Let a_n = 98
⇒ a + (n – 1)d = 98 
⇒ 14 + 7(n – 1) = 98
n = 13
Hence, there are 13 two-digit numbers which are divisible by 7.

Question. How many two digits numbers are divisible by 3?
Answer : 2-digit numbers divisible by 3 are 12, 15, 18, ..., 99 which is in AP.
So, a_n = 99, d = 15 – 12 = 3
Now, a_n = a + (n – 1) d 
⇒ 99 = 12 + (n – 1) 3
⇒ 87 = 3 (n – 1)
⇒ 29 = n – 1
⇒ n = 30
Thus, 30, 2-digit numbers are divisible by 3.

Question. Find the number of natural numbers between 102 and 998 which are divisible by 2 and 5 both.
Answer : 110, 120, 130, … , 990
a_n = 990
⇒ 110 + (n − 1) × 10 = 990
∴ n = 89

Question. Which term of the AP: 3, 14, 25, 36, ... will be 99 more than its 25^th term?
Answer : Let an be the term which is 99 more than 25^th term of given AP.
ATQ, a_n = a₂₅ + 99
⇒ a + (n – 1)d = a + 24d + 99 
⇒ 11(n – 1) = 24 × 11 + 99
⇒ n = 34
Hence, 34^th is the required term.

Question. Which term of the progression 20, 19 (1/4),18 (1/2), 17 (3/4), , , ,... is the first negative term?
Answer : Here d = −3/4
Let the nth term be first negative term.
∴ 20 + (n - 1) (-3/4)  < 0 fi 3n > 83
⇒ n > 27(2/3)
Hence, 28^th term is first negative term.

Question. Find the middle term of the AP: 6, 13, 20, ..., 216.
Answer : Given AP is 6, 13, 20, ..., 216
n^th term, a_n = 216
⇒ a + (n – 1)d = 216
⇒ 6 + 7(n – 1) = 216
⇒ 7n = 217
⇒ n = 31 
Since, the number of terms in AP are 31, so, the middle most term is 16th term.
[ ∵ middle term = (31+1)/2 = 16^th term ]
∴ 16^th term, a₁₆ = a + 15d = 6 + 15 × 7 = 111.

Question. The 4^th term of an AP is zero. Prove that the 25th term of the AP is three times its 11th term. 
Answer : a₄ = a + (4 – 1)d
0 = a + 3d
⇒ a = –3d            [∵ Given, a₄ = 0]
Now a₂₅ = a + (25 – 1)d = a + 24d
= –3d + 24d = 21d = 3 × 7d
Hence, a₂₅ = 3 × a₁₁
[∵ Since a₁₁ = a + (11 – 1)d = –3d + 10d = 7d] 

Question. In an AP, the first term is 12 and the common difference is 6. If the last term of the AP is 252, find its middle term.
Answer : Let an = 252 = last term
⇒ a + (n – 1)d = 252
⇒ 12 + (n – 1)6 = 252
⇒ n = 41
∴  Since number of terms is odd, so only one middle term.
Now, middle term = (41+1/2)
= 21^st term
∴  21^st term, a₂₁ = a + 20d
= 12 + 20 × 6
= 132
= middle term value.

Question. Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.
Answer : Numbers between 101 and 999 which are divisible by both 2 and 5 (i.e., by 10) are 110, 120, 130, ... 990.
Now, a_n = a + (n – 1)d
⇒ 990 = 110 + (n – 1)10
⇒ n = 89
∴ Natural numbers which are divisible by 2 and 5 both are 89.

Please refer to attached file for CBSE Class 10 Mathematics Worksheet - Arithmetic Progression