Practice Test: Arithmetic Progressions


25 Questions MCQ Test Mathematics (Maths) Class 10 | Practice Test: Arithmetic Progressions


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This mock test of Practice Test: Arithmetic Progressions for Class 10 helps you for every Class 10 entrance exam. This contains 25 Multiple Choice Questions for Class 10 Practice Test: Arithmetic Progressions (mcq) to study with solutions a complete question bank. The solved questions answers in this Practice Test: Arithmetic Progressions quiz give you a good mix of easy questions and tough questions. Class 10 students definitely take this Practice Test: Arithmetic Progressions exercise for a better result in the exam. You can find other Practice Test: Arithmetic Progressions extra questions, long questions & short questions for Class 10 on EduRev as well by searching above.
QUESTION: 1

If the sum of first n terms of an AP be 3n2 – n and it's common difference is 6, then its first term is :

Solution:
QUESTION: 2

If 7th and 13th terms of an A.P. be 34 and 64, respectively, then it's 18th term is :

Solution:
QUESTION: 3

The sum of all 2-digit odd positive numbers is :

Solution:

Here a = 11 and d = 2, tn= 99, n = ?
Sum of the n terms = (n/2)[2a+(n -1)d]
But tn = a + (n -1)d
⇒ 99 = 11+ (n-1)2
⇒ 99 -11 = (n-1)2
⇒ 88/2 = (n-1)
∴ n = 45.
subsitute n = 45  in sum of the n terms we obtain
⇒ s45 = (45/2)(2×11 + (45 -1)2)
⇒ s45 = (45/2)(110)
⇒ s45 = 45×55.
⇒  s45 = 2475.
∴ sum of all two digit odd positive numbers = 2475.

QUESTION: 4

The fourth term of an A.P. is 4. Then the sum of the first 7 terms is :

Solution:

QUESTION: 5

In an A.P., s1 = 6, s7 = 105, then sn : sn-3 is same as :

Solution:
QUESTION: 6

In an A.P. s3 = 6, s6 = 3, then it's common difference is equal to :

Solution:



 

QUESTION: 7

The number of terms common to the two A.P. s 2 + 5 + 8 + 11 + ...+ 98 and 3 + 8 + 13 + 18 +...+198

Solution:

For first A.P
2+5+8+11+......+98
a=2,an​=98,d=3
an​=a+(n−1)d
98=2+(n−1)3
98=2+3n−3
3n=99
n=33
Number of term =33
For first A.P
3+8+13+18+......+198
a=3,an​=198,d=5
an​=a+(n−1)d
198=3+(n−1)5
198=3+5n−5
5n=200
n=40
No of terms =40
Common terms=40−33=7

QUESTION: 8

(p + q)th and (p – q)th terms of an A.P. are respectively m and n. The pth term is :

Solution:

l=a+(n-1)d
(p+q)th term is m
m=a+(p+q-1)d
m=a+pd+qd-d  ….1
(p-q)th term is n
n=a+(p-q-1)d
n=a+pd-qd-d   ….2
Adding 1 and 2
m+n=2a+2pd-2d
m+n=2(a+pd-d)
½(m+n=a+(p-1)d
So pth term is ½(m+n)

QUESTION: 9

The first, second and last terms of an A.P. are a,b and 2a. The number of terms in the A.P. is :

Solution:

A.P : a , b , . . . . . . . . . . . . . .2a

1st term= a1 = a

2nd term = a2= b

nth term = an = 2a

d = a2 - a1 = b-a

an = a1 + (n-1)d = a + (n-1)(b-a) = 2a

(n-1)(b-a) = a

(n-1) = a / (b-a)

n = a / (b-a) + 1 = b / ( b -a )

Sn = n / 2 * ( a1 + an) = b / 2(b-a) * ( a + 2a) = 3ab / 2(b-a)

QUESTION: 10

Let s1, s2, s3 be the sums of n terms of three series in A.P., the first term of each being 1 and the common differences 1, 2, 3 respectively. If s1 + s3 = λs2, then the value of λ is :

Solution:
QUESTION: 11

Sum of first 5 terms of an A.P. is one fourth of the sum of next five terms. If the first term = 2, then the common difference of the A.P. is :

Solution:
QUESTION: 12

If x,y,z are in A.P., then the value of (x + y – z) (y + z – x) is equal to :

Solution:
QUESTION: 13

The number of numbers between 105 and 1000 which are divisible by 7 is :

Solution:
QUESTION: 14

If the numbers a,b,c,d,e form an A.P. then the value of a – 4b + 6c – 4d + e is equal to :

Solution:
QUESTION: 15

If sn denotes the sum of first n terms of an A.P., whose common difference is d, then sn – 2sn-1 + sn-2 (n >2) is equal to :

Solution:

sn​−2sn−1​+sn−2 ​= (sn​−sn−1​)−(sn−1​−sn−2​)

= an​−an−1 ​[∵(sn​−sn−1​)= an​]

= [a+(n−1)d]−[a+(n−2)d]

= a+nd−d−a−nd+2d

= d

QUESTION: 16

The sum of all 2-digited numbers which leave remainder 1 when divided by 3 is :

Solution:
The 2-digit number which when divided by 3 gives remainder 1 are: 10, 13, 16, ...97
Here a = 10, d = 13 - 10 = 3
tn = 97
nth term of an AP is tn = a + (n – 1)d
97 = 10 + (n – 1)3
⇒ 97 = 10 + 3n – 3
⇒ 97 = 7 + 3n
⇒ 3n = 97 – 7 = 90
Therefore, n = 90/3 = 30
Recall sum of n terms of AP, 
Sn = n/2[2a + (n-1)d]

S30 = 30/2[2(10) + (30-1)3]
= 15[20 + 87] = 15 x 107 = 1605
Hence sum of 2-digit numbers which when divided by 3 yield 1 as remainder is 1605.
QUESTION: 17

The first term of an A.P. of consecutive integers is p2 + 1. The sum of 2p + 1 terms of this series can be expressed as :

Solution:
QUESTION: 18

If the sum of n terms of an AP is 2n2 + 5n, then its nth term is –

Solution:
QUESTION: 19

If the last term of an AP is 119 and the 8th term from the end is 91 then the common difference of the AP is –

Solution:
QUESTION: 20

If {an} = {2.5, 2.51, 2.52,...} and {bn} = {3.72, 3.73, 3.74,...} be two AP's, then a100005 – b100005 =

Solution:

Observing both the AP’s we see that the common difference of both the AP’s is same ,so difference between their corresponding terms will be same ie,a1-b1=2.5-3.72=-1.22
a2-b2=2.51-3.73=-1.22
 So , a100005-b100005=-1.22

QUESTION: 21

If A1 and A2 be the two A.M.s between two numbers a and b, then (2A1 – A2) (2A2 – A1) is equal to :

Solution:

QUESTION: 22

 are in A.P., then n is equal to 

Solution:
QUESTION: 23

 where Sn denotes the sum of the first n terms of an A.P., then the common difference of the A.P. is

Solution:
QUESTION: 24

If a,b,c are positive reals, then least value of (a + b + c) (1/a+1/b+1/c) is :

Solution:
QUESTION: 25

The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.

Solution:

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