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This mock test of RS Aggarwal Test: Arithmetic Progressions for Class 10 helps you for every Class 10 entrance exam.
This contains 10 Multiple Choice Questions for Class 10 RS Aggarwal Test: Arithmetic Progressions (mcq) to study with solutions a complete question bank.
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QUESTION: 1

The 11th and 13th terms of an AP are 35 and 41 respectively, its common difference is

Solution:

11th term is 35 so a+10d = 35 => equation 1

13th term is 41 so a+12d = 41 => equation 2

solve both equations by elimination method

we get common difference as 3

QUESTION: 2

An AP consists of 31 terms. If its 16th term is m, then sum of all the terms of this AP is

Solution:

S_{31} = 31/2(2a + 30d)

a_{16} = a + 15d = m

⇒ S_{31} = (31/2) x 2(a + 15d) ⇒ S_{31} = 31 m

QUESTION: 3

Which term of the AP: 18, 23, 28, 33,... is 98?

Solution:

98=a+(n-1)d

98=18+(n-1)5

80=(n-1)5

16=n-1

17=n

this explains that 98 is 17th term of the AP

QUESTION: 4

The 16th term of the

Solution:

A= 15

d = a_{2} - a_{1}

d = 25 / 2 - 15

d = ( 30 - 25 / 2 )

d = - 5/ 2

16th term of an AP = a + 15 d

15 +( 15 × ( -5 / 2 ))

15 + (-75 / 2 )

= - 45 / 2

QUESTION: 5

The list of numbers -10, -6, -2, 2,... is

Solution:

a_{1}=-10

a_{2}=-6

so the common difference

d=(-6-(-10)) = 4

QUESTION: 6

The sum of first 16 terms of the AP : 10, 6, 2,... is

Solution:

The formula for the sum of first n terms of a AP is n/2(2a+(n-1)d)

here n=16 a=10 d=6-10=2-6=-4

so sum of first 16 terms=16/2(2·10+(16-1)·-4)

=8(20+15·-4)

=8(20-60)

=8(-40)

=-320

QUESTION: 7

The sum of first five multiples of 3 is

Solution:

First five multiple of 3 are:

3, 6, 9, 12, 15

Here first term, *a* = 3

common difference, *d* = 6 – 3 = 3

Number of term, *n* = 5

⇒ *s*_{5} = 9 × 5 = 45

QUESTION: 8

Two APs have same common difference. The first term of one of these is -1 and that of the other is - 8. Then the difference between their 4th term is

Solution:

**Correct Answer :- c**

**Explanation : **ad_{1} = -1

ad_{2} = -8

4th term of ad_{1} is

a_{4} = a + 3d

a_{4} = -1 + 3d.....(1)

Similarly ad_{2} is

a_{4} = -8 + 3d.......(2)

Subtracting (2) from (1), we get

-1 + 3d -(-8 + 3d)

=> 7

QUESTION: 9

The 21st term of the AP whose first two terms are -3 and 4, is

Solution:

a_{1}=-3,

a_{2}=4,

d=4-(-3)=7,

a_{21}=?,

a_{21}=a+(n-1)d

=-3+(21-1)7

= -3+(20)(7)

= -3+140

=137

QUESTION: 10

If 18, a, b, -3 are in AP, then a + b is equal to

Solution:

we know

AM is ( a + b)/2

(a+b) = [ {18+(-3) }/2] X 2 =15

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