RS Aggarwal Test: Arithmetic Progressions

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

S₃₁ = 31/2(2a + 30d)
a₁₆ = a + 15d = m
⇒ S₃₁ = (31/2) x 2(a + 15d) ⇒ S₃₁ = 31 m

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

A= 15
d = a₂ - a₁
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

a₁=-10
a₂=-6
so the common difference
d=(-6-(-10)) = 4

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

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

⇒ s₅ = 9 × 5 = 45

Correct Answer :- c

Explanation : ad₁ = -1

ad₂ = -8

4th term of ad₁ is 

a₄ = a + 3d

a₄ = -1 + 3d.....(1)

Similarly ad₂ is

a₄ = -8 + 3d.......(2)

Subtracting (2) from (1), we get

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

=> 7

a₁=-3,
a₂=4,
d=4-(-3)=7,
a₂₁=?,
a₂₁=a+(n-1)d
=-3+(21-1)7
= -3+(20)(7)
= -3+140
=137

We know

AM  is ( a + b)/2

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