Question

15.The sum of first n terms of a certain series is given as 3n2 2n . Show that the series
is an arithmetic series

Answer 1

SOLUTION :
GIVEN : Sn = 3n² - 2n

Put n = 1
S₁ = 3(1)² - 2(1)
S₁= 3 - 2
S₁ = 1
S₁ = a = 1
a= 1(first term )

Put n = 2
Sn = 3n² - 2n
S₂ = 3(2)² - 2(2)
S₂ = 3(4) - 4
S₂ = 12 - 4
S₂ = 8

a₂ = S₂ - S₁                               [ an = sn - s(n-1)]
a2 = 8 - 1 = 7
a2 = 7

Put n = 3
Sn = 3n² - 2n
S3 = 3(3)² - 2(3)
S3 = 3(9) - 6
S3 = 27- 6
S3 = 21

a3 = S3 - s2
a3 = 21 - 8 = 13
a3 = 13

Terms of the series are a1,a2,a3 -
1,7,13….

a2 - a1 = 7 -1 = 6                             [ d = a2 - a1]
a3 - a2 = 13 - 7 = 6

Since the difference of any two consecutive terms is same. So the series is an arithmetic series .

HOPE THIS WILL HELP YOU...