Question

Find the sum of first n terms of the series a, 3a,5a......

Answer 1

first term= a
common difference, d= 3a-a = 2a

Sum of n terms, Sn = (n/2)[2a+(n-1)d]
Putting the values
Sn= (n/2)[2a+(n-1)(2a)]

Sn = (n/2)[2a+2an-2a]

Sn= (n/2)[2an] = (n)(an)

Sn= an²



So sum of n terms is an²



hope this helps you

Answer 2

Sn=n/2(2a+(n-1)d)
Sn=n/2(2a+2an-2a)
Sn=n(an)
Sn=an^2