# sum of first fifteen terms of AP is 570 and its 12th term is 62 .find sum of first 30 terms and sum of Nth term

Suppose a be the first term and d be the common difference of an A.P.
So, 12th term =

${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, nth term of an AP, a= a + (n - 1) d
Sn=n2[2a+(n1)d]Sn=n2[2a+(n1)d]
Sum of first 15 terms =

Solving (i) and (ii) we get;

So from (i) we have;

So sum of 30 terms =
And sum of nth term = $\frac{n}{2}\left[2a+\left(n-1\right)d\right]$

• 0

sum of nth term u could give two ..equations according to the question first equation is:

1)n/2*(2a+(n-1)*d)

2)n/2*(a+a1) a1=nth term..

• -1
What are you looking for?