Question

the sum of n terms of an A.P. is 3n^2+2n, then find its n th term.
answer-we know that
tn=s(n)-s(n-1)
=(3n^2+2n)-[3(n-1)^2+2(n-1)]
=(3n^2+2n)-[3n^2-6n+3+2n-2]
=3n^2+2n-[3n^2-6n+2n+3-2]
=3n^2+2n-(3n^2-4n+1]
=3n^2+2n-3n^2+4n-1
=6n-1 ans

Answer 1

or, we know that

Sn = n/2 (2a + (n-1)d) = 3n^2+2n
an + n(n-1)/2 d = 3n^2+2n
an + d/2 n^2 - d/2 n = 3n^2+2n
d/2 n^2 + (a - d/2)n = 3n^2+2n

d/2 = 3 so d=6
(a-3)=2 so a=5

Tn = 5 + 6(n-1) = 6n-1