Question

sum of first n terms of an AP is given by sn2=2n2+3n. find the sixteenth term of the AP.

Answer 1

$sn = {2n}^{2} + 3n \\ \frac{n}{2}(2a + (n - 1)d) = n(2n + 3) \\ 2a + (n - 1)d = 4n + 6 \\ 2a + (n - 1)d = 10 - 4 + 4n \\ 2a + (n - 1)d = 10 - 4(n - 1)$
Comparing both sides, we get
2a=10
a=5 and d - 4

T16 = a +15d
T16 = 5 + 15(-4)
T16 = 5 - 60
T16 = - 55