Question

The sum of first n terms of an AP is given by Sn= 2n square +3n. Find the sixteenth term of the AP.

Answer 1

this is your answer to your questions

Answer 2

Answer:

$16^{th}\: term \: in \: A.P = a_{16}=64$

Step-by-step explanation:

$Given\:Sum\: of \: first\:n\\terms \: S_{n}=2n^{2}+3n$

$Let \: n^{th} \: term \: in \: A.P\\=a_{n}$

$a_{n}=S_{n}-S_{n-1}\\=2n^{2}+3n-[2(n-1)^{2}+3(n-1)]\\=2n^{2}+3n-[2(n^{2}-2n+1)+3n-3]\\=2n^{2}+3n-(2n^{2}-4n+2+3n-3)\\=2n^{2}+3n-2n^{2}+4n-2-3n+3\\=4n+1$

$Now,16^{th}\: term \: in \: A.P = a_{16}=4\times 16+1\\=64+1\\=65$

Therefore,

$16^{th}\: term \: in \: A.P = a_{16}=65$

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