Question

10. The sum of first n terms of an A.P. is 2n^2+ n. Find nth term and common difference of the
A.P.​

Answer 1

Answer:

$\sf{The \ n^{th} \ term \ is \ -1+4n \ and \ common}$

$\sf{difference \ is \ 4.}$

Given:

$\sf{In \ an \ AP,}$

$\sf{\leadsto{Sum \ of \ n \ terms=2n^{2}+n}}$

To find:

$\sf{n^{th} \ term \ and \ common \ difference. }$

Solution:

$\sf{S_{n}=2n^{2}+n}$

$\sf{Hence,}$

$\sf{S_{1}=2(1)^{2}+1}$

$\sf{\therefore{S_{1}=2+1=3}}$

$\sf{\therefore{t_{1}=a=3}}$

$\sf{Also,}$

$\sf{S_{2}=2(2)^{2}+2}$

$\sf{\therefore{S_{2}=8+2=10}}$

$\sf{But, \ t_{2}=S_{2}-S_{1}}$

$\sf{\therefore{t_{2}=10-3=7}}$

$\sf{Here, \ t_{1}=3 \ and \ t_{2}=7}$

$\boxed{\sf{Common \ difference(d)=t_{2}-t_{1}}}$

$\sf{\therefore{Common \ difference(d)=7-3=4}}$

$\boxed{\sf{t_{n}=a+(n-1)d}}$

$\sf{\therefore{t_{n}=3+(n-1)\times4}}$

$\sf{\therefore{t_{n}=3-4+4n}}$

$\sf{\therefore{t_{n}=-1+4n}}$

$\sf\purple{\tt{\therefore{The \ n^{th} \ term \ is \ -1+4n \ and \ common}}}$

$\sf\purple{\tt{difference \ is \ 4.}}$

Answer 2

Given ,

• The sum of first n terms of an AP is 2(n)² + n

We know that ,

$\boxed{ \sf{ a_{n} = s_{n} - s_{n - 1}}}$

Thus ,

First term = 3

Second term = 10 - 3 = 7

Therefore ,

• The common difference of given AP will be 4

Now , the nth term of an AP is given by

$\boxed{ \sf a_{n} = a + (n - 1)d}$

Thus ,

$\sf \mapsto a_{n} = 3 + (n - 1)4 \\ \\ \sf \mapsto 3 + 4n - 4 \\ \\ \sf \mapsto 4n - 1$

Therefore ,

The nth term of given AP is 4n - 1