Question

If the sum of the first n terms of an AP is n^2+3n, then find the 20th term of the series.​

Answer 1

Answer:

$\sf{The \ 20^{th} \ term \ is \ 42 \ in \ series.}$

Given:

$\sf{Sum \ of \ n \ terms \ of \ an \ AP \ is}$

$\sf{n^{2}+3n}$

To find:

$\sf{The \ 20^{th} \ term \ of \ the \ series.}$

Solution:

$\sf{Method \ (I)}$

$\sf{S_{n}=n^{2}+3n}$

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

$\sf{\therefore{S_{1}=4}}$

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

$\sf{\therefore{t_{1}=a=4}}$

$\sf{S_{2}=2^{2}+3(2)}$

$\sf{\therefore{S_{2}=4+6}}$

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

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

$\sf{\therefore{4+t_{2}=10}}$

$\sf{\therefore{t_{2}=10-4}}$

$\sf{\therefore{t_{2}=6}}$

$\sf{d=t_{2}-t_{1}}$

$\sf{\therefore{d=6-4}}$

$\sf{\therefore{d=2}}$

$\sf{Here,}$

$\sf{First \ term(a)=4,}$

$\sf{Common \ difference (d)=2}$

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

$\sf{\therefore{t_{20}=4+(20-1)\times2}}$

$\sf{\therefore{t_{20}=4+38}}$

$\sf{\therefore{t_{2}=42}}$

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$\sf{Method (II)}$

$\sf{\underline{Concept:}}$

$\sf{Difference \ between \ S_{n+1} \ and \ S_{n}}$

$\sf{will \ be \ of \ t_{n+1}.}$

$\boxed{\sf{t_{n+1}=S_{n+1}-S_{n}}}$

$\sf{\therefore{t_{20}=S_{20}-S_{19}}}$

$\sf{But, \ S_{n}=n^{2}+3n}$

$\sf{\therefore{t_{20}=[20^{2}+3(20)]-[19^{2}+3(20)]}}$

$\sf{\therefore{t_{20}=[400+60]-[361+57]}}$

$\sf{\therefore{t_{20}=460-418}}$

$\sf{\therefore{t_{20}=42}}$

$\sf\purple{\tt{\therefore{The \ 20^{th} \ term \ is \ 42 \ in \ series.}}}$

Answer 2

Given ,

The sum of first n terms of an AP is

• (n)² + 3n

We know that ,

$\boxed{ \tt{a_{n} = S_{n} - S_{n - 1} }}$

Thus , the 20th term of AP will be

$\sf \implies a_{20} = S_{20} - S_{19}$

$\sf \implies a_{20} = {(20)}^{2} + 3(20) - \{ {(19)}^{2} + 3(19) \}$

$\sf \implies a_{20} = 400 + 60 - 361 - 57$

$\sf \implies a_{20} = 460 - 418$

$\sf \implies a_{20} = 42$

Hence , the 20th term of given AP is 42