The sum of first $n$ terms of an A.P. is $5n^2 + 3n$. If its $m$th term is 168, find the value of $m$. Also, find the 20th term of this A.P.

Given:

The sum of first $n$ terms of an A.P. is $5n^{2} +3n$ and its $m$th term is 168.

To do:

We have to find the value of $m$ and $20^{th}$ term of the given A.P.

Solution:

$S_{n} =5n^{2} +3n$

For $n=1,\ S_{1} =5\times 1^{2} +3\times 1=5+3=8$

Therefore, first term $a=8$

For $n=2,\ S_{2} =5\times 2^{2} \ +3\times 2=20+6=26$

$\therefore$ Second term of the A.P.$=S_{2} -S_{1}$

$=26-8$

$=18$

Common difference of the A.P., $d=$second term $-$ first term

$=18-8=10$

We know that,

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

$a_{m}=8+( m-1) \times 10$

$168=8+10m-10$

$10m=168+2$

$10m=170$

$m=17$

$a_{20}=8+( 20-1) \times 10$

$=8+19\times 10$

$=8+190$

$=198$

Therefore, the value of $m$ is $17$ and the $20^{th}$ term of the given A.P. is $198$.

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Updated on: 10-Oct-2022

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