The sum of first $m$ terms of an A.P. is $4m^2 – m$. If its $n$th term is 107, find the value of $n$. Also, find the 21st term of this A.P.

Given:

The sum of first $m$ terms of an A.P. is $4m^2 – m$ and its $n$th term is 107.

To do:

We have to find the value of $n$ and $21^{st}$ term of the given A.P.

Solution:

$S_{m} =4m^{2} -m$

For $m=1,\ S_{1} =4\times 1^{2} -1=4-1=3$

Therefore, first term $a=3$

For $n=2,\ S_{2} =4\times 2^{2} -2=16-2=14$

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

$=14-3$

$=11$

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

$=11-3=8$

We know that,

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

$a_{n}=3+( n-1) \times 8$

$107=3+8n-8$

$8n=107+5$

$8n=112$

$n=14$

$a_{21}=3+( 21-1) \times 8$

$=3+20\times 8$

$=3+160$

$=163$

Therefore, the value of $n$ is $14$ and the $21^{st}$ term of the given A.P. is $163$. 

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Updated on: 10-Oct-2022

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