If the 10th term of an A.P. is 21 and the sum of its first ten terms is 120, find its $n$th term.

Given:

The 10th term of an A.P. is 21 and the sum of its first ten terms is 120

To do:

We have to find the $n$th term of the A.P.

Solution:

Let the first term of the A.P. be $a$ and the common difference be $d$.

We know that,

nth term of an A.P. $a_n=a+(n-1)d$

Therefore,

$a_{10}=a+(10-1)d$

$21=a+9d$

$a=21-9d$......(i)

Sum of $n$ terms of an A.P. $S_n=\frac{n}{2}(2a+(n-1)d)$

$S_{10}=\frac{10}{2}(2a+(10-1)d$

$120=5(2a+9d)$

$120=10a+45d$

$120=10(21-9d)+45d$     (From (i))

$120=210-90d+45d$

$90d-45d=210-120$

$45d=90$

$d=\frac{90}{45}$

$d=2$

This implies,

$a=21-9(2)$

$=21-18$

$=3$

nth term of the A.P. $a_n=a+(n-1)d$

$=3+(n-1)2$

$=3+2n-2$

$=2n+1$

Hence, the nth term of the given A.P. is $2n+1$.   

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Updated on: 10-Oct-2022

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