If the sum of a certain number of terms starting from first term of an A.P. is $25, 22, 19, …,$ is 116. Find the last term.

Given:

The sum of a certain number of terms starting from first term of an A.P. is $25, 22, 19, …,$ is 116. 

To do:

We have to find the last term.

Solution:

Here \( a=25 \) and \( d=22-25=3 \)

We know that,

\( S_{n}=\frac{n}{2}[2 a+(n-1) d] \)
\( \Rightarrow 116=\frac{n}{2}[25 \times 2+(n-1)(-3)] \)
\( 232=n(50-3 n+3) \)

\(  232=(53-3 n) n \)
\(  232=53 n-3 n^{2} \)

\( 3 n^{2}-53 n+232=0 \)
\( 3 n^{2}-24 n-29 n+232=0 \)

$3 n(n-8)-29(n-8)=0$

$(n-8)(3 n-29)=0$
$n-8=0$ or $3n-29=0$

$n=8$ or $3 n=29$

$n=8$ or $n=\frac{29}{3}$ which is not possible

\( \therefore \) Number of terms \( =8 \)

This implies,

\( l=a_{8}=a+(n-1) d=25+(8-1) \times (-3) \)
$=25+7(-3)$

$=25-21$

$=4$

The last term is $4$.

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Updated on: 10-Oct-2022

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