Find the sum of last ten terms of the A.P.: $8, 10, 12, 14,…, 126$.

Given:

Given A.P. is $8, 10, 12, 14,…, 126$.
To do:
We have to find the sum of the last ten terms of the A.P.: $8, 10, 12, 14,…, 126$.

Solution:

For finding the sum of the last ten terms, we can write the given A.P. in reverse order.

This implies, the A.P. now becomes,

$126, 124,........, 14, 12, 10, 8$
Here,

First term \( (a)=126, \) common difference \( (d)=124-126=-2 \)

We know that,

${S}_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\therefore \mathrm{S}_{10}=\frac{10}{2}[2 a+(10-1) d]$
$= 5[2(126)+9(-2)]$

$=5(252-18)$
$=5 \times 234$

$=1170$

The sum of the last ten terms of the A.P.: $8, 10, 12, 14,…, 126$ is $1170$.

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Updated on: 10-Oct-2022

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