Find the sum of last ten terms of the AP: $ 8,10,12, \cdots, 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 $1170$.