Find the sum of last ten terms

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

i.e., 126,124,122… 12,10,8

Here, first term (a) = 126, common difference, (d) = 124-126=-2

$\therefore \quad S_{10}=\frac{10}{2}[2 a+(10-1) d] \quad\left[\because S_{n}=\frac{n}{2}[2 a+(n-1) d]\right]$

$=5\{2(126)+9(-2)\}=5(252-18)$

$=5 \times 234=1170$