Question

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

Answer 1

The given A.P is 8, 10, 12, 14,...., 126.
a = 8 and d = 2.
When this A.P is reversed, we get the A.P.
126, 124, 122, 120,....
So, first term becomes 126 and common difference −2.
The sum of first 10 terms of this A.P is as follows:
$$\begin{gathered} S_{10}=\frac{10}{2}\left[2\times 126+9\left(-2\right)\right] \\ =5\left[234\right] \\ =1170 \end{gathered}$$