Find the sum of last 8 terms of the A.P.

-12, -10, -8, …….., 58.

Sum of last 8 terms of A.P. -12, -10, -8, …….., 58 = Sum of first 8 terms of A.P. 58, 56, 54, ………, -10, -12.

In A.P. 58, 56, 54, ………, -10, -12,

$S = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] = \dfrac{8}{2}[2 \times 58 + (8 - 1) \times (-2)] \\[1em] = 4[116 - 14] \\[1em] = 4 \times 102 \\[1em] = 408.$

Hence, sum = 408.