Find the sum of the first 22 terms of the A.P. : 8, 3, -2, ……..

In the above A.P. a = 8 and d = 3 - 8 = -5.

$S = \dfrac{n}{2}(a + l) \\[1em] = \dfrac{n}{2}(a + a_{22}) \\[1em] = \dfrac{n}{2}(a + a + (22 - 1)d) \\[1em] = \dfrac{22}{2}(8 + 8 + 21 \times -5) \\[1em] = 11(16 - 105) \\[1em] = 11 \times -89 \\[1em] = -979.$

Hence, sum = -979.