Question

find four consecutive terms in an A.P whose sum is 88 andof the 1st andthe 3rd term is 40

Answer 1

Given,
$a1 + a3 = 40$
$= > (a + (1- 1)d) + (a + (3 - 1)d) =40 \\ = > a + a + 2d = 40 \\ = > 2a + 2d = 40 \\ = > a + d = 20 - - - - - - - > 1$
Also given,
$S4 = 88 \\as \: we \: know \: that \\ Sn = \frac{n}{2}(2a + (n - 1)d) \\ = > 88 = 2(2a + 3d) \\ = > 2a + 3d = 44 - - - - - - - > 2$


eq.1 x 2 - eq.2,
$2a + 2d = 40 \\ 2a + 3d = 44 \\ - \: \: \: - \: \: \: \: \: \: \: \: - \\ - - - - - - - \\ - d = - 4 \\ = > d = 4 \\ = > a = 16. \\ - - - - - - - -$




A. P,
16,20,24,28,.......



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