Question

solve the question attached

Answer 1

Let a,b,c,d are in GP and common ratio is r.

Then,

$a = a \\ \\ b = ar \\ \\ c = a {r}^{2} \\ \\ d = a {r}^{3} \\ \\ (a + b) = (a + ar) = a(1 + r) \\ \\(b + c) = (ar+ a {r}^{2} ) = ar(1 + r) \\ \\ (c + d) = (a {r}^{2} + a {r}^{3} ) = {a}^{2} (1 + r) \\ \\ since \: {(b + c)}^{2} = {a}^{2} {r}^{2} {(1 + r)}^{2} \\ \\ and \\ \\ (a + b)(c + d) = {a}^{2} {r}^{2} {(1 + r)}^{2} \\ \\ \\ \\ therefore \\ {(b + c)}^{2} = (a + b)(c + d)$

It is the condition of GP.

Hence. (a+b), (b+c) and (c+d) are in GP


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