If a, b, c are in A.P. and a, b, d are in G.P., show that a, (a − b), (d − c) are in G.P.

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Solution

$a, b and c are in A . P . ∴ 2 b = a + c . . . . . . . (i) Also, a, b and d are in G . P . ∴ b^{2} = a d . . . . . . . (ii) Now, {(a - b)}^{2} = a^{2} - 2 a b + b^{2} \Rightarrow {(a - b)}^{2} = a^{2} - a (a + c) + a d [Using (i) and (ii)] \Rightarrow {(a - b)}^{2} = a^{2} - a^{2} - a c + a d \Rightarrow {(a - b)}^{2} = a d - a c \Rightarrow {(a - b)}^{2} = a (d - c) Therefore, a, (a - b) and (d - c) are in G . P .$

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