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, c are in A.P.

$\Rightarrow 2 b = a + c$ ...... (1)

a, b, d are in GP

$\Rightarrow b^{2} = a d$ ....... (2)

Now

$(a - b)^{2} = a (d - c)$ [Using (2)]

$a^{2} - 2 a b = - a c$

(a^2 - 2ab = ab - ac)

$a (a - b) = a (b - c)$

$a - b = a - c$

$2 b = a + c$

$a + c = a + c$ [Using equation (1)]

LHS = RHS

$\Rightarrow a, (a - b), (d - c)$ are in G.P.

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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.

a, b, c are in A.P. ⇒2b=a+c ...... (1) a, b, d are in GP ⇒b2=ad ....... (2) Now (a−b)2=a(d−c) [Using (2)] a2−2ab=−ac (a^2 - 2ab = ab - ac) a(a−b)=a(b−c) a−b=a−c 2b=a+c a+c=a+c [Using equation (1)] LHS = RHS ⇒a,(a−b),(d−c) are in G.P.