If a, b, c are in A.P. and a, b, d are in G.P., then prove that a, a - b, d - c are in G.P.

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Solution

Here a ; b, c are in A.P.
$2 b = a + c$ ........... (i)

and a, b, d are in G.P., so

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

Now,

$(a - b)^{2} = a^{2} + b^{2} - 2 a b$

$= a^{2} + a d - a (a + c)$

Using equation (i) and (ii)

$= a^{2} + a d - a^{2} - a c$

$= a d - a c$

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

$\frac{(a - b)}{a} = \frac{(d - c)}{a - b}$

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

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If a, b, c are in A.P. and a, b, d are in G.P., then prove that a, a - b, d - c are in G.P.

Here a ; b, c are in A.P. 2b=a+c ........... (i) and a, b, d are in G.P., so b2=ad ......... (ii) Now, (a−b)2=a2+b2−2ab =a2+ad−a(a+c) Using equation (i) and (ii) =a2+ad−a2−ac =ad−ac (a−b)2=a(d−c) (a−b)a=(d−c)a−b ⇒a,(a−b),(d−c) are in G.P.