If a, b, c are in AP, and a, b, d are in GP, show that a,

Solution:

To prove: $a,(a-b)$ and $(d-c)$ are in GP.

Given: a, b, c are in AP, and a, b, d are in GP

Proof: As a,b,d are in GP then

$\mathrm{b}^{2}=\mathrm{ad} \ldots$ (i)

As a, b, c are in AP

2b = (a + c) … (ii)

Considering a, (a – b) and (d – c)

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

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

From eqn. (i) and (ii)

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

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

$=a d-a c$

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

From the above equation we can say that a, (a – b) and (d – c) are in GP.