If second, third and sixth terms of an A.P. are consecutive terms of a G.P.,

Solution:

Here, second term, $a_{2}=a+d$

Third term, $a_{3}=a+2 d$

Sixth term, $a_{6}=a+5 d$

As, $a_{2}, a_{3}$ and $a_{6}$ are in G.P.

$\therefore$ First term of G.P. $=a_{2}=A=a+d$

Second term of G.P. $=A r=a+2 d$

Third term of G.P. $=A r^{2}=a+5 d$

$\therefore(a+2 d)^{2}=(a+d) \times(a+5 d)$

$\Rightarrow a^{2}+4 a d+4 d^{2}=a^{2}+6 a d+5 d^{2}$

$\Rightarrow 2 a d+d^{2}=0$

$\Rightarrow d(2 a+d)=0$

$\Rightarrow d=0$ or $2 a+d=0$

But, $d=0$ is not possible.

$\therefore d=-2 a$

$\therefore r=\frac{\mathrm{a}+2 \mathrm{~d}}{\mathrm{a}+\mathrm{d}}$

$\Rightarrow r=\frac{a+2(-2 a)}{a+(-2 a)}$

$\Rightarrow r=\frac{3}{1}=3$