Question

If $2^{nd},3^{rd}$ and $6^{th}$ of an $A.P.$ are three consecutive terms of a $G.P.$, then common ratio of the $G.P.$ is

• A. $\frac{5}{4}$
• B. $\frac{9}{4}$
• C. $\frac{2}{9}$
• D. $\frac{1}{2}$

Answer 1

Answer: (A)

$\frac{5}{4}$

$a-2d,a-d,a,a+d,a+2d,a+3d$ are 6 terms of arithmetic progression

so $\frac{a}{(a-d)}=\frac{(a+3d)}{a}=r$

so $a²=(a+3d)(a-d)$ i.e. $2ad-3d²=0$

so $d=0$ or $d=2a/3$

in the first case $d=0$

ratio is $r=\frac{a}{(a-0)}=1$

in the second case $d=2a$

so $r=\frac{a}{(a-2a/3)}=3$

so ratio is $r=3$