Question

If $3^{rd},7^{th}$ and ${12}^{th}$ terms of an AP are three consecutive terms of GP, then the common ratio of the GP is :

• A. $\frac{5}{4}$
• B. $\frac{9}{4}$
• C. $\frac{2}{9}$
• D. $\frac{1}{2}$
• E. $\frac{12}{7}$

Answer 1

The correct option is A $\frac{5}{4}$

Let first term and common difference of an AP are $a$ and $d.$
Thus $T_3=a+2d$,
$T_7=a+6d$
and $T_{12}=a+11d$
But $T_3,T_7$ and $T_{12}$ are consecutive terms of an GP.
Thus $(a+6d{)}^2=(a+2d\left)\right(a+11d)$
$\Rightarrow a^2+36d+12ad=a^2+13ad+22d^2$
$\Rightarrow ad=14d^2$
$\Rightarrow a=14d$
Therefore, terms are $14d+2d,14d+6d,14d+11d$
i.e., $16d,20d,25d$
Hence, the common ratio is $\frac{20d}{16d}=\frac{5}{4}$.