Question

If $a_1,a_2,a_3,....$ are in A.P. such that $\frac{a_4}{a_7}=\frac{3}{2}$, then the ${13}^{th}$ of the A.P. is

• A. $\frac{3}{2}$
• B. $0$
• C. $12a_1$
• D. $14a_1$

Answer 1

The correct option is B $0$

It is given that $a_1,a_2,a_3,.....$ are in A.P such that $\frac{a_4}{a_7}=\frac{3}{2}$.

We know that the general term of an arithmetic progression with first term $a$ and common difference $d$ is $T_n=a+(n-1)d$, therefore,

$\frac{a_4}{a_7}=\frac{3}{2}\Rightarrow \frac{a+(4-1)d}{a+(7-1)d}=\frac{3}{2}\Rightarrow \frac{a+3d}{a+6d}=\frac{3}{2}\Rightarrow 2(a+3d)=3(a+6d)$

$\Rightarrow 2a+6d=3a+18d\Rightarrow 3a-2a=6d-18d\Rightarrow a=-12d.........\left(1\right)$

Now, the $13$th term of the A.P is as follows:

$a_{13}=a+(13-1)d=a+12d=-12d+12d=0\left(using\right(1\left)\right)$

Hence, the $13$th term of the A.P. is $0$.