Question

In a certain A.P. the ${32}^{nd}$ term is twice the ${12}^{th}$ term.If ${70}^{th}$ term $=m{31}^{st}$ term.Find $m$

Answer 1

Solution:-

Let us assume,

First term $=a$

Common difference $=d$

Given that ${32}^{nd}$ term is twice the ${12}^{th}$ term.

$a_{32}=2a_{12}$

As we know that,

$a_n=a+(n-1)d$

$\therefore a+31d=2(a+11d)$

$\Rightarrow a+31d=2a+22d$

$\Rightarrow a=9d⟶\left(1\right)$

Now,

Given that $a_{70}=m{a}_{31}$

$m=?$

$\therefore a_{31}=a+30d$

From ${eq}^n\left(1\right)$, we have

$\Rightarrow a_{31}=9d+30d=39d$

Now,

$a_{70}=a+69d$

From ${eq}^n\left(1\right)$, we have

$\Rightarrow a_{70}=9d+69d$

$\Rightarrow a_{70}=78d$

$\Rightarrow a_{70}=2\times 39d$

$\Rightarrow a_{70}=2a_{31}(\because a_{31}=39d)$

$\therefore m=2$

Hence, the correct answer is $2$.