Question

The ${10}^{th}$ term of an $A.P$ is $52$ and ${16}^{th}$ term is $82$. Find the ${32}^{th}$ term.

Answer 1

Let first term is $a$ and common difference is $d$ then

$T_{10}=a+9d=52$ ......(1)

$T_{16}=a+15d=82$ ......(2)

Subtracting equation(1) from equation(2), we get

$6d=30$

$d=5$

Substituting in value of $d$ in equation(1), we get

$a+45=52$

$a=7$

Now,

$T_{32}=a+31d$

On substituting the values of $a$ and $d$, we get

$T_{32}=7+31\times 5$

$T_{32}=7+155=162$

Hence, ${32}^{th}$ term is $162.$