Question

State whether True or False. The ${31}^{st}$ term of an AP whose ${11}^{th}$ term is 38 and ${16}^{th}$ term is 73 is 178.

• A. 150
• B. 178

Answer 1

The correct option is A

150

We are given that $a_{11}=38$ and $a_{16}=73$ where, $a_{11}$ is the ${11}^{th}$ term and $a_{16}$ is the ${16}^{th}$ term of an AP.

Using formula $a_n=a+(n-1)d$ to find $n^{th}$ term of arithmetic progression, we get
$38=a+(11-1)d$
$73=a+(16-1)d$
$\Rightarrow 38=a+10d$ ....(1)
$\Rightarrow 73=a+15d$ .....(2)
These are equations consisting of two variables. Let’s solve them using substitution method.
We have,
$38=a+10d$
$\Rightarrow$ $a=38-10d$

Let’s put value of a in the equation (2), we get
$73=38-10d+15d$
$\Rightarrow 35=5d$
Therefore Common difference is, d = 7.
Putting value of d in equation (1), we get
$38=a+70$
$\Rightarrow a=-32$
Therefore, common difference is $d=7$ and
First term is $a=-32$
Using formula $a_n=a+(n-1)d$ to find $n^{th}$ term of arithmetic progression, we get
$a_{31}=-32+(31-1)\left(7\right)$
$\Rightarrow$$-32+210=178$
Therefore, ${31}^{st}$ term of AP is 178.