Question

If $t_{11}$ and $t_{16}$ for an $A.P.$ are respectively $38$ and $73$, then $t_{31}$ is $........$

• A. $178$
• B. $177$
• C. $176$
• D. $175$

Answer 1

The correct option is A $178$

Given:
$t_{11}=38,t_{16}=73$

$n^{th}$ term of an A.P. is given as,

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

According to the given condition,

$\Rightarrow a+10d=38$ $...\left(1\right)$

$\Rightarrow a+15d=73$ $...\left(2\right)$

where $a$ is the first term and $d$ is the common difference of given AP.
On subtracting $\left(1\right)$ from$\left(2\right)$, we get,

$5d=35$

$\Rightarrow d=7$

substituting the value of $d$ in $\left(1\right)$, we get

$a+10\times 7=38\Rightarrow a+70=38$

$\Rightarrow a=-32$

Therefore,

$t_{31}=a+30d=-32+30\times 7=-32+210=178$
Hence, option $\left(A\right)$ is correct.