Question

Find the ${31}^{st}$ term of an A. P. whose ${10}^{th}$ term is $38$ and ${16}^{th}$ term is $74$.

Answer 1

Let first term of AP is $a$ and common difference is $d$.
Use formula of $t_n=a+(n-1)d$.......(i)
Given, ${10}^{th}$ term is $38$ and ${16}^{th}$ term is $74$.
So, substitute $n=10$ and $16$ in equation(i), we get
$t_{10}=a+(10-1)d=38$ and $t_{16}=a+(16-1)d=74$
$\Rightarrow a+9d=38$ and $a+15d=74$
Substitute $a=38-9d$ in $a+15d=74$, we get
$\Rightarrow 38-9d+15d=74$
$\Rightarrow 38+6d=74$
$\Rightarrow 6d=74-38$
$\Rightarrow d=\frac{36}{6}$
$\therefore d=6$
Substitute $d=6$ in $a=38-9d$, we get
$\Rightarrow a=38-9\left(6\right)$
$\Rightarrow a=38-54$
$\Rightarrow a=-16$
Thus, ${31}^{th}$ term is $t_{31}=a+30d$
Substitute, $a=-16$ and $d=6$ in above expression.
$t_{31}=-16+30\left(6\right)$
$=-16+180$
$\therefore t_{31}=164$