Question

Find the ${31}^{st}$ term of an AP whose ${11}^{th}$ term is $38$ and the ${16}^{th}$ term is $73$.

Answer 1

Given that, $a_{11}=38$ $a_{16}=73$

We know that,

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

$a_{11}=a+(11-1)d$

$38=a+10d$ ... (i)

Similarly,

$a_{16}=a+(16-1)d$

$73=a+15d$ ... (ii)

On subtracting (i) from (ii), we get

$35=5d$

$\therefore d=7$

From equation (i),

$38=a+\left(10\right)\left(7\right)$

$\Rightarrow 38-70=a$

$\therefore a=-32$

Now $a_{31}=a+(31-1)d$

$=-32+30\left(7\right)$

$=-32+210$

$=178$

Hence, ${31}^{st}$ term is $178.$