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

Solution:

Given that,

$a_{11}=38$

$a_{16}=73$

We know that,

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

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

$38=a+10 d$ ..(1)

Similarly,

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

$73=a+15 d$

$73=a+15 d$ ...(2)

On subtracting (1) from (2), we obtain

$35=5 \mathrm{~d}$

$d=7$

From equation (1),

$38=a+10 \times(7)$

$38-70=a$

$a=-32$

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

$=-32+30(7)$

$=-32+210$

$=178$

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