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

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 .
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