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

Given:

The 11th term of an AP is 38 and the 16th term is 73.

To do:

We have to find the 31st term of this AP.

Solution:

Let the first term of the A.P. be $a$ and the common difference be $d$.

We know that,

nth term of an A.P. $a_n=a+(n-1)d$

Therefore,

$a_{11}=a+(11-1)d=38$

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

$a_{16}=a+(16-1)d=73$

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

Subtracting (i) from (ii), we get,

$a+15d-a-10d=73-38$

$5d=35$

$d=\frac{35}{5}$

$d=7$

This implies,

$a+10d=38$

$a+10(7)=38$

$a=38-70$

$a=-32$

Therefore,

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

$=a+30d$

$=-32+30(7)$

$=210-32$

$=178$

The 31st term of the AP is $178$.

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