Question

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

Answer 1

Answer:

General term of an AP is ,

$\bf{a_n=a+(n-1)d}$

Given  11th term is 38

$\implies a_{11}=38\\\\\implies a+(11-1)d=38\\\\\implies a+10d=38...(1)$

Also  16th term is 73.

$\implies a_{16}=73\\\\\implies a+15d=73...(2)$

Now solve (2) - (1) , we get ,

$(a+15d)-(a+10d)=73-38\\\\\implies 5d=35\\\\\implies d =7$

sub. d value in (2) , we get ,

$\implies a+15(7)=73\\\\\implies a =73-105\\\\\implies a=-32$

Now , 31st term is ,

$a_{31}\\\\\implies a+30d\\\\\implies -32+30(7)\\\\\implies -32+210\\\\\implies \bold{\bf{\red{178}}}$