Question

Find the 31st term of an A.P. whose 11th term is 38 and 6th term is 73.

Answer 1

Given

$a_{11}=38$

$a+(n-1)d=38$

$a+(11-1)d=38$

$a+10d=\mathrm{38........}\left(1\right)$

$a_6=73$

$a+(n-1)d=38$

$a+(6-1)d=73$

$a+5d=\mathrm{73........}\left(2\right)$

Subtracting (2) from (1), we get

$a+10d-a-5d=38-73$

$5d=-35$

$d=-7$

Putting (1), we get

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

$a-70=38$

$a=108$

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

$=108+(31-1)(-7)$

$=108-210$

$=-102$