Question

The ${17}^{th}$ term from the end of the A.P. $-36,-31,-26,...,79$ is

• A. $9$
• B. $4$
• C. $-6$
• D. $-1$

Answer 1

The correct option is D $-1$

Here,
First term $a=-36$
Common difference $d=-31+36=5$
Last term $l=79$

Hence ${17}^{th}$ term from the end
$=l-(n-1)d=79-(17-1)5=-1$

Alternate Solution:
$-36,-31,-26,...,79$ is in A.P.
$\therefore 79,74,...,-26,-31,-36$ will be also in A.P. with first term $79$ and common difference $-5.$
$\therefore T_{17}=79+(17-1)(-5)=-1$