Question

In an A.P., $6th$ term =$11th$ times of $11th$ term, its $17th$ term is?

• A. $1$
• B. $-1$
• C. $5$
• D. $Noneofthese$

Answer 1

The correct option is D $Noneofthese$

$11\left(a_{11}\right)=17\left(a_{17}\right)$

$11(a+10d)=17(a+16d)$

$11a+110d=17a+272d$

$11a-170=272d-110d$

$6a=162d$

$a=\frac{162}{-6}=-27d$

$a_{28}=a+(28-1)d$

$-27+27=0$