Question

The $4^{th}$ term from the end of the AP: $11,8,5,...,-49$ is

• A. $-37$
• B. $-40$
• C. $-43$
• D. $-58$

Answer 1

Answer: (B)

$-40$

In the given A.P. $a=11,d=-3$, last term $\left(l\right)=-49$,
then $l=a+(n-1)d$
$\Rightarrow -49=11+(n-1)(-3)$
$\Rightarrow 3n=63\Rightarrow n=21$
Hence, there are $21$ terms in AP.
The $4^{th}$ term from the last will be 18th term from beginning.
So, $a_{18}=a+(18-1)d$
$\Rightarrow 11+17(-3)$
$\Rightarrow 11-51=-40$
SO $4^{th}$ term from the end of the AP: $11,8,5,...,-49$ is $-40$.