Question

The $4^{th}$ term from the end of an AP is $–11,–8,–5,\dots ..,49$ is
$A.37$
$B.40$
$C.43$
$D.58$

Answer 1

Given A.P is, $–11,–8,–5,\dots ..,49$

$\Rightarrow a=-11,l=49$ and,

$d=t_2-t_1$

$d=-8-(-11)$

$d=3$

We know that, the nth term of an AP from the end is;
$a_n=l-(n-1)d$ ---(1)

$\Rightarrow a_4=49-(4-1)3$

$\Rightarrow a_4=49-9$

$\Rightarrow a_4=40$
Therefore, The $4^{th}$ term from the end is $40$.

Hence, Option $B$ is correct.