Question

Find ${20}^{th}$ term from the end of an A.P $3,7,11,..............407$.

Answer 1

Given AP is $3,7,11,........407$

We know $t_n=a+(n-1)d$
$\Rightarrow 407=3+(n-1)4$
$\Rightarrow n=102$
Number of terms $=102$
If we start from last, then first term $=407$
Common difference $=-4$
${20}^{th}$ term $=407+(20-1)(-4)$
$=407-76$

$=331$