Question

Find the ${12}^{th}$ term from the end of $A.P.-2,-4,-6,....,-100$.

Answer 1

$-2,-4,-6,..........-100$

this is an A.P with $a=-2,d=-2$

now ${12}^{th}$ term from end is

$a_{12}=a+\left(n-1\right)d$

where $a=-100,d=2$

$a_{12}=-100+\left(12-1\right)2$

$a_{12}=-100+22$

$a_{12}=-78$