Answer
$a_{9}= 20$
Work Step by Step
$a_{n}$ of the arithmetic progression is $a_{n} = a_{1} + (n-1)d$
Second term of arithmetic progression is $a_{2} =-1$
$a_{2} = a_{1} + (2-1)d$
$a_{2} = a_{1} + d$
$ a_{1} + d =-1$ Equation $(1)$
Fourth term of arithmetic progression is $a_{4} =5$
$a_{4} = a_{1} + (4-1)d$
$a_{4} = a_{1} + 3d$
$ a_{1} +3d =5$ Equation $(2)$
Subtract Equation $(1)$ from Equation $(2)$
$ a_{1} +3d -(a_{1} + d ) = 5 - (-1)$
$ a_{1} +3d -a_{1} - d = 5 +1 $
$2d=6$
$d=3$
Substituting $d$ value in Equation $(1)$
$ a_{1} + d =-1$
$ a_{1} + 3 =-1$
$ a_{1} =-1-3$
$ a_{1} =-4$
Using $ a_{1} $ , $d$ values and $n=9$ ,
$a_{9} = a_{1} + (9-1)d$
$a_{9} = -4 + (8)3$
$a_{9} =-4+24$
$a_{9} = 20$
