Answer
$a_{10} = -19$
Work Step by Step
$a_{n}$ of the arithmetic sequence is $a_{n} = a_{1} + (n-1)d$
Given $a_{3} =2$
$a_{17} = -40$
$a_{3} =a_{1} + (3-1)d$
$a_{3} =a_{1} + 2d$
$a_{1} + 2d = 2$ Equation $(1)$
Similarly,
$a_{17} =a_{1} + (17-1)d$
$a_{17} =a_{1} + 16d$
$a_{1} + 16d= -40$ Equation $(2)$
Subtract Equation $(1)$ from Equation $(2)$
$a_{1} + 16d - (a_{1} + 2d) = -40 -2 $
$a_{1} + 16d - a_{1} - 2d = -42 $
$14d = -42$
$d = -3$
Substituting $d$ value in Equation $(1)$
$a_{1} + 2d = 2$
$a_{1} + 2(-3) = 2$
$a_{1} -6 = 2$
$a_{1} = 8$
Using $ a_{1} $ , $d$ values and $n=10$ ,
$a_{10} =a_{1} + (10-1)d$
$a_{10} =a_{1} + 9d$
$a_{10} =8 +9(-3)$
$a_{10} =8 -27$
$a_{10} =-19$
