# Find the 10th term of the AP -3, -1, 1, 3, ....

$(a)\;10\qquad(b)\;3\qquad(c)\;-3\qquad(d)\;15$

Toolbox:
• In an Arithmetic progression (AP) , $T_{n} = a + (n-1)d$
In an Arithmetic progression (AP), $T_{n} = a + (n-1)d$
Here,  $a = -3, d = 2$, we need to find the 10th term, $n = 10$
$T_{10} = -3 + (10-1) 2 = -3 + 9 X 2 = -3 +18 = 15$