Find the number of term in the Arithmetic Progression(A.P) 2, -9, -20,...-141.

A.

11

B.

12

C.

13

D.

14

Correct answer is D

T1, T2, T3

2, -9, -20 .... -141

l = a + (n - 1)d

first term, a = 2

common difference d = T3 - T2

= T2 - T1

= -20 - (-9) = -9 -2

= -20 + 9

= -9 -2

= -20 + 9

= -11

-11 = -11

d = -1

last term l = -141

-141 = 2 + (\(\cap\) - 1) (-11)

-141 = 2 + (-11 \(\cap\) + 11)

= 2 - 11\(\cap\) + 11

-141 = 13 - 11\(\cap\)

-141 - 13 = -11\(\cap\)

-154 = -11\(\cap\)

\(\cap\) = \(\frac{-154}{-11}\)

\(\cap\) = 14