Correct Answer - Option 2 : 8

**Concept:**

The **sum of n numbers in the A.P**. is

\(\rm S = {n\over2}[2a+(n-1)d]\)

\(\rm S = {n\over2}[a+l]\)

The **nth term of the A.P** = a + (n - 1)d

Where a, l and d is the first, last number and common difference of the A.P. respectively

__Calculation:__

Let the first term of A.P. = a, d = 4

Given Last term = a + 4(n - 1) = 31

4(n - 1) = 31 - a

n - 1 = \(\rm 31 - a\over4\)

n = \(\rm 31 - a\over4\) + 1

n = \(\rm 35 - a\over4\)

Given S = 136

\(\rm 136 = {{35-a\over4}\over2}[a+31]\)

1088 = (35 - a)(a + 31)

1088 = -a^{2} + 4a + 1085

a^{2} - 4a + 3 = 0

(a - 1)(a - 3) = 0

a = 1 or 3

a cannot be equal to 1 as n will be in fraction, which is not possible.

∴ a = 3 and n = \(\rm 35 - 3\over4\) = **8**