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 = -a2 + 4a + 1085
a2 - 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
