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Sum of nth terms in A.P. is 136, common difference 4 and last term is 31. Find the value of n.
1. 7
2. 8
3. 9
4. 10

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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

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