Question

The sum of terms of an A. P is 136,the common difference 4,anf last term 31 find n?

Answer 1

Answer:

8

Step-by-step explanation:

Given,

Sum of 'n' terms of an A.P(S_n) = 136

Common difference(d) = 4

Last term(l) = 31

To Find :-

value of 'n'

How To Do :-

As they given the value of last term(l) we need to substitute in the formula of 'n' term of an A.P then we will get one equation with first term(a). We need to substitute the value of 'a' in the formula of sum of 'n' terms of an A.P after by simplifying we will get the value of 'n'.

Formula Required :-

Sum of 'n' terms of an A.P :-

$S_n=\dfrac{n}{2}[2a+(n-1)d]$

'n' term of an A.P :-

$a_n=a+(n-1)d$

Solution :-

Last term = 31

→ 31 = a + (n - 1)4

31 = a + 4n - 4

a = 31 - 4n + 4

a = 35 - 4n

[ Let it be equation - 1]

$136=\dfrac{n}{2}[2a+(n-1)4]$

[∴ S_n = n/2[2a + (n - 1)d]

Substituting value of 'a'

$136=\dfrac{n}{2}[2(35-4n)+4n-4]$

136 × 2 = n[70 - 8n + 4n - 4]

272 = n [66 - 4n]

272 = 66n - 4n²

4n² - 66n + 272 = 0

2( 2n² - 33n + 136 ) = 0

2n² - 33n + 136 = 0

2n² - 16n - 17n +  136 = 0

2n(n - 8) - 17(n - 8) = 0

(n - 8) (2n - 17) = 0

→ n = 8 , 17/2

∴ n = 8 because the value of 'n' can'nt be fractional.