Answer: 5th term.
Explanation:
The sum of nth term of Arithmetic Progression
Sn = n/2{ (2a + (n - 1)d , where a = first term, d is the common difference of the progression. From the question, Sn = 95, a = 27
But to find d, always subtract the second term from the first term.
So from here
d = 23 - 27
= -4
Now , solving the question , we have
95 = n/2 { (2(27) + (n - 1) x (-4)
95 = n/2 { 54 -4n + 4 }
Multiply both side by 2 to make it a linear expression
190 = n { 58 - 4n }
Open the bracket by multiplying by n .
190 = 58n - 4n², now, rearrange the expression to give a quadratic expression/ equation.
4n² - 58n + 190 = 0
Using any method to solve
4n² - 38n - 20n + 190 = 0
By factor
2n(2n - 19) - 10(2n - 19) = 0
(2n - 19) is common
(2n - 19)(2n - 10) = 0
solving for n, the root
n = 19/2 or 5
Therefore, n could = 19/2 or 5.
Now check the roots or results to see the one the gives ,95.
Using 5
5/2( 2 x27 + 4(-4)}
5/2 (54 -16)
5/2 x 38
= 190/2
= 95.
So 19/2 is term extraneous.
Therefore, the 5th term of that progression will give 95.