The sum of n terms of an A.P. is given by (5n2 + 7n + 4)/4. What will ...
Let sum of 4 terms and sum of 5 terms be S4 and S5 respectively.
S5 = [5(5)2 + 7(5) + 4]/4 = 41
S4 = [5(4)2 + 7(4) + 4]/4 = 28
5th term of an A.P. = S5 - S4 = 13
Hence, option 2.
The sum of n terms of an A.P. is given by (5n2 + 7n + 4)/4. What will ...
To find the fifth term of an arithmetic progression (A.P.), we need to use the formula for the sum of the first n terms of an A.P.:
Sn = (n/2)(2a + (n-1)d)
Where Sn is the sum of the first n terms, a is the first term, and d is the common difference.
Given that the sum of the first n terms is (5n^2 + 7n + 4)/4, we can equate this expression to Sn:
(5n^2 + 7n + 4)/4 = (n/2)(2a + (n-1)d)
Simplifying this equation, we get:
5n^2 + 7n + 4 = n(2a + (n-1)d)
Expanding the right side, we have:
5n^2 + 7n + 4 = 2an + nd - d
Rearranging the terms, we get:
5n^2 + (7 - 2a)n + (4 + d) = 0
This is a quadratic equation in terms of n. Since it is given that this equation is true for all n, the coefficient of n^2 must be zero. Therefore, we have:
5 = 0
This is not possible, so there must be an error in the given equation for the sum of the first n terms. Without the correct equation, we cannot determine the fifth term of the A.P.