In an arithmetic sequence, the first term is 8 and the common difference is 1/4. If the sum of the first 2n terms is equal to the sum of the next n terms, find n.
+128475 The (2n)th term is 8 + (1/4)(2n - 1)
The sum of the first 2n terms =
(2n/2) ( 8 + 8 + (1/4)(2n - 1)]
(n) [16 + (1/4)(2n - 1)] (1)
And the sum of the next n terms after this =
Sum of first (2n + n) terms - Sum of first (2n) terms =
(3/2)[n] [ 16 + (1/4)(3n - 1)] - (n) [16 + (1/4)(2n - 1)] (2)
Set (1) = (2) and we have that
(n) [16 + (1/4)(2n - 1)] = (3/2)[n] [ 16 + (1/4)(3n - 1)] - (n) [16 + (1/4)(2n - 1)] simplify
(2n)[ 16 + (1/4)(2n - 1)] = (3/2)[n][ 16 + (1/4)(3n - 1)]
32n + (1/2)[n] (2n - 1) = 24n + (3/8)[n] (3n - 1)
32n + n^2 - (1/2)n = 24n + (9/8)n^2 - (3/8)n
(1/8)n^2 + (1/8)n - 8n = 0
n^2 + n - 64n = 0
n^2 - 63n = 0
n(n - 63) = 0
Set each factor to 0 and solve for n
n = 0 reject
n = 63
