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.

Guest Oct 11, 2019

edited by
Guest
Oct 11, 2019

#1**+1 **

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

CPhill Oct 11, 2019