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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.

 Oct 11, 2019
edited by Guest  Oct 11, 2019
 #1
avatar+104962 
+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

 

 

cool cool cool

 Oct 11, 2019
 #2
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My apologies, but what happens to the (1/8) during, (1/8)n^2 + (1/8)n - 8n  =  0, before becoming, n^2 + n - 64n  = 0.

Guest Oct 12, 2019
 #3
avatar+104962 
+1

I just multiplied through by  8  on both sides.........this eliminates those pesky fractions !!!

 

 

 

 

cool cool cool

CPhill  Oct 12, 2019

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