SOLUTION: in an arithmetic series t10 is 7 times the value of t3. The sum of the first 30 terms is 2460. determine the value of t6.

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Question 1161060: in an arithmetic series t10 is 7 times the value of t3. The sum of the first 30 terms is 2460. determine the value of t6.
Answer by greenestamps(13200) About Me  (Show Source):
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Let the first term be a(1) and the common difference be d.

(1) a(10) is 7 times a(3):

a%5B1%5D%2B9d+=+7%28a%5B1%5D%2B2d%29
6a%5B1%5D%2B5d+=+0

(2) The sum of the first 30 terms is 2460:

30%28a%5B1%5D%2Ba%5B30%5D%29%2F2+=+2460
30%28%28a%5B1%5D%29%2B%28a%5B1%5D%2B29d%29%29+=+4920
2a%5B1%5D%2B29d+=+164

Solve the pair of equations from (1) and (2):

6a%5B1%5D%2B87d+=+492
6a%5B1%5D%2B5d+=+0
82d+=+492
d+=+6

The common difference d is 6; use it to find a(1):

6a%5B1%5D%2B5%286%29+=+0
a%5B1%5D+=+-30%2F6+=+-5

The first term is -5; the common difference is 6.

The 6th term is the first term, plus the common difference 5 times:

t(6) = -5+5(6) = -5+30 = 25