Question 254675: Three numbers form an arithmetic sequence, common difference being 11. If first number is decreased by 6, second number decreased by 1, and third number doubled, resulting numbers form a geometric sequence. determine the numbers that form the arithmetic sequence
Answer is -26, -15, -4, 14, 25, 36
Answer by drk(1908)   (Show Source):
You can put this solution on YOUR website! Let the first term be A
Let the second term be A + 11
Let the third term be A + 22
we know that common difference, d, is 11.
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A - 6, A + 10, 2(A+22)
is now a geometric sequence. They have common ratios which can be expressed as
(A+10)/(A-6) = (2A+44)/(A+10)
Cross multiplying, we get
(A+10)^2 = (2A+44)(A-6)
A^2 + 20A + 100 = 2A^2 + 32A -264
set = 0 as
A^2 + 12A -364 = 0
(A - 14)(a + 26) = 0
A = 14 or A = -26
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If A = 14, then the arithmetic sequence is
14, 25, 36
and the geometric sequence is
8, 24, 72
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If A = -26, then the arithmetic sequence is
-26, -15, -4
and the geometric sequence is
-32, -16, -8
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