Question 572590: find the first 3 terms of a geometric progression whose sum is 42 and whose product is 512
Answer by htmentor(1343) (Show Source):
You can put this solution on YOUR website! find the first 3 terms of a geometric progression whose sum is 42 and whose product is 512
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The nth term of a geometric sequence is
a(n) = ar^(n-1) where a = the first term, r=the common ratio
The 1st 3 terms are a, ar, and ar^2
The sum of these terms is 42:
a + ar + ar^2 = 42 = a(1+r+r^2) [1]
The product is 512:
a*ar*ar^2 = 512 = (ar)^3 [2]
The last equation gives r = 8/a. Substitute this value in [1]:
a(1 + 8/a + 64/a^2) = 42
Simplifying gives a^2 - 34a + 64 = 0
Factor:
(a-32)(a-2) = 0
This gives a=2, a=32
Using the 1st solution, we have r=4
So one such progression satisfying the conditions is a(n) = 2(4)^(n-1),
which gives 2,8,32 as the 1st 3 terms
[If the other solution a=32 were used, the 1st 3 terms would be reversed(r=1/4): 32,8,2]
Either way, the sum and products are the same.
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