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Help with mats? [Geometrical Progression]?
So I got this Geometrical Progression, right? I know the sum of the first three numbers which is 21 and the sum of the last three numbers which is 168. This only led me to fining out the difference between the first and last numbers. Now what?
1 Answer
- BrianLv 78 years agoFavorite Answer
The nth term of a GP is a(n) = a*r^(n-1), where a is the first term and r is the common
ratio. We are given that the first three terms add to 21 and the last three add to 168.
So a + ar + a*r^2 = 21 and a*r^(n-3) + a*r^(n-2) + a*r^(n-1) = 168.
Thus a*(1 + r + r^2) = 21 and a*r^(n-3) * (1 + r + r^2) = 168.
Therefore r^(n-3) = 168/21 = 8. At this point we have to make an educated guess.
If r = 2 and n = 6 then 2^(6-3) = 2^3 = 8, and so a*(1 + 2 + 2^2) = 21 ----> a = 3.
Thus the GP is 3, 6, 12, 24, 48, 96, the first three terms adding to 21 and the
last three adding to 168 as required.