Question 1007853: find three consecutive term of a G.P whose sum is 21 and whose product is 64
Answer by fractalier(6550) (Show Source):
You can put this solution on YOUR website! Call the first term a and the ratio r. Then our three terms are
a, ar and ar^2. The facts in the problem state that
a + ar + ar^2 = 21 and
a(ar)(ar^2) = 64
a^3*r^3 = 64
(ar)^3 = 64
ar = 4 (or r = 4/a)
so that
a + 4 + ar^2 = 21
a + ar^2 = 17
a + a(4/a)^2 = 17
a + 16/a = 17
a^2 + 16 = 17a
a^2 - 17a + 16 = 0
(a - 16)(a - 1) = 0
a = 16 or a = 1 so that
r = 1/4 or r = 4
and our three terms are
1, 4, 16 or 16, 4, 1
|
|
|