# MATH

three consecutive terms of a geomentric progression series have product 343 and sum 49/2.
fine the numbers.
HOW WILL ONE SOLVE THAT?
THANKS

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the terms would be a, ar, and ar^2

so a(ar)(ar^2) = 343
a^3 r^3 = 343
(ar)^3 = 343
ar = 7 or a = 7/r

a + ar + ar^2 = 49/2
a(1 + r + r^2) = 49/2
(7/r)(1 + r + r^2) = 49/2
7/r + 7 + 7r = 49/2
times 2r
14 + 14r + 14r^2 = 49r
14r^2 - 35r + 14 = 0
2r^2 - 5r + 2 = 0
(2r - 1)(r - 2) = 0
r = 1/2 or r = 2

if r = 1/2, a = 7/(1/2) = 14
if r = 2, a = 7/2

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