Asked by Archiebald
three consecutive terms of a geomentric progression series have product 343 and sum 49/2.
fine the numbers.
HOW WILL ONE SOLVE THAT?
THANKS
fine the numbers.
HOW WILL ONE SOLVE THAT?
THANKS
Answers
Answered by
Reiny
just use your definitions:
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
state your conclusion
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
state your conclusion
Answered by
Obed
How did the 2r come about?
Answered by
Thomas
Is that all