The sum of 4th and 6th terms of a geometric series is 80.If the product of the 3rd and 5th term is 256, determine the first term and common ratio

Answer 1

The sum of 4th and 6th terms of a geometric series is 80
---> ar^3 + ar^5 = 80
ar^3(1 + r^2) = 80 or ar^3 = 80/(1+r^2)

the product of the 3rd and 5th term is 256
----> (ar^2)(ar^4) = 256
a^2 r^6 = 256

divide:
ar^3(1 + r^2)/(a^2 r^6) = 80/256
(1+r^2)/ar^3 = 5/16 , but ar^3 = 80/(1+r^2)

(1+ r^2)^2 / 80 = 5/16
(1 + r^2)^2 = 400/16 = 25
1+r^2 = ± 5

r^2 = 4 or r^2 = -6 , the last part is not real

r^2 = 4
r = ± 2

if r = ± 2, sub into a^2 r^6 = 256
64a^2 = 256
a^2 = 4
a = ± 2

so our sequence could be
2, 4, 8, 16, 32, 64, 128, ....
or
2, -4, 8, -16, 32, -64, 128, ..., but that does not satisfy the first condition
or
-2, -4, -8, -16, -32, -64, ... , but that does not satisfy the first condition
or
-2, 4, -8, 16, -32, 64, 128

so a = 2, r = 2
or
a = -2, r = -2