Question

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 first term and common ratio.

Answer 1

ar^3 + ar^5 = 80
ar^2 * ar^4 = 256
Note that
80 = 16+64 = 2*(2^3+2^5)
256 = 2^8 = 2^2 * 2^6
so it looks like
a = 2, r=2
and the GP is 2,4,8,16,32,64,...

Answer 2

let t equal the 3rd term

t r + t r^3 = t r (1 + r^2) = 80

t * t r^2 = (t r)^2 = 256 ... t r = 16

substituting ... 16 (1 + r^2) = 80 ... 1 + r^2 = 5 ... r = 2 ... t = 8

t = (1st) * r^2 ... 1st = 8 / (2^2) = 2

Answer 3

If oobleck had not noticed that particular sum of the two powers of 2 ...

ar^3 + ar^5 = 80 ---> ar^3( 1 + r^2) = 80
ar^2 * ar^4 = 256 ---> a^2 r^6 = 256 ---- > a r^3 = 16

sub that into the first:
16(1+ r^2) = 80
1+r^2 = 8
r^2 = 4
r = ± 2

If r = 2, a(8) = 16, a = 2
if r + -2, a(-8) = 16, a = -2

for r = 2 and a = 2, the sequence is : 2, 4, 8, 16, 32, 64, ....
for r = -2 and a = -2, the sequence is: -2, 4, -8, 16, -32, 64, ...

Answer 4

Not understandable clearly

Answer 5

i didn`t get the point clearly