Question

In geometric sequence tn find:
common ratio r if S2 = 8, S4 = 32

Answer 1

immediately you noticed that 8 and 32 are powers of 2
2 4 8 16 32 64 128 256 ......

Answer 2

or...
stubbornly following the formulas:
ar = 8
ar^3 = 32
divide them
ar^3/(ar) = 32/8
r^2 = 4
r = ± 2
in ar = 8
a = 4
your sequence could be
4, 8, 16, 32, ...
or
4, -8, 16, -32, ...

Answer 3

cool mathhelper :) I never thought of that!

Answer 4

but it's Sum 2 = 8, sum 4 = 32

Answer 5

No, adding d each step is an arithmetic sequence.
You are looking for a GEOMETRIC sequence where each term is a constant , r , TIMES the previous term.
Google math is fun geometric sequence

Answer 6

it goes
term 0 = 2 * 2^0 = 2
term 1 = 2 * 2 = 4
term 2 = 4 * 2 = 8 which is 2*2*2
term 3 = 8 * 2 = 16 which is 2*2*2*2
term 4 = 16 * 2 = 32 which is 2*2*2*2*2

Answer 7

Ok then, should have been more explicit.
if Sum(2) = 8 then a(r^2 - 1)/(r-1) = 8 , where r ≠ 1
if sum(4) = 32, then a(r^4 -1)/(r-1) = 32
again, divide these two equations ...
(r^4 - 1)/(r^2 - 1) = 4
r^4 -1 = 4r^2 - 4
r^4 - 4r^2 + 3 = 0
(r^2 - 3)(x^2 - 1) = 0
r = ± √3 or r = ± 1 , but if r ≠ 1
r = -1, ± √3

checking:
if r = -1
a(r^2 - 1)/(r-1) = 8
a(1-1)/(-2) = 8
no solution for r = -1

if r = + √3
a(r^2 - 1)/(r-1) = 8
a(3-1)/(√3-1) = 8
a = 8(√3-1)/2 = 4(√3-1) , ok
if r = -√3
a(r^2 - 1)/(r-1) = 8
a(3-1)/(-√3-1) = 8
a = 8(-√3-1)/(2) = 4(-√3 - 1) , that works too

r = ± √3

Answer 8

Oh, sorry. I did not understand the question.