Can someone help me, this is Geometric series a1+a5=1285 a2*a4=6400
Let r be the common ratio
Let the first term be a, the second term, ar, the fourth term,ar^3 and the fifth term, ar^4
So we have
a + ar^4 = 1285 → ar^4 = 1285 - a (1)
ar * ar^3 = 6400 → a (ar^4) = 6400 (2)
Subbing (1) into (2) we have
a (1285 - a) = 6400
1285a - a^2 = 6400 rearrange as
a^2 - 1285a + 6400 = 0 factor
(a - 1280) (a - 5) = 0
So....setting each factor to 0 and solving for a produces a = 1280 or a = 5
If a = 1280, we have
1280r^4 = 1285 - 1280
1280r^4 = 5
r^4 = 5 /1280 = 1 /256 → r = 1/4
So....one possible series is
1280, 320, 80, 20, 5
And if a = 5, we have
5r^4 = 1285 - 5
5r^4 = 1280
r^4 = 256 → r = 4
And another possible series is
5, 20, 80, 320, 1280
a + ar^4 = 1285, where r = Common ratio.
ar^1 * ar^3 = 6400, solve for a, r
By simple iteration:
a = 5 First term
r = 4 Common ratio. So that:
5, 5*4^1, 5*4^2, 5*4^3, 5*4^4..........etc.
5, 20, 80, 320, 1,280.......etc.
Let r be the common ratio
Let the first term be a, the second term, ar, the fourth term,ar^3 and the fifth term, ar^4
So we have
a + ar^4 = 1285 → ar^4 = 1285 - a (1)
ar * ar^3 = 6400 → a (ar^4) = 6400 (2)
Subbing (1) into (2) we have
a (1285 - a) = 6400
1285a - a^2 = 6400 rearrange as
a^2 - 1285a + 6400 = 0 factor
(a - 1280) (a - 5) = 0
So....setting each factor to 0 and solving for a produces a = 1280 or a = 5
If a = 1280, we have
1280r^4 = 1285 - 1280
1280r^4 = 5
r^4 = 5 /1280 = 1 /256 → r = 1/4
So....one possible series is
1280, 320, 80, 20, 5
And if a = 5, we have
5r^4 = 1285 - 5
5r^4 = 1280
r^4 = 256 → r = 4
And another possible series is
5, 20, 80, 320, 1280