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# Help

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Can someone help me, this is Geometric series a1+a5=1285 a2*a4=6400

Aug 22, 2017
edited by Guest  Aug 22, 2017

#2
+100516
+1

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

Aug 23, 2017

#1
0

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.

Aug 22, 2017
#2
+100516
+1

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

CPhill Aug 23, 2017
#3
0

Thanks

Guest Aug 23, 2017