Question

The sum of the last three terms of a G.P having n terms is 1024 times the sum of the first three terms of the progression. If the third term is 5, find the last term.

Answer 1

the third term is 5 ----> a r^2=5
a = 5/r^2

1024(a + ar + ar^2) = ar^(n-3) + ar^(n-2) + ar^(n-1)
divide by a
1024(1 + r + r^2) = r^(n-3) + r^(n-2) + r^(n-1)
1024(1 + r + r^2) = r^(n-3)(1 + r + r^2)
1024 = r^(n-3)

we know: 2^10 = 1024 OR 4^5 = 1024
case1
let r = 2 , then
n-3 = 10
n=13

a = 5/2^2 = 5/4

sequence is 5/4, 5/2, 5, 10, 20, 40, 80, 160, 320, 640, 1280, 2560, 5120

The last term is ar^12 = 5120

check: 1024(5/4 + 5/2 + 5) = 8960
1280+2560+5120 = 8960

case2 , r = 4, n = 8
terms are 5/16, 5/4, 5, 20, 80, 320, 1280, 5120
1024(5/16 + 5/4 + 5) = 6720
320+1280+5120 = 6720

both cases work,
anyway, the last term in each case is 5120

Answer 2

S(n)-S(n-3) = 1024S(3)

(a(r^n-1)-a(r^(n-3)-1)/(r-1) = 1024(r^3-1)/(r-1)

(r^n-1)-(r^(n-3)-1) = 1024(r^3-1)
r^(n-3)(r^3-1) = 1024(r^3-1)
r^(n-3) = 1024 = 2^10
So, let's say r=2 and n=13

Then the sum of the last 3 terms is
2^11+2^12+2^13 = 2^10(2+4+8)

and the sum of the 1st 3 terms is 2+4+8 = 14

But we want the 3rd term to be 5, so a = 5/14

and the 13th term is 5/14 * 2^13 = 5*4096/7

Answer 3

I botched it at the end. I'm sure you can see where.

Good work, Reiny, as usual!