infinite GP series?
In a certain infinite geometric progression , the sum of any three consecutive terms equals 7 times the sum of all the terms that follow them. Find the ratio of any term of the progression to the sum of all the terms that follow it.
(1)8:1
(2)4:1
(3)3:1
(4)2:1
(5)1:1
any shortcuts here ? back solving ? tricks ?
How do i choose the correct answer ?
i know sum of infinite GP = a/(1-r) but thats not useful here.
someone please advice
thanks
Comments
Let the G.P. be a + ar + ar^2 + ar^3 + ...
a + ar + ar^2 = 7ar^3/(1-r)
1 + r + r^2 = 7r^3/(1-r)
(1-r)(1+r+r^2) = 7r^3
1 - r^3 = 7r^3
1 - 8r^3 = 0
8r^3 = 1
r^3 = 1/8
r = 1/2
The ratio of successive terms is 1:1/2 or 2:1
Edit:
Sum of terms following ar^n will be:
ar^(n+1)/(1-r)
= a*2^(n+1)/(1-2)
= a*2^n
The ratio of any term and the sum of successive terms will be 1:1.
let's take any term of the GP as ar^(n-1).
the sum of all the terms after it is ar^n/ 1-r
we need to find their ratio:
ar^n-1 : ar^n/1-r
=r^n-1: r^n/1-r
=1: r/1-r
=1-r : r
=1:2r...........(1)
now, we need to find r.
given that
a + ar + ar^2 = 7* ar^3/1-r
a( 1+r+r^2) = 7ar^3/1-r
1+r+r^2 = 7r^3/1-r
1-r(1+r+r^2) = 7r^3
1 + r + r^2 - r - r^2 - r^3 = 7r^3
1- r^3= 7r^3
1= 8r^3
r^3= 1/8
r= 1/2.
substituting this in (1), we have
1: 2(1/2)
= 1:1.
therefore, the ratio between any term and the sum of all the terms following it is 1:1.
The sum to infinity is useful, try considering the sum of the first three terms compared to seven times (the sum to infinity minus the first three terms).
Beyond that, you're on your own with your homework. What? Did you really think I was going to do it all for you?
1/(1-(-1/2)) =2/3