A geometric series has first term a and common ratio r where |r| > 1. The sum of the first n terms of the series is denoted by Sn.
Given that S4 = 10 x S2.
a) Find the value of r.
I've tried finding the formula for the sum of the first four terms and equating that to ten times the sum of the first two terms so im then left with two unknowns (r and a) and i thought that i'd be able to solve it, but i cant because of the powers on some of the r's.
Could someone please explain how this can be done?
As in the formula for geometric series is = a(r^n -1) over (r-1) so there are going to be r's to the power of 4 and 2, cos thats what n is, or am i totally wrong?
And to DeanK2 i double checked and on the sheet it says |r| > 1
As in the formula for geometric series is = a(r^n -1) over (r-1) so there are going to be r's to the power of 4 and 2, cos thats what n is, or am i totally wrong?
And to DeanK2 i double checked and on the sheet it says |r| > 1
Here's what I got.
a-ar^4 = 10(a - ar^2) (Because S4 = 10S2 and the "1-r"s cancel) 1 - r^4 = 10 - 10r^2 (The a terms cancel) r^4 - 10r^2 = -9 (Rearrange) Let b=r^2 b^2-10b+9=0 (Substitute in "b" and rearrange) Therefore, b=1 and b=9 r=1, r=3, r=-1, r=-3 r=3 because r=1, r=-1, r=-3 are not within the range.
Yeah i got that far too and then cross multiplied the (r-1) , but my teacher said that if |r| > 1 then to use the other formula a(r^n -1) over (r-1). so then i was left with ar^4 - ar + a = 10ar^3 - 10ar + 10a. but i cant see how to work it out from that
a-ar^4 = 10(a - ar^2) (Because S4 = 10S2 and the "1-r"s cancel) 1 - r^4 = 10 - 10r^2 (The a terms cancel) r^4 - 10r^2 = -9 (Rearrange) Let b=r^2 b^2-10b+9=0 (Substitute in "b" and rearrange) Therefore, b=1 and b=9 r=1, r=3, r=-1, r=-3 r=3 because r=1, r=-1, r=-3 are not within the range.
ah, ok thanks very much, i can see how that would work. doest that mean that wether or not |r| </> 1 you still use the formula a(1-r^n) over (1-r)?