# geometric progression problem

• Oct 21st 2008, 04:30 AM
listeningintently
geometric progression problem
hello :D

here's the question I need a hand with

Find the first three terms of a geometric progression where all terms are positive, S4 (the sum to four terms) = 21 2/3 and the sum to infinity = 27

thanks in advance, all help is much appreciated
• Oct 21st 2008, 05:04 AM
Find the first three terms of a geometric progression where all terms are positive, S4 (the sum to four terms) = 21 2/3 and the sum to infinity = 27

The formulas used for GP
Sn=a(1-(r^n))/(1-r) for n terms...(2)

and for infinite terms
Si=a/(1-r) for 0<r<1.......(1)

a=1st term
r=constant ratio
use it in above
put (1) in (2) for n= 4 to get r and thus a
further use (2) to find the terms reqd (Cool)

• Oct 21st 2008, 05:07 AM
Soroban
Hello, listeningintently!

Quote:

Find the first three terms of a geometric progression
where all terms are positive, $\displaystyle S_4 \:=\:\frac{65}{3}$, and $\displaystyle S_{\infty} = 27$

You're expected to know these formulas:

. . Sum of the first $\displaystyle n$ terms: .$\displaystyle S_n \;=\;a\frac{1-r^n}{1-r}$

. . Sum to infinity: .$\displaystyle S_{\infty} \:=\:\frac{a}{1-r}$

The sum of the first four terms is $\displaystyle \tfrac{65}{3}\!:\quad a\frac{1-r^4}{1-r} \:=\:\frac{65}{3}$ .[1]

The sum to infinity is 27: .$\displaystyle \frac{a}{1-r} \:=\:27 \quad\Rightarrow\quad a \:=\:27(1-r)$ .[2]

Substitute [2] into [1]: .$\displaystyle 27(1-r)\cdot\frac{1-r^4}{1-r} \;=\;\frac{65}{3} \quad\Rightarrow\quad 27(1-r^4) \:=\:\frac{65}{3}$

. . $\displaystyle 1 - r^4 \:=\:\frac{65}{81} \quad\Rightarrow\quad r^4 \:=\:\frac{16}{81}\quad\Rightarrow\quad r \:=\:\sqrt[4]{\frac{16}{81}} \quad\Rightarrow\quad \boxed{r \:=\:\tfrac{2}{3}}$

Substitute into [2]: .$\displaystyle a \:=\:27\left(1-\tfrac{2}{3}\right) \:=\:9 \quad\Rightarrow\quad \boxed{a \:=\:9 }$

Therefore, the first three terms are: .$\displaystyle a, \,ar,\, ar^2 \;=\;9,\:6,\:4$