geometric progression problem

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October 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
October 21st 2008, 05:04 AM
ADARSH

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)
October 21st 2008, 05:07 AM
Soroban

Hello, listeningintently!

Quote:

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

You're expected to know these formulas:

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

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

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

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

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

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

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

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