hello
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
hello
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
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
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You're expected to know these formulas:Find the first three terms of a geometric progression
where all terms are positive, $\displaystyle S_4 \:=\:\frac{65}{3}$, and $\displaystyle S_{\infty} = 27$
. . 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: .$\displaystyle a, \,ar,\, ar^2 \;=\;9,\:6,\:4$