Infinite GP

Algebra Level 3

If each term of an infinite geometric progression is thrice the sum of the terms following it, then what is the common ratio of the geometric progression?

The answer is 0.25.

2 solutions

Chew-Seong Cheong
Aug 15, 2015

Let the GP be:

$\begin{aligned} S & = a + ar + ar^2 + ... \\ & = a + r(a + ar + ar^2 + ...) \\ & = \color{#3D99F6}{a} + rS \\ & = \color{#3D99F6}{3rS} + rS \quad \quad \small \color{#3D99F6} {[\text{Given}]} \\ \Rightarrow S & = 4rS \\ \Rightarrow r & = \frac{1}{4} = \boxed{0.25} \end{aligned}$

short, simple and elegant solution. upvoted BTW it is 0.25 not 0.24

Ravneet Singh - 5 years, 10 months ago

Thanks for upvoting and spotting the error. Amended.

Chew-Seong Cheong - 5 years, 10 months ago

Similar solution:

Given,

$a_{n} = 3(a_{n+1} + a_{n+2} + a_{n+3} + ...)$ $\Rightarrow a_{n} = 3(\frac{a_{n+1}}{1-r})$ $\Rightarrow a_{n} = 3(\frac{a_{n}r}{1-r})$ $\Rightarrow 1 = \frac{3r}{1-r}$ $\Rightarrow r = \frac{1}{4} = 0.25$

Paras Lehana - 3 years, 8 months ago

Brian Charlesworth
Jul 19, 2015

Note first that since each term of the G.P. is equal to thrice the sum of the terms following it, the ratio $r$ of the progression must be such that $|r| \le 1.$

So if the $n$ th term of the G.P. is $ar^{n-1}$ for some constant $a$ then we have that, for any $n \ge 1,$

$ar^{n-1} = 3 \times \displaystyle\sum_{k=n}^{\infty} ar^{k}$

$\Longrightarrow 1 = 3 \times \displaystyle\sum_{k=1}^{\infty} r^{k} = 3 \times \dfrac{r}{1 - r} \Longrightarrow 1 - r = 3r \Longrightarrow r = \dfrac{1}{4} = \boxed{0.25}.$

hey Brian nice solution. How were you able to make the assumption in the first two lines?

thanks in advance

guido barta - 5 years, 5 months ago

It's been a while since I wrote this solution, but I think that my intent was to simply establish that the G.P. as described did indeed converge. To this end, since we can assume that each term is finite, and since the sum of all the terms other than the first is one-third that of the first term, we know that the sum of all the terms is finite, i.e., the infinite series converges, (which in turn implies that $|r| \lt 1$ since these two conditions are equivalent).

Brian Charlesworth - 5 years, 5 months ago

Infinite GP

The answer is 0.25.

2 solutions

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