In an infinite geometric progression each term is equal to twice the sum of all the terms that follow it. Find its sum given its first two terms.

Question

In an infinite geometric progression each term is equal to twice the sum of all the terms that follow it.

If the sum of first two terms is 12 what is the sum of entire progression?

I.e., given that $a+ar=12,$ how to approach the question further?

Answer 1

The entire progression is $\frac{a}{1-r}$, as we know. Now $a(1+r)=12$, so that $$ \frac{a}{1-r}=\frac{a(1+r)}{1-r^2}=\frac{12}{1-r^2}. $$ Now use that each term is twice the sum of all the terms that follow it, to conclude that $r=\frac{1}{3}$. Hence the entire progression is $$ \frac{12}{1-\frac{1}{9}}=\frac{27}{2}. $$

Answer 2

HINT Your progression is basically $$ a + ar + ar^2 + \ldots = \frac{a}{1-r} $$ so if the first term is twice the sum of the rest, you get $$ a/2 = ar + ar^2 + \ldots = r \left(a + ar + ar^2 + \ldots\right) = \frac{ar}{1-r} $$ which gives you the second equation. Cancel $a$ to get $$ \frac{1}{2} = \frac{r}{1-r}. $$

Answer 3

Note that $a=2(ar+ar^2+\dots)$ and $ar=2(ar^2+ar^3+\dots)$ from the given condition applied to $a$ and $ar$

So we get $a-2ar=ar$ so that $a=3ar$ so $a=9, ar=3$ and the rest of the progression sums to $ar^2+ar^3+\dots =\frac {ar}2=\frac 32$