Choose an infinite GP from given terms for a particular sum Is it possible to choose an infinite GP from amongst the terms 1, 1/2, 1/4, 1/8, 1/16 ... with a sum a) 1/5? b) 1/7 ? 
My approach was simply choosing terms and probable ratios to match the sum...I would be grateful if you could reveal any trick for this question?

Since all terms are of the form $\frac{1}{2^n}$, any geometric progression will be with ratio $\frac{1}{2^k}$
The summation of a geometric series is:
$$\sum\limits_{k=0}^\infty ar^k = \frac{a}{1-r}$$
The question then is what choice of $a$ (initial term) and what choice of $r$ (ratio) will give the desired results.
Since $a$ and $r$ will both be of the form $\frac{1}{2^n}$ we have the sum must be of the form
$$\frac{\frac{1}{2^m}}{1-\frac{1}{2^n}} =\frac{2^{n-m}}{2^n-1}$$
for some given $n$ and $m$
The question is, can you find non-negative integers $n$ and $m$ such that $\frac{2^{n-m}}{2^n-1}$ is equal to $\frac{1}{5}$?  is equal to $\frac{1}{7}$?  If not, then why?  If yes, which?

For $\frac{1}{5}$, the question becomes whether we can find $a,b$ such that $5\cdot 2^a = 2^b-1$.  Notice that for $a\geq 1$ and $b\geq 1$ the left side is even while the right side is odd.  If $b=0$ the right side is zero and the left side is positive.  If $a=0$, the left side is equal to $5$, but no integer $b$ satisfies $2^b-1=5$
Thus $\frac{1}{5}$ is impossible.
For $\frac{1}{7}$, we ask if we can find $a,b$ such that $7\cdot 2^a=2^b-1$.  From the same logic, we see that they cannot both be positive and $b$ cannot be zero.  This leaves $a=0$ and we find that $b=3$ works.
This corresponds back to our $n$ and $m$ from before implying the geometric progression $\frac{1}{8},\frac{1}{64},\frac{1}{512},\dots$ has sum equal to $\frac{1}{7}$
