Problem 11.1. The sum of the first $n$ terms of an arithmetic progression with first term $m$ and difference 2 is equal to the sum of the first $n$ terms of a geometric progression with first term $n$ and ratio 2 .

a) Prove that $m+n=2^{m}$;

b) Find $m$ and $n$, if the third term of the geometric progression is equal to the 23 -rd term of the arithmetic progression.

Emil Kolev
11.1. a) Using the formulas for the sums of arithmetic and geometric progressions we obtain the equality

$$
\frac{n[2 m+2(n-1)]}{2}=n\left(2^{m}-1\right)
$$

whence $m+n=2^{m}$.

b) It follows that $4 n=m+44$. Using a), we obtain $2^{m+2}=44+5 m$. It is easy to see that $m=4$ is a solution. If $m4$ then it follows by induction that $2^{m+2}>44+5 m$. Therefore $m=4$ and $n=12$.