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Problem 1365

AIME late

Algebra Difficulty 5.6 Prove it Bulgarian Mathematical Competitions · Bulgaria

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.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution:

a) Using the formulas for the sums of arithmetic and geometric progressions we obtain the equality
$\frac{n[2m + 2(n-1)]}{2} = n\left(2^{m} - 1\right)$
whence $m+n=2^{m}$ .

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

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.