In the sequence 1, 3, 5, 7, ..., k, ..., 57, the sum of the numbers up to k, excluding k, is equal to the sum of the numbers from k up to 57, also excluding k. What is k?

Question

Let $a_n$ be the $n^{\text{th}}$ term of a decreasing infinite geometric progression. If $a_1 + a_2 + a_3 = 52$ and $a_1a_2 + a_2a_3 + a_3a_1 = 624$, then the sum of this geometric progression is:

Answer 1

Step 1: The sum of the first $n$ odd numbers is always $n^2$. Here $57$ is the $29$th odd number, so the total sum is $29^2=841$.
Step 2: If $k$ is the $m$-th odd number, the terms before it sum to $(m-1)^2$, and this must equal half of everything except $k$: \[ (m-1)^2=\frac{841-k}{2}, \quad k=2m-1. \]
Step 3: Substituting and simplifying gives $m^2-m-420=0$, which solves to $m=21$.
Step 4: So $k$ is the $21$st odd number: \[ \boxed{k=2(21)-1=41} \]

In the sequence 1, 3, 5, 7, ..., k, ..., 57, the sum of the numbers up to k, excluding k, is equal to the sum of the numbers from k up to 57, also excluding k. What is k?

Show Hint

Correct Answer: 41

Solution and Explanation

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