Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 3.3 AMC 10/12 Find the answer

Given the sequence $\{a\_n\}$ , where $a\_n=32$ , and the sum of the first $n$ terms is $S\_n=63$ .

1. If the sequence $\{a\_n\}$ is an arithmetic sequence with a common difference of $11$ , find $a\_1$ .
2. If the sequence $\{a\_n\}$ is a geometric sequence with the first term $a\_1=1$ , find the sum of the first $m$ terms, $T\_m$ , of the sequence $\{a \;\_{ n }^{ 2 }\}$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

1. From the given information, we have: $\frac{n(a\_1+a\_n)}{2}=S\_n=63$ , and $a\_1+11(n-1)=a\_n=32$ . Solving these equations simultaneously, we get: $\boxed{a\_1=10}$ , $\boxed{n=3}$ or $\boxed{a\_1=1}$ , $\boxed{n=\frac{42}{11}}$ (discarded as $n$ must be an integer).

2. From the given information, we have: $a\_1q^{n-1}=32$ and $\frac{a\_1(1-q^{n})}{1-q}=63$ . Solving these equations simultaneously, we get: $\boxed{q=2}$ , $\boxed{n=6}$ . Therefore, the sequence $\{a\_n^2\}$ is a geometric sequence with the first term $1$ and a common ratio of $4$ . Hence, $\boxed{T\_m=\frac{1-4^{m}}{1-4}=\frac{4^{m}-1}{3}}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.