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Let {gₙ } be a geometric sequence, with only positive terms. Let xₙ =lngₙ ​ . Prove that {xₙ } is an arithmetic sequence. Hint: a sequence x ₙ​ is arithmetic if there exists an a and a d such that xₙ =a+(n−1)d. So you should prove that those valid of a and d exist.

User Rvange
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Sure, let's go through it step by step.

First, consider the geometric sequence where we denote the first term as 'a' and the common ratio as 'r'. The general form of any term in a geometric sequence is written as gₙ = a * (r ^(n - 1)). Here, a stands for the first term, r is the common ratio, and n denotes the n-th term.

Now, let's take the natural logarithm (ln) of both sides of the equation. This gives us ln(gₙ) = ln(a * r^(n - 1)).

Make use of the logarithmic properties to rewrite the right-hand side of the equation. It can simplify to ln(a) + (n - 1) * ln(r) after applying the logarithmic identity ln(ab) = ln(a) + ln(b).

Observe that the simplified equation is in the form of a + (n - 1) * d, which matches exactly the generic structure of an arithmetic sequence xₙ = a + (n - 1) * d. Here, 'a' is the first term and 'd' is the common difference of the arithmetic sequence.

Hence, it's concluded that the sequence xₙ = ln(gₙ) is an arithmetic sequence. In this case, our first term 'a' is ln(a), and the common difference 'd' is ln(r), where 'a' and 'r' are the first term and the common ratio of the original geometric sequence respectively.

So, we have successfully demonstrated that if we have a geometric sequence with positive terms, and we transform it by taking the natural logarithm of each term, the result is an arithmetic sequence. In the resulting arithmetic sequence, the first term equals the natural logarithm of the first term in the original geometric sequence and the common difference equals the natural logarithm of the common ratio in the original geometric sequence.

User Statey
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