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Prove 2^a , 2^b , 2^c forms a geometric sequence when a , b , c forms an arithmetic sequence.

User Erric
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Answer: Given that a, b, and c form an arithmetic sequence, so they are equally spaced, meaning the difference between each of them is constant. Let's denote the common difference between a, b, and c as d.

So, we have:

a = b - d

c = b + d

Using these two equations, we can prove that 2^a, 2^b, and 2^c form a geometric sequence.

We have:

2^a = 2^b / 2^d

2^c = 2^b * 2^d

Since the product of two powers of the same base is equivalent to the power of the base with the sum of the exponents, we have:

2^a * 2^c = 2^b * 2^b * 2^d * 2^d = 2^(b+b+d+d) = 2^(2b+2d)

So, we have:

2^a * 2^c = 2^(2b+2d) = 2^(2b) * 2^(2d) = 2^b * 2^b * 2^d * 2^d = 2^b * 2^c

This means that 2^a, 2^b, and 2^c form a geometric sequence, with the common ratio of 2^d.

Explanation:

User Mehmet Sahin
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