213k views
1 vote
Prove 2^a , 2^b , 2^c forms a geometric sequence when a , b , c forms an arithmetic sequence.

User Erric
by
8.0k points

1 Answer

3 votes

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
by
7.2k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories