Arithmetic Sequence for homework,im also new to this website

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The discussion centers on proving that the terms 2ab/(a+b), b, and 2bc/(b+c) form an arithmetic sequence given that a, b, and c are consecutive numbers in a geometric sequence. Participants emphasize the importance of proper notation, particularly the placement of parentheses, to avoid confusion in mathematical expressions. The solution involves showing that the differences between the terms are equal, leading to the conclusion that b² = ac. Some contributors suggest leveraging the properties of geometric sequences, specifically expressing b and c in terms of a and a common ratio. The conversation highlights the need for clarity in mathematical communication and rigorous proof in sequence relationships.
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Homework Statement



a, b and c are consecutive numbers in a geometric sequence, where a+b ≠ 0 and b+c ≠ 0

* "/" means divide *

Show that 2ab/a+b, b and 2bc/b+c are consecutive terms in an arithmetic sequence

The Attempt at a Solution



i know this has something to do with it...

a,b,c GP

Show 2ab/a+b, b, 2bc/b+c ...

T1, T2, T3 of AP

T2-T1=T3-T2

b-(2ab/a+b)=(2bc/b+c) - b

b2 = ac
 
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Digital Genius said:

Homework Statement



a, b and c are consecutive numbers in a geometric sequence, where a+b ≠ 0 and b+c ≠ 0

* "/" means divide *

Show that 2ab/a+b, b and 2bc/b+c are consecutive terms in an arithmetic sequence

If you mean 2ab/(a+b), b , 2bc/(b+c) write it that way. Parentheses matter.

The Attempt at a Solution



i know this has something to do with it...

a,b,c GP

Show 2ab/a+b, b, 2bc/b+c ...

T1, T2, T3 of AP

T2-T1=T3-T2

b-(2ab/a+b)=(2bc/b+c) - b

b2 = ac

You haven't used the fact that a,b,c are a GP. That means ##b=ar^2,~c=ar^3##. Use that. Or you can divide that last equation by ab, assuming neither are zero.
 
Digital Genius said:

Homework Statement



a, b and c are consecutive numbers in a geometric sequence, where a+b ≠ 0 and b+c ≠ 0

* "/" means divide *

Show that 2ab/a+b, b and 2bc/b+c are consecutive terms in an arithmetic sequence

The Attempt at a Solution



i know this has something to do with it...

a,b,c GP

Show 2ab/a+b, b, 2bc/b+c ...

T1, T2, T3 of AP

T2-T1=T3-T2

b-(2ab/a+b)=(2bc/b+c) - b

b2 = ac

What you wrote means
\frac{2ab}{a} + b \text{ and } \frac{2bc}{b}+c
Is that what you wanted, or did you want
\frac{2ab}{a+b} \text{ and } \frac{2bc}{b+c}?
Parentheses are important: if you mean ##\frac{A}{B+C}## you need to write it in text as A/(B+C).
 
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