1. ## challenging sequences problem

Suppose that a, b, and c are three consecutive terms in a geometric sequence. Show that 1/a+b, 1/2b, and 1/c+b are three consecutive terms in an arithmetic sequence.

so far i have:
c / b = b / a
since they are consecutive terms in a geometric sequence.
then i cross multiplied so i have ac = b^2
i don't know what to do after this. please help or give me a hint if you know how to do this problem.

2. Originally Posted by oblixps
Suppose that a, b, and c are three consecutive terms in a geometric sequence. Show that 1/a+b, 1/2b, and 1/c+b are three consecutive terms in an arithmetic sequence.

so far i have:
c / b = b / a
since they are consecutive terms in a geometric sequence.
then i cross multiplied so i have ac = b^2
i don't know what to do after this. please help or give me a hint if you know how to do this problem.
See if you can show that

$\displaystyle \frac{1}{2b} - \frac{1}{a+b} = \frac{1}{c+b} - \frac{1}{2b}$.

3. i don't know how to get to
with just what i have now: ac = b^2
or am i missing some info?

4. Originally Posted by oblixps
i don't know how to get to
with just what i have now: ac = b^2
or am i missing some info?
Try "solving" the equation and see if it reduces to $\displaystyle ac = b^2$. Then see if you can reverse your steps.

5. Hello, oblixps!

Suppose that $\displaystyle \{a, b, c\}$ are three consecutive terms in a geometric sequence.

Show that $\displaystyle \left\{\frac{1}{a+b}, \frac{1}{2b}, \frac{1}{b+c}\right\}$ are three consecutive terms in an arithmetic sequence.

I tried various approaches . . . awkward has the best one.

Show that the three fractions have a common difference:

. . $\displaystyle \frac{1}{2b} - \frac{1}{a+b} \;=\;\frac{1}{b+c} - \frac{1}{2b}$ .[1]

Since $\displaystyle \{a,b,c\}$ form a geometric sequence,

. . we have: .$\displaystyle \begin{Bmatrix}a &=& a \\ b &=&ar \\ c &=&ar^2 \end{Bmatrix}$

Substitute these into [1].

6. ok so i made the fractions have common denominators
a-b / 2b(a+b) = b-c / 2b(b+c)

then i cross multiplied so now i have:
(b+c)(a-b)(2b) = (b-c)(a+b)(2b)

i multiplied everything out and canceled terms so i have:
2b^3 - 2abc = 0
2b(b^2 - ac) = 0
b^2 - ac = 0
b^2 = ac

is this right? because it seems kind of weird to factor and use the zero product property since either factor can be 0. if this is so, how are you supposed to do this problem without going backwards? unless you must go backwards to solve this problem?

7. Originally Posted by oblixps
ok so i made the fractions have common denominators
a-b / 2b(a+b) = b-c / 2b(b+c)

then i cross multiplied so now i have:
(b+c)(a-b)(2b) = (b-c)(a+b)(2b)

i multiplied everything out and canceled terms so i have:
2b^3 - 2abc = 0
2b(b^2 - ac) = 0
b^2 - ac = 0
b^2 = ac

is this right? because it seems kind of weird to factor and use the zero product property since either factor can be 0. if this is so, how are you supposed to do this problem without going backwards? unless you must go backwards to solve this problem?
You have the right idea. (Or you can follow Soroban's method, which will also work.) Just write down your equations in the reverse order you have listed them above. Instead of canceling terms, you will need to add some terms to both sides of the equation.

If you are worried about the possibility that a+b, 2b, or c+b could be zero, you are quite right to do so. That they are non-zero should be added as an assumption, although I guess you could say it's implied by the problem statement.

8. thanks to both of you!
i understand now

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### prove that three consecutive terms of an arithmetic progression are also consecutive terms of a geometric progression

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