# challenging sequences problem

• February 25th 2009, 02:39 PM
oblixps
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. (Nod)
• February 25th 2009, 03:21 PM
awkward
Quote:

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. (Nod)

See if you can show that

$\frac{1}{2b} - \frac{1}{a+b} = \frac{1}{c+b} - \frac{1}{2b}$.
• February 25th 2009, 03:51 PM
oblixps
i don't know how to get to http://www.mathhelpforum.com/math-he...a32f5883-1.gif
with just what i have now: ac = b^2
or am i missing some info?
• February 25th 2009, 04:17 PM
awkward
Quote:

Originally Posted by oblixps
i don't know how to get to http://www.mathhelpforum.com/math-he...a32f5883-1.gif
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 $ac = b^2$. Then see if you can reverse your steps.
• February 25th 2009, 04:18 PM
Soroban
Hello, oblixps!

Quote:

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

Show that $\left\{\frac{1}{a+b}, \frac{1}{2b}, \frac{1}{b+c}\right\}$ 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:

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

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

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

Substitute these into [1].

• February 25th 2009, 05:59 PM
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?
• February 26th 2009, 01:31 PM
awkward
Quote:

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.
• February 26th 2009, 04:41 PM
oblixps
thanks to both of you!
i understand now (Nod)