Geometric Progression Question

**dagamer** · January 30th 2009, 08:52 PM

Question :
If a,b,c are in Geometric Progression , prove that (1/a+b)+(1/b+c) = (1/b) .

I found that b^2=ac , and thus c = (b^2)/a and a = (b^2)/c here .
I managed to solve this problem , by substituting "c" in the equation as , (b^2)/a , and simplifying , after which the left hand side finally equated to 1/b.

I ran into a doubt , why do I need to substitute only for "c" in the equation to equate ? Why cant one substitute for both "a" , "c" or "a" , "c" , "b" simultaneously ?
I found that , once I substitute for more than once variable in the equation , the equation no longer equates . This maybe a stupid doubt , but please help me clear that abstract gap in my mind .

Thanks so much .

**Rincewind** · January 30th 2009, 09:12 PM

Perhaps an alternative method would clear up the confusion. Since they are a geometric they differ by a multiplicative constant, lets call it $r$ . That is,

$b = ar, \qquad c = br = ar^2$

So $a+b = a(1+r)$ and $b+c = ar(1+r),$

therefore

$ \begin{array}{rcl} \dfrac{1}{a+b} + \dfrac{1}{b+c} & = & \dfrac{1}{a(1+r)} + \dfrac{1}{ar(1+r)} \\ & = & \dfrac{r}{ar(1+r)} + \dfrac{1}{ar(1+r)} \\ & = & \dfrac{1 + r}{ar(1+r)} \\ & = & \dfrac{1}{ar} \\ & = & \dfrac{1}{b}. \end{array} $

Hope this helps.

**dagamer** · January 31st 2009, 12:10 AM

Thanks Rincewind , that really helped me figure it out.

Can anyone post and explain the derivation for 'n' terms of a Geometric Progression as well ? ( And if possible , for infinite terms too ? )

Thanks .

**Soroban** · January 31st 2009, 06:44 AM

Hello, dagamer!

Can anyone explain the derivation for $n$ terms of a GP?
(And, if possible, for infinite terms, too?)

. . .We have: . . $S \;=\; a + ar + ar^2 + ar^3 + \hdots + ar^{n-1}$ . . . . .[1]

Multiply by $r\!:\;\;rS \;=\; \quad\;\; ar + ar^2 + ar^3 + \hdots + ar^{n-1} + ar^n$ .[2]

Subtract [2] from [1]: . $S - rS \;=\;a - ar^n \quad\Rightarrow\quad S(1-r) \;=\;a(1-r^n)$

. . Therefore: . $\boxed{S \;=\;a\,\frac{1-r^n}{1-r}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

For an infinite series . . .

. . .We have: . . $S \;=\; a + ar + ar^2 + ar^3 + \hdots$ . [1]

Multiply by $r\!:\;\;rS \;=\; \quad\;\; ar + ar^2 + ar^3 + \hdots$ . .[2]

Subtract [2] from [1]: . $S - rS \;=\;a \quad\Rightarrow\quad S(1-r) \:=\:a$

. . Therefore: . $\boxed{S \;=\;\frac{a}{1-r}}\;\;\text{ for }|r| < 1$

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