Geometric and Arithmetric progression

**Tangera** · May 6th 2008, 07:43 AM

Thank you for helping!!

Q: The first 3 terms of a convergent geometric progression are coincidentally also the 16th, 4th and 1st terms of an arithmetic progression respectively. FInd the ration of the sum to infinity for the geometric progression to its first term.

Thank you very much.

**Soroban** · May 6th 2008, 01:34 PM

Hello, Tangera!

Another challenging problem . . .

The first 3 terms of a G.P. are the 16th, 4th and 1st terms of an A.P.
Find the ratio of the sum to infinity for the G.P. to its first term.

For the G.P., let $a$ = first term, $r$ = common ratio.
For the A.P., let $b$ = first term, $d$ = common difference.

We have: . $\begin{array}{|c|c|} \text{G.P.} & \text{A.P.} \\ \hline<br /> a & b + 15d \\ ar & b + 3d \\ ar^2 & b \end{array}\quad\Rightarrow\quad\begin{array}{cccc }a &=& b+15d & {\color{blue}[1]}\\ ar &=& b+3d & {\color{blue}[2]}\\ ar^2 &=& b & {\color{blue}[3]} \end{array}$

$\begin{array}{ccccccccc}\text{Subtract [1] - [2]:} & a - ar &=& 12d & \Rightarrow & a(1-r) &=& 12d & {\color{blue}[4]}\\ \text{Subtract [2] - [3]:} & ar-ar^2 &=& 3d & \Rightarrow & ar(1-r) &= & 3d & {\color{blue}[5]} \end{array}$

$\text{Divide }[5] \div [4]\!:\;\;\frac{ar(1-r)}{a(1-r)} \:=\:\frac{3d}{12d} \quad\Rightarrow\quad r \:=\:\frac{1}{4}$

Substitute into [4]: . $a\left(1-\frac{1}{4}\right) \:=\:12d\quad\Rightarrow\quad a \:=\:16d$

The G.P. has the sum: . $S \;=\;\frac{a}{1-r} \;=\;\frac{16d}{1-\frac{1}{4}} \;=\;\frac{64d}{3}$

Therefore, the ratio is: . $\frac{S}{a} \;=\;\frac{\frac{64d}{3}}{16d} \;=\;\boxed{\frac{4}{3}}$

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