1. ## Ap/gp

The question goes:

The first, second and fourth terms of a convergent geometric progression are consecutive terms of an arithmetic progression. Prove that the common ratio of the geometric progression is (-1 + sq root 5)/2.

The first term of the geometric progression is positive. Show that the sum of the first 5 terms of this progression is greater than the nine tenths of the sum to infinity.

I tried but I couldnt get the first part. As for the second part, I don't really understand the 'nine tenths' part?

Thanks in advance if you could help!

2. Hello, margaritas!

Here's the first part . . .

The first, second and fourth terms of a convergent geometric progression
are consecutive terms of an arithmetic progression.
Prove that the common ratio of the geometric progression is $\displaystyle \frac{-1 + \sqrt{5}}{2}$

The first, second and fourth terms of a GP are: $\displaystyle a,\:ar,\:ar^3$

Three consecutive terms of an AP are: $\displaystyle a,\:a+d,\:a+2d$

[Yes, I let $\displaystyle a$ be the first term of both progressions.]

Then we have: .$\displaystyle \begin{array}{cc}ar = a+ d\quad\Rightarrow\quad ar - a = d\quad\Rightarrow\quad a(r - 1) = d \\ ar^3 = a + 2d\quad\Rightarrow\quad ar^3 - a = d\quad\Rightarrow\quad a(r^3 - 1) = d\end{array}\,\begin{array}{cc}(1)\\(2)\end{array}$

Divide (2) by (1): .$\displaystyle \frac{a(r^3 - 1)}{a(r - 1)} \:=\:$ $\displaystyle \frac{2d}{d}\quad\Rightarrow\quad \frac{a(r-1)(r^2+r+1)}{a(r-1)}\:=\:\frac{2d}{d}$

And we have: .$\displaystyle r^2 + r + 1 \:=\:2\quad\Rightarrow\quad r^2 + r - 1\:= \:0$

The Quadratic Formula gives us: .$\displaystyle r\:=\:\frac{-1\pm\sqrt{1^2 - 4(1)(-1)}}{2(1)}\;=\;\frac{-1\pm\sqrt{5}}{2}$

Therefore, the ratio is: .$\displaystyle \boxed{r \:= \:\frac{-1+\sqrt{5}}{2}}$

Note: We are told that the geometric series is convergent.
. . . . .Hence: $\displaystyle |r| < 1$

3. Oh now I get it, thanks so much for your help Soroban!