Geometric Progression

**azuki** · January 20th 2009, 12:02 AM

A geometric series has first term 1 and the common ratio r is positive. The sum of the first 5 terms is twice the sum of the terms from the 6th to the 15th inclusive. Prove that r^5 = 1/2 (sqrt3 - 1).

I manage to get up to 2(r^5 - 1) = r^15 - r^5.

Please help.

**AlvinCY** · January 20th 2009, 12:40 AM

Originally Posted by azuki

A geometric series has first term 1 and the common ratio r is positive. The sum of the first 5 terms is twice the sum of the terms from the 6th to the 15th inclusive. Prove that r^5 = 1/2 (sqrt3 - 1).

I manage to get up to 2(r^5 - 1) = r^15 - r^5.

Please help.

Sum of a geometric series: $S_n=\frac{a(r^n-1)}{r-1}$

$S_5=\frac{a(r^5-1)}{r-1}$

$S_{15}=\frac{a(r^{15}-1)}{r-1}$

The sum of the first 5 terms is twice the sum of the terms from the 6th to the 15th inclusive implies:

$S_5 = 2S_{15} - 2S_5$

$3S_5 = 2S_{15}$

$\frac{3a(r^5-1)}{r-1}=\frac{2a(r^{15}-1)}{r-1}$

$3a(r^5-1)=2a(r^{15}-1)$

$3(r^5-1)=2(r^{15}-1)$

$3(r^5-1)=2[(r^5)^3-1]$ (note: the expression in the brackets is now a difference of two CUBES, so we can use: $a^3-b^3=(a-b)(a^2+ab+b^2)$

So $(r^5)^3-1=(r^5-1)(r^{10}+r^5+1)$

$3(r^5-1)=2(r^5-1)(r^{10}+r^5+1)$

$3=2(r^{10}+r^5+1)$

$\frac{3}{2}=r^{10}+r^5+1$

$0=r^{10}+r^5-\frac{1}{2}$

$0=2r^{10}+2r^5-1$

$0=2(r^5)^2+2r^5-1$

Let $X=r^5$

$0=2X^2+2X-1$

Using the quadratic formula ( $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ ),

$X=\frac{-2\pm\sqrt{4-4\times2\times-1}}{4}$

$X=\frac{-2\pm\sqrt{12}}{4}$

$X=\frac{-2\pm2\sqrt{3}}{4}$

$X=\frac{-1\pm\sqrt{3}}{2}$

Putting $X=r^5$ back in:

$r^5=\frac{-1\pm\sqrt{3}}{2}$

But the question said that $r$ is a positive number, so $r^5$ must ALSO be positive, therefore $r^5=\frac{-1+\sqrt{3}}{2}$

Proving that $r^5=\frac{1}{2}(\sqrt{3}-1)$

**azuki** · January 20th 2009, 05:32 AM

Why sum of the first 5 terms is twice the sum of the terms from the 6th to the 15th inclusive implies:

S_5 = 2S_15 – 2S_5

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