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# Algebra

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Let
a + ar + ar^2 + ar^3 + \dotsb
be an infinite geometric series.  The sum of the series is $4.$  The sum of the cubes of all the terms is $10.$ Find the common ratio.

Feb 15, 2024

#1
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The sum of the infinite geometric series listed is $$\frac{a}{1-r}$$, and this equals four, so $$\frac{a}{1-r}=4$$.

The series made up of all the cubes of these terms is $${a}^3+{a}^{3}{r}^{3}+{a}^3{r}^{6}+{a}^{3}{r}^{9}...$$. Because $$r<1$$, then $${r}^{3}<1$$. So applying our formula again, $$\frac{{a}^3}{1-{r}^3}=10$$. Now we have $$a=4(1-r)$$, and $${a}^{3}=10(1-{r}^{3})$$. We can substitute the first equation in the second one, $$64{(1-r)}^{3}=10(1-{r}^{3})$$

$$64(1-3r+3{r}^{2}-{r}^{3})=10-10{r}^{3}$$

$$54-192r+192{r}^{2}-54{r}^{3}=0$$.

$$9-32r+32{r}^{2}-9{r}^{3}=0$$

This is factored into

$$-(r-1)(9{r}^{2}-23r+9)=0$$.

We know r is not 1, so

$$9{r}^2-23r+9=0$$.

We solve and discard one solution, getting​$$\frac{23-\sqrt{205}}{18}$$as the common ratio.

Feb 15, 2024
edited by hairyberry  Feb 15, 2024
#2
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$$4 = a + ar + ar^2 + ar^3 + \dotsb$$

$$10 = a^3 + a^3r^3 + a^3r^6 + \dotsb$$  Find $$r$$

Infinite geometric series formula: a1 [first term] divided by (1 - r) [1 - common ratio] = $$a_1\over{(1 - r)}$$

Using this piece of information, we can rewrite the equations from earlier:

$$4 = {a\over{(1 - r)}}$$$$4 - 4r = a$$

$$10 = {a^3\over{1 - r^3}}$$$$10 - 10r^3=a^3=(4-4r)^3=-64r^3+192r^2-192r+64$$

$$10 - 10r^3 = -64r^3+192r^2-192r+64$$$$54r^3-192r^2+192r-54=0$$$$54r^3-54 - 192r^2+192r=0$$

Factor by grouping: This part is hard...

$$54(r^3-1)-192r(r-1)=0$$

$$r^3-1=(r-1)(r^2+r+1)$$; remember this!!! Factoring difference/sum of cubes is important!!

$$54(r-1)(r^2+r+1)-192(r-1)=0$$$$(r - 1)(54r^2+54r-138)=0$$

We know the common ratio cannot be 1, because then the series would not converge.

$$54r^2+54r-138=0$$; divide by 54: $$r^2+r-{23\over9}=0$$

$$r = {-1 \pm \sqrt{1+{92\over9}} \over 2}={-1\pm{\sqrt{101}\over3}\over2}=-{1\over2}\pm{\sqrt{101}\over6}$$