# Problem with Geometric Progression

Question: An infinite geometric progression has a finite sum. Given that the first term is 18 and that the sum of the first 3 terms is 38. Calculate the values of (i)the common ratio, (ii) the sum to infinity.

What I'm doing:

Sum $s= \cfrac{a(r^n -1)}{r-1}$

$38 = 18\cfrac{r^3 -1}{r-1}$

$38r - 38 = 18r^3 - 18$

$18r^3 - 38r + 20 = 0$

???

• Seems to be geometric progression, but title says arithmetic. Oct 16, 2015 at 14:35

## 3 Answers

i)

The factor theorem tells us that $r-1$ is a factor. So you can use long division to get the quadratic factor: $$\require{enclose} \begin{array}{r} 18r^2+18r-20 \\[-3pt] r-1 \enclose{longdiv}{18r^3+0r^2-38r+20} \\[-3pt] \underline{18r^3-18r^2}\phantom{2} \\[-3pt] -20r+20 \\[-3pt] \underline{0} \end{array}$$ $$18r^3 - 38r + 20 = 0$$ $$\implies (r-1)(18r^2+18r-20)=0$$ $$\implies (r-1)\left(r-\frac23 \right)\left(r+\frac53 \right)=0$$ $$\implies r=1,-\frac53,\frac23$$ but since $-1 \lt r \lt 1 \implies r=\frac23$

ii)

$S_{\infty}=\cfrac{a}{1-r}=\cfrac{18}{1-\frac23}=54$

• I dont really get how you broke down $18r^3 - 38r + 20 = 0$ into $\implies (r-1)(18r^2+18r-20)=0$ .... Oct 16, 2015 at 17:02
• @thezaynthe I used long division Oct 16, 2015 at 17:03
• @thezaynthe I added some detail on the long division Oct 16, 2015 at 17:14
• thanks! I remember learning how to do this about 6 or 7 months ago for my Cambridge exams.. Oct 16, 2015 at 17:22
• @thezaynthe Do you understand the factor theorem that I used to deduce $r-1$ is a factor? If not I can explain it in my answer also. Oct 16, 2015 at 17:24

You have not used the fact that $r-1$ is a factor of your polynomial, which will get you a quadratic. For three terms it is easier to write $18(1+r+r^2)=38$ and you start with the quadratic. To have a finite sum you need $|r|\lt 1$, which gives you which root to use.

$\textbf{hint}$

The first $3$ terms are $$a+ar+ar^2 = 38$$ where $a=18$ now we have to sum to infinity we utilize a well known fact $$\sum_{n=0}^{\infty}ar^n = \frac{a}{1-r}$$ where $|r|<1$ which is important for the first calculation for finding $r$.

• thanks I havent started solving the b) part yet... though shouldnt a + ar + ar^2 = 38 (what you did) be equal to [ 18 (r^3 -1)] / (r-1)] (what Im doing) (And Im still in high school and my math sucks:) does that E thing stand for sum? Oct 16, 2015 at 14:45
• As @RossMillikan pointed out you have a root that corresponds to $1-r$ so if you use long division you will find that $$\frac{r^3-1}{r-1} = 1+r+r^2$$ you will yield the desired result. I think its easier to use the three terms of the series rather than the $n$ sum. Oct 16, 2015 at 14:48
• further more you can see (try it yourself - I wish I tried doing such things back then) if you can generalize to $$\frac{r^n-1}{r-1}=?$$ Oct 16, 2015 at 14:49
• thanks. btw, how do you do that separate line thingy? Im still kinda knew to stackexchange Oct 16, 2015 at 14:55
• use double $\$\$ signs. Also look at this mathjax tutorial Oct 16, 2015 at 14:57