# Geometric Progression rejection

• May 8th 2011, 05:06 AM
Punch
[SOLVED] Geometric Progression rejection
A geometric progression has first term $\displaystyle a$ and common ratio $\displaystyle r$. The sum of the first three terms is $\displaystyle 4.88$ and the sum to infinity is $\displaystyle 10$.

Find the values of $\displaystyle a$ and $\displaystyle r$.

$\displaystyle S_3=4.88=\frac{a(1-r^3)}{1-r}$ ------(1)
$\displaystyle S_\infty=10=\frac{a}{1-r}$ ------(2)

(2)-(1): $\displaystyle \frac{a}{1-r}-\frac{a(1-r^3)}{1-r}=10-4.88$

$\displaystyle \frac{ar^3}{1-r}=5.12$

$\displaystyle ar^3=5.12-5.12r$

$\displaystyle ar^3+5.12r-5.12=0$ -----(3)

From (2), $\displaystyle a=10-10r$ -----(4)

Sub (4) into (3): $\displaystyle (10-10r)r^3+5.12r-5.12=0$

$\displaystyle 10r^3-10r^4+5.12r-5.12=0$

$\displaystyle r=-0.4+0.6928, -0.4-0.6928, 1, 0.8$

$\displaystyle =0.2928, -1.093, 1, 0.8$

I understand why i have to reject $\displaystyle -1.093$ and $\displaystyle 1$. But i don't understand why $\displaystyle 0.2928$ is rejected too
• May 8th 2011, 05:12 AM
Prove It
Notice that $\displaystyle \displaystyle S_3 = \frac{a(1 - r^3)}{1 - r} = \frac{a}{1 - r}(1 - r^3) = S_{\infty}(1 - r^3)$. Solve for $\displaystyle \displaystyle r$.