Geometric Progression rejection

**Punch** · May 8th 2011, 05:06 AM

A geometric progression has first term $a$ and common ratio $r$ . The sum of the first three terms is $4.88$ and the sum to infinity is $10$ .

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

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

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

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

$ar^3=5.12-5.12r$

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

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

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

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

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

$=0.2928, -1.093, 1, 0.8$

I understand why i have to reject $-1.093$ and $1$ . But i don't understand why $0.2928$ is rejected too

**Prove It** · May 8th 2011, 05:12 AM

Notice that $\displaystyle S_3 = \frac{a(1 - r^3)}{1 - r} = \frac{a}{1 - r}(1 - r^3) = S_{\infty}(1 - r^3)$ . Solve for $\displaystyle r$ .

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[SOLVED] Geometric Progression rejection

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