# Find first term and common ratio of Geometric progression

• April 10th 2011, 10:53 PM
mathguy80
Find first term and common ratio of Geometric progression
Hi All,

I need a hint for this one. I think I may be missing something here.

Find the first term and common ratio if,

$
33S_5 = S_{10}
$

$
S_{10} = S_9 + 5
$

• April 10th 2011, 11:19 PM
veileen
Who are $S_5, S_9, S_{10}$? Write their formula using the first term (let's say $b_1$) and the ration (q), then you'll have a system of equations in two unknowns.
• April 11th 2011, 04:07 AM
Sambit
I suppose $S_5,S_9,S_{10}$ are, respectively, the sum of first 5,9 and 10 terms--right?

If this is the case and first term= $a$, ratio= $r$, then the sum of the GP series upto first $n$ terms will be

$S_n=a+ar+ar^2+ar^3+...+ar^{n-1}$

$=a[1+r+r^2+r^3+....+r^{n-1}]$

$=a\frac{r^n-1}{r-1}$

So your first equation reduces to $33a\frac{r^5-1}{r-1}=a\frac{r^{10}-1}{r-1}$ and second equation reduces to $a\frac{r^{10}-1}{r-1}=a\frac{r^9-1}{r-1}+5$

Solve these two equations to get $a$ and $r$.
• April 11th 2011, 08:44 AM
bjhopper
geometric progressions
Hi mathguy80,
Sambits first equation can be simplified to produce a quadratic equation in one unknown (r) SUbstitute r in the second to find a.

bjh
• April 11th 2011, 09:44 PM
mathguy80
Thanks guys, That clears things up. The quadratic equation was the key. I was a little confused by the large power and thought I wouldn't be able to factorize. Substituting, $r^5 = p$ made it into a simple quadratic to solve! :)