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,

$\displaystyle

33S_5 = S_{10}

$

$\displaystyle

S_{10} = S_9 + 5

$

Thanks for your help!

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- Apr 10th 2011, 10:53 PMmathguy80Find 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,

$\displaystyle

33S_5 = S_{10}

$

$\displaystyle

S_{10} = S_9 + 5

$

Thanks for your help! - Apr 10th 2011, 11:19 PMveileen
Who are $\displaystyle S_5, S_9, S_{10}$? Write their formula using the first term (let's say $\displaystyle b_1$) and the ration (q), then you'll have a system of equations in two unknowns.

- Apr 11th 2011, 04:07 AMSambit
I suppose $\displaystyle S_5,S_9,S_{10}$ are, respectively, the sum of first 5,9 and 10 terms--right?

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

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

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

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

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

Solve these two equations to get $\displaystyle a$ and $\displaystyle r$. - Apr 11th 2011, 08:44 AMbjhoppergeometric 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 - Apr 11th 2011, 09:44 PMmathguy80
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,$\displaystyle r^5 = p$ made it into a simple quadratic to solve! :)