:eek: really having touble with all that need urgent help!!!

regards,

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- May 22nd 2007, 09:22 PMcarlasaderA Level Maths part2
:eek: really having touble with all that need urgent help!!!

regards, - May 22nd 2007, 09:52 PMJhevon
1)

a geometric series is one of the form: $\displaystyle \sum_{n=1}^{\infty} ar^{n-1}$

where a is the first term, r is the common ratio and n is the current number of the term.

a)

we are given that the second term is 80, $\displaystyle \Rightarrow ar = 80$

we are also told that the fifth term is 5.12, $\displaystyle \Rightarrow ar^4 = 5.12$

$\displaystyle \Rightarrow \frac {ar^4}{ar} = \frac {5.12}{80}$

$\displaystyle \Rightarrow r^3 = 0.064$

$\displaystyle \Rightarrow r = 0.4 \mbox{ or } \frac {4}{10}$

since $\displaystyle ar = 80 \mbox{ we have } a = \frac {80}{r} = \frac {80}{ \frac {4}{10}} = 200$

so the first term is 200, and the common ration is $\displaystyle \frac {4}{10}$

b)

For a geometric series with $\displaystyle \left| r \right| < 1$, the sum to infinity is given by: $\displaystyle S_{ \infty} = \frac {a}{1 - r}$

So here, $\displaystyle S_{ \infty} = \frac {200}{1 - \frac {4}{10}} = 333 \frac {1}{3}$

c)

The sum of the first $\displaystyle n$ terms of a geometric series is given by $\displaystyle S_{n} = \frac {a(1 - r^n)}{1 - r}$

So $\displaystyle S_{ \infty} - S_{14} = 333 \frac {1}{3} - \frac {200(1 - 0.4^{14})}{1 - 0.4} = 0.0008947 = 9 \times 10^{-4}$ - May 23rd 2007, 05:23 AMcarlasaderthanks
can u please help me with the rest of the homework...anyone :)

- May 27th 2007, 10:33 AMJhevon
i believe the other questions were answered in your subsequent posts