A Level Maths part2

**carlasader** · May 22nd 2007, 09:22 PM

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

regards,

**Jhevon** · May 22nd 2007, 09:52 PM

Originally Posted by carlasader

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

regards,

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

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, $\Rightarrow ar = 80$
we are also told that the fifth term is 5.12, $\Rightarrow ar^4 = 5.12$

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

$\Rightarrow r^3 = 0.064$

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

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

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

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

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

c)
The sum of the first $n$ terms of a geometric series is given by $S_{n} = \frac {a(1 - r^n)}{1 - r}$
So $S_{ \infty} - S_{14} = 333 \frac {1}{3} - \frac {200(1 - 0.4^{14})}{1 - 0.4} = 0.0008947 = 9 \times 10^{-4}$

**carlasader** · May 23rd 2007, 05:23 AM

can u please help me with the rest of the homework...anyone

**Jhevon** · May 27th 2007, 10:33 AM

i believe the other questions were answered in your subsequent posts

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A Level Maths part2

thanks

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