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A geometric series has a constant ratio of $\frac{1}{2}$ and a sum to infinity of 6.

1. Calculate the first term of the series.
2. Calculate the $8^{\text {th }}$ term of the series.
3. Given: $\sum_{k=1}^{n} 3(2)^{1-k}=5,8125 \quad$ Calculate the value of $n$.
4. If $\sum_{k=1}^{20} 3(2)^{1-k}=p$, write down $\sum_{k=1}^{20} 24(2)^{-k}$ in terms of $p$.
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1.

$S_{\infty}=\frac{a}{1-r}$
$\therefore 6=\frac{a}{1-\frac{1}{2}}$
$\therefore a=3$

2.

$\therefore T_{8}=3\left(\frac{1}{2}\right)^{8-1}=\frac{3}{128}$ or $0,023($ to 3 d.p. $)$

3.

$\sum_{k=1}^{n} 3(2)^{1-k}=5,8125$
$\therefore 3(2)^{1-1}+3(2)^{1-2}+3(2)^{1-3}+\ldots . .=5,8125$
$\therefore 3+\frac{3}{2}+\frac{3}{4}+\ldots . .=5,8125$
$\therefore$ a geometric series with $a=3$ and $r=\frac{1}{2}$ $\therefore S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}$
$\therefore 5,8125=\frac{3\left(1-0,5^{n}\right)}{0,5}$
$\therefore \frac{5,8125 \times 0,5}{3}=1-0,5^{n}$
$\therefore 0,5^{n}=1-\frac{5,8125 \times 0,5}{3}$
$\therefore n=\log _{0,5}\left(1-\frac{5,8125 \times 0,5}{3}\right)$
$\therefore n=5$

4.

$\sum_{k=1}^{20} 24(2)^{-k}$
$=\sum_{k=1}^{20}(4)(3)(2)(2)^{-k}$
$=\sum_{k=1}^{20}(4) 3(2)^{1-k}$
$=(4) \sum_{k=1}^{20} 3(2)^{1-k}$
$=4 p$
by Diamond (88,180 points)

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