1.
A hard rubber ball is dropped from a roof of a building that is \(7\) metres high. The ball rises to \(57\%\) of the height from which it fell after each bounce. What height does the ball reach after \(5\) bounces?
Step 1. The initial height is \(7\text{ m}\), so \(a=7\). Step 2. The ball rises to \(57\%\) of its previous height after each bounce. \[ r=0.57 \] Step 3. After the first bounce, the bounce height is the second term. After the fifth bounce, it is the sixth term. Step 4. Use the geometric sequence formula. \[ a_n=ar^{n-1} \] Step 5. Substitute \(n=6\), \(a=7\), and \(r=0.57\). \[ a_6=7(0.57)^{6-1} \] Step 6. Calculate. \[ a_6=7(0.57)^5\approx 0.4211 \] Therefore, after \(5\) bounces, the ball reaches about \(0.42\text{ m}\). Answer: A
2.
How many ancestors does a person have in \(15\) generations if your parents are considered first generation \((2\) ancestors\()\) and your grandparents are considered second generation \((4\) ancestors\()\)?
Step 1. The number of ancestors forms a geometric sequence. \[ 2,4,8,\cdots \] Step 2. Identify the first term, common ratio, and number of generations. \[ a=2,\quad r=2,\quad n=15 \] Step 3. Since the question asks how many ancestors a person has in \(15\) generations total, use the geometric series sum formula. \[ S_n=\frac{a(1-r^n)}{1-r} \] Step 4. Substitute the values. \[ S_{15}=\frac{2(1-2^{15})}{1-2} \] Step 5. Simplify. \[ \begin{aligned} S_{15}&=\frac{2(1-32768)}{-1}\\&=\frac{2(-32767)}{-1}\\&=65534\end{aligned} \] Therefore, the person has \(65534\) ancestors in \(15\) generations. Answer: B
3.
Determine the common ratio of the following geometric sequence: \(-8,64,-512,4096,\cdots\)
Step 1. The common ratio of a geometric sequence is found by dividing a term by the previous term. \[ r=\frac{t_n}{t_{n-1}} \] Step 2. Use the first two terms. \[ r=\frac{64}{-8}=-8 \] Step 3. Check with the next terms. \[ \frac{-512}{64}=-8,\quad \frac{4096}{-512}=-8 \] Therefore, the common ratio is \(-8\). Answer: D
4.
If the sum of an infinite geometric series is \(63\) and the first term is \(21\), determine the common ratio.
Step 1. Use the infinite geometric series formula. \[ S_{\infty}=\frac{a}{1-r} \] Step 2. Substitute \(S_{\infty}=63\) and \(a=21\). \[ 63=\frac{21}{1-r} \] Step 3. Solve for \(r\). \[ \begin{aligned} 63(1-r)&=21\\63-63r&=21\\-63r&=21-63\\-63r&=-42\\r&=\frac{-42}{-63}\\r&=\frac{2}{3}\end{aligned} \] Therefore, the common ratio is \(\frac{2}{3}\). Answer: D
5.
Find the exact value of the sum for the following series: \(\sum_{i=2}^{\infty}3\left(\frac{2}{3}\right)^i\)
Step 1. This is an infinite geometric series. \[ \sum_{i=2}^{\infty}3\left(\frac{2}{3}\right)^i \] Step 2. The common ratio is: \[ r=\frac{2}{3} \] Step 3. Find the first term of the series by substituting \(i=2\). \[ a=3\left(\frac{2}{3}\right)^2=3\left(\frac{4}{9}\right)=\frac{4}{3} \] Step 4. Use the infinite geometric series formula. \[ S=\frac{a}{1-r} \] Step 5. Substitute \(a=\frac{4}{3}\) and \(r=\frac{2}{3}\). \[ \begin{aligned} S&=\frac{\frac{4}{3}}{1-\frac{2}{3}}\\&=\frac{\frac{4}{3}}{\frac{1}{3}}\\&=\frac{4}{3}\times 3\\&=4\end{aligned} \] Therefore, the exact sum is \(4\). Answer: B
6.
Find the common ratio for the following series: \(\sum_{i=2}^{\infty}3\left(\frac{2}{3}\right)^i\)
Step 1. Look at the geometric series. \[ \sum_{i=2}^{\infty}3\left(\frac{2}{3}\right)^i \] Step 2. In a geometric series, the common ratio is the value being repeatedly multiplied. Step 3. The base inside the power is \(\frac{2}{3}\). Therefore, the common ratio is \(\frac{2}{3}\). Answer: D
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