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Find the 12th term of a GP with common ratio 2, whose 8th term is 192.
Answer:
$t_8 = ar^7 = 192$. Since $r = 2$, we have $a(2^7) = 192 \implies 128a = 192 \implies a = 1.5$.
$t_{12} = ar^{11} = 1.5 \times 2^{11} = 1.5 \times 2048 = \mathbf{3072}$.
(Alternatively, $t_{12} = t_8 \times r^4 = 192 \times 16 = 3072$).
Find the $10^{th}$ and $n^{th}$ terms of the GP: 5, 25, 125, ...
Answer:
First term $a = 5$, common ratio $r = 5$.
$10^{th}$ term: $t_{10} = ar^9 = 5(5^9) = \mathbf{5^{10}}$.
$n^{th}$ term: $t_n = ar^{n-1} = 5(5^{n-1}) = \mathbf{5^n}$.
A sequence is given by the recursive rule $t_1 = 2, t_{n+1} = 3t_n - 2$ for $n \ge 1$. Which term of the sequence is 730?
Answer:
Let's generate the sequence:
$t_1 = 2$
$t_2 = 3(2) - 2 = 4$
$t_3 = 3(4) - 2 = 10$
$t_4 = 3(10) - 2 = 28$
$t_5 = 3(28) - 2 = 82$
$t_6 = 3(82) - 2 = 244$
$t_7 = 3(244) - 2 = 730$
So, 730 is the $7^{th}$ term.
Which term of the GP: 2, 6, 18, ... is 4374? Write the explicit formula as well as the recursive formula for the $n^{th}$ term.
Answer:
Here $a = 2, r = 3$. Set $2(3^{n-1}) = 4374 \implies 3^{n-1} = 2187$. Since $3^7 = 2187$, $n-1 = 7 \implies n = 8$. It is the $8^{th}$ term.
Explicit formula: $\mathbf{t_n = 2 \times 3^{n-1}}$.
Recursive formula: $\mathbf{t_1 = 2, t_n = 3t_{n-1}}$ for $n \ge 2$.
A ball is dropped from a height of 80 metres. After hitting the ground, it bounces back to 60% of the height from which it fell. It continues bouncing in this way.
(i) What height does the ball reach after the 5th bounce?
(ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the $6^{th}$ time?
Answer:
(i) Height after 5th bounce = $80 \times (0.6)^5 = 80 \times 0.07776 = \mathbf{6.2208}$ metres.
(ii) The total distance includes the initial drop (80m) plus the up-and-down distance of the 5 bounces.
Total Distance = $80 + 2 \times 80(0.6^1 + 0.6^2 + 0.6^3 + 0.6^4 + 0.6^5)$.
The geometric series sum is $0.6 \frac{1 - 0.6^5}{1 - 0.6} = 0.6 \frac{1 - 0.07776}{0.4} = 1.5 \times 0.92224 = 1.38336$.
Total Distance = $80 + 160(1.38336) = 80 + 221.3376 = \mathbf{301.3376}$ metres.
Which term of the sequence $2, \sqrt{2}, 1, \dots$ is $1/128$?
Answer:
This is a GP with $a = 2$ and $r = \frac{1}{\sqrt{2}}$.
Set $t_n = 2 \left(\frac{1}{\sqrt{2}}\right)^{n-1} = \frac{1}{128}$.
$\left(\frac{1}{\sqrt{2}}\right)^{n-1} = \frac{1}{256}$.
Since $256 = 2^8 = (\sqrt{2})^{16}$, we have $\left(\frac{1}{\sqrt{2}}\right)^{n-1} = \left(\frac{1}{\sqrt{2}}\right)^{16}$.
Therefore, $n-1 = 16 \implies n = 17$. It is the $17^{th}$ term.
Fig. 8.12 shows Stages 0 to 3 of the Sierpiński square carpet. Stage 0 is a square sheet of paper. To construct Stage 1, each side of the square is trisected... The centre square is then removed and the 8 smaller squares are retained...
(i) How many red squares are there in Stages 0 to 3?
(ii) Can you predict the number of red squares in Stages 4 and 5?
(iii) Can you find a rule for the number of red squares at the $n^{th}$ stage? Write the explicit formula as well as the recursive formula.
(iv) Suppose the area of the square in Stage 0 is 1 square unit. What is the explicit and recursive formula for the area of the red region at the $n^{th}$ stage? What happens to this area as n increases?
Answer:
(i) Stage 0: 1, Stage 1: 8, Stage 2: 64, Stage 3: 512.
(ii) Stage 4: $8^4 = \mathbf{4096}$, Stage 5: $8^5 = \mathbf{32768}$.
(iii) Explicit formula: $\mathbf{t_n = 8^n}$. Recursive formula: $\mathbf{t_0 = 1, t_n = 8t_{n-1}}$.
(iv) In each stage, the area is multiplied by $\frac{8}{9}$ because 1 out of every 9 squares is removed. Explicit formula for area: $\mathbf{A_n = (\frac{8}{9})^n}$. Recursive formula: $\mathbf{A_0 = 1, A_n = \frac{8}{9} A_{n-1}}$. As n increases, the area gets closer and closer to 0.
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