Question 1: i. The first three terms of a geometric progression are $n-2$, $n$, and $n+4$ respectively. We need to find the value of $n$ and the common ratio. ii. We need to find the geometric mean of 81 and 121. Question 2: Given the geometric progression 9, 3, 1, ..., we need to find the common ratio, the 5th term, and the sum of the first 8 terms. Question 3: It is given that $a$ varies directly as $b$ and inversely as the square of $c$, and $a=9$ when $b=12$ and $c=2$. We need to find the constant of variation $k$, the value of $a$ when $b=16$ and $c=4$, and the value of $c$ when $b=25$ and $a=3$.

1. Problem Description

Question 1:

i. The first three terms of a geometric progression are $n-2$, $n$, and $n+4$ respectively. We need to find the value of $n$ and the common ratio.

ii. We need to find the geometric mean of 81 and

1

2

1.

Question 2:

Given the geometric progression 9, 3, 1, ..., we need to find the common ratio, the 5th term, and the sum of the first 8 terms.

Question 3:

It is given that $a$ varies directly as $b$ and inversely as the square of $c$, and $a=9$ when $b=12$ and $c=2$. We need to find the constant of variation $k$, the value of $a$ when $b=16$ and $c=4$, and the value of $c$ when $b=25$ and $a=3$.

2. Solution Steps

Question 1:

i. a. In a geometric progression, the ratio between consecutive terms is constant. Therefore, we have

$\frac{n}{n-2} = \frac{n+4}{n}$

$n^2 = (n-2)(n+4)$

$n^2 = n^2 + 4n - 2n - 8$

$n^2 = n^2 + 2n - 8$

$0 = 2n - 8$

$2n = 8$

$n = 4$

i. b. The common ratio $r$ is $\frac{n}{n-2}$. Substituting $n=4$, we get

$r = \frac{4}{4-2} = \frac{4}{2} = 2$

ii. The geometric mean of two numbers $x$ and $y$ is $\sqrt{xy}$. So, the geometric mean of 81 and 121 is

$\sqrt{81 \times 121} = \sqrt{81} \times \sqrt{121} = 9 \times 11 = 99$

Question 2:

i. The common ratio $r$ is the ratio between consecutive terms.

$r = \frac{3}{9} = \frac{1}{3}$

ii. The $n$th term of a geometric progression is given by $ar^{n-1}$, where $a$ is the first term.

The 5th term is $9 \times (\frac{1}{3})^{5-1} = 9 \times (\frac{1}{3})^4 = 9 \times \frac{1}{81} = \frac{1}{9}$

iii. The sum of the first $n$ terms of a geometric progression is given by

$S_n = \frac{a(1-r^n)}{1-r}$

The sum of the first 8 terms is

$S_8 = \frac{9(1-(\frac{1}{3})^8)}{1-\frac{1}{3}} = \frac{9(1-\frac{1}{6561})}{\frac{2}{3}} = \frac{9(\frac{6560}{6561})}{\frac{2}{3}} = \frac{9 \times 6560 \times 3}{6561 \times 2} = \frac{27 \times 6560}{13122} = \frac{177120}{13122} = \frac{88560}{6561} = \frac{29520}{2187} = \frac{9840}{729} = \frac{3280}{243}$

Question 3:

Given that $a$ varies directly as $b$ and inversely as the square of $c$, we have

$a = k \frac{b}{c^2}$

where $k$ is the constant of variation.

a. We are given that $a=9$ when $b=12$ and $c=2$. Substituting these values into the equation, we get

$9 = k \frac{12}{2^2}$

$9 = k \frac{12}{4}$

$9 = 3k$

$k = 3$

b. We want to find the value of $a$ when $b=16$ and $c=4$. We know that $k=3$, so

$a = 3 \frac{16}{4^2} = 3 \frac{16}{16} = 3 \times 1 = 3$

c. We want to find the value of $c$ when $b=25$ and $a=3$. We know that $k=3$, so

$3 = 3 \frac{25}{c^2}$

$1 = \frac{25}{c^2}$

$c^2 = 25$

$c = \pm 5$

3. Final Answer

Question 1:

i. a. $n = 4$

i. b. $r = 2$

ii. $99$

Question 2:

i. $r = \frac{1}{3}$

ii. $\frac{1}{9}$

iii. $\frac{3280}{243}$

Question 3:

a. $k = 3$

b. $a = 3$

c. $c = \pm 5$