Question 1: i. The terms $T_1$, $T_2$, and $T_3$ of a geometric progression are $n-2$, $n$, and $n+4$ respectively. Find: a. The value of $n$. b. The common ratio. ii. Find the geometric mean of 81 and 121. Question 2: Given the geometric progression 9, 3, 1, ..., find: i. The common ratio. ii. The 5th term. iii. 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 $c = 2$. Find: a. The value of $k$. b. Value of $a$ when $b = 16$ and $c = 4$. c. Values of $c$ when $b = 25$ and $a = 3$.

1. Problem Description

Question 1:

i. The terms $T_1$, $T_2$, and $T_3$ of a geometric progression are $n-2$, $n$, and $n+4$ respectively. Find:

a. The value of $n$.

b. The common ratio.

ii. Find the geometric mean of 81 and

1

2

1.

Question 2:

Given the geometric progression 9, 3, 1, ..., find:

i. The common ratio.

ii. The 5th term.

iii. 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 $c = 2$. Find:

a. The value of $k$.

b. Value of $a$ when $b = 16$ and $c = 4$.

c. Values of $c$ when $b = 25$ and $a = 3$.

2. Solution Steps

Question 1:

i.

a. In a geometric progression, the ratio of consecutive terms is constant. Thus, we have:

$\frac{T_2}{T_1} = \frac{T_3}{T_2}$

Substituting the given values:

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

Cross-multiplying:

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

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

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

$0 = 2n - 8$

$2n = 8$

$n = 4$

b. The common ratio $r$ is given by:

$r = \frac{T_2}{T_1} = \frac{n}{n-2}$

Since $n = 4$,

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

ii. The geometric mean of two numbers $x$ and $y$ is $\sqrt{xy}$.

Therefore, the geometric mean of 81 and 121 is:

$\sqrt{81 \times 121} = \sqrt{9^2 \times 11^2} = 9 \times 11 = 99$

Question 2:

Given the geometric progression 9, 3, 1, ...

i. The common ratio $r$ is given by:

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

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

The 5th term is:

$T_5 = 9 \times (\frac{1}{3})^{5-1} = 9 \times (\frac{1}{3})^4 = 9 \times \frac{1}{81} = \frac{1}{9}$

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:

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

Question 3:

It is given that $a$ varies directly as $b$ and inversely as the square of $c$.

$a = \frac{kb}{c^2}$, where $k$ is the constant of proportionality.

$a = 9$ when $b = 12$ and $c = 2$.

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

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

$9 = 3k$

$k = 3$

a. The value of $k$ is

3.

b. $a = \frac{3b}{c^2}$

When $b = 16$ and $c = 4$,

$a = \frac{3 \times 16}{4^2} = \frac{48}{16} = 3$

c. $a = \frac{3b}{c^2}$

When $b = 25$ and $a = 3$,

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

$3 = \frac{75}{c^2}$

$c^2 = \frac{75}{3} = 25$

$c = \sqrt{25} = 5$

3. Final Answer

Question 1:

i.

a. $n = 4$

b. $r = 2$

ii. $99$

Question 2:

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

ii. $T_5 = \frac{1}{9}$

iii. $S_8 = \frac{3280}{243}$

Question 3:

a. $k = 3$

b. $a = 3$

c. $c = 5$