Question 1: i. Given a geometric progression with terms $T_1 = n-2$, $T_2 = n$, and $T_3 = n+4$, find the value of $n$ and the common ratio. ii. Find the geometric mean of 81 and 121. Question 2: Given the geometric progression 9, 3, 1, ..., find the common ratio, the 5th term, and the sum of the first 8 terms. Question 3: Given that $a$ varies directly as $b$ and inversely as the square of $c$, and $a = 9$ when $b = 12$ and $c = 2$, find the value of $k$ in the relationship $a = \frac{kb}{c^2}$, the value of $a$ when $b = 16$ and $c = 4$, and the values of $c$ when $b = 25$ and $a = 3$.

1. Problem Description

Question 1:

i. Given a geometric progression with terms $T_1 = n-2$, $T_2 = n$, and $T_3 = n+4$, find the value of $n$ and the common ratio.

ii. Find the geometric mean of 81 and

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2

1.

Question 2:

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

Question 3:

Given that $a$ varies directly as $b$ and inversely as the square of $c$, and $a = 9$ when $b = 12$ and $c = 2$, find the value of $k$ in the relationship $a = \frac{kb}{c^2}$, the value of $a$ when $b = 16$ and $c = 4$, and the values 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:

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

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

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

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

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

$0 = 2n - 8$

$2n = 8$

$n = 4$

i. b. The common ratio $r = \frac{T_2}{T_1} = \frac{n}{n-2} = \frac{4}{4-2} = \frac{4}{2} = 2$.

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

Geometric mean of 81 and 121 is $\sqrt{81 \times 121} = \sqrt{9^2 \times 11^2} = 9 \times 11 = 99$.

Question 2:

i. The common ratio $r$ of the geometric progression 9, 3, 1, ... is $\frac{3}{9} = \frac{1}{3}$.

ii. The $n$th term of a geometric progression is given by $ar^{n-1}$, where $a$ is the first term. Here, $a = 9$ and $r = \frac{1}{3}$.

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 $S_n = \frac{a(1-r^n)}{1-r}$. Here, $a = 9$, $r = \frac{1}{3}$, and $n = 8$.

$S_8 = \frac{9(1-(\frac{1}{3})^8)}{1-\frac{1}{3}} = \frac{9(1-\frac{1}{6561})}{\frac{2}{3}} = \frac{9(\frac{6560}{6561})}{\frac{2}{3}} = 9 \times \frac{6560}{6561} \times \frac{3}{2} = \frac{27 \times 6560}{2 \times 6561} = \frac{27 \times 3280}{6561} = \frac{88560}{6561} = \frac{3280}{243}$

Question 3:

Given $a = \frac{kb}{c^2}$.

a. We are given $a = 9$, $b = 12$, and $c = 2$.

$9 = \frac{k \times 12}{2^2} = \frac{12k}{4} = 3k$

$k = \frac{9}{3} = 3$

b. We have $k = 3$, $b = 16$, and $c = 4$.

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

c. We have $k = 3$, $b = 25$, and $a = 3$.

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

$3c^2 = 75$

$c^2 = 25$

$c = \pm 5$

3. Final Answer

Question 1:

i. a. $n = 4$

i. b. $r = 2$

ii. Geometric mean = 99

Question 2:

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

ii. 5th term = $\frac{1}{9}$

iii. Sum of first 8 terms = $\frac{3280}{243}$

Question 3:

a. $k = 3$

b. $a = 3$

c. $c = \pm 5$