a) Given that the terms $T_1, T_2,$ and $T_3$ of a geometric progression (GP) are $x, x+2,$ and $x+3$ respectively, we need to find: i) The value of $x$. ii) The common ratio. iii) The sum to infinity. b) Find the geometric mean of 81 and 121.

1. Problem Description

a) Given that the terms $T_1, T_2,$ and $T_3$ of a geometric progression (GP) are $x, x+2,$ and $x+3$ respectively, we need to find:

i) The value of $x$.

ii) The common ratio.

iii) The sum to infinity.

b) Find the geometric mean of 81 and

1

2

1.

2. Solution Steps

a) i) In a GP, the ratio between consecutive terms is constant. Therefore,

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

Substituting the given values, we have:

$\frac{x+2}{x} = \frac{x+3}{x+2}$

Cross-multiplying gives:

$(x+2)(x+2) = x(x+3)$

$x^2 + 4x + 4 = x^2 + 3x$

$4x - 3x = -4$

$x = -4$

a) ii) The common ratio $r$ is given by $\frac{x+2}{x}$.

Substituting $x = -4$, we get:

$r = \frac{-4+2}{-4} = \frac{-2}{-4} = \frac{1}{2}$

a) iii) The sum to infinity of a GP is given by the formula:

$S_\infty = \frac{a}{1-r}$

where $a$ is the first term and $r$ is the common ratio.

In this case, $a = x = -4$ and $r = \frac{1}{2}$.

$S_\infty = \frac{-4}{1 - \frac{1}{2}} = \frac{-4}{\frac{1}{2}} = -4 \times 2 = -8$

b) The geometric mean of two numbers $a$ and $b$ is given by $\sqrt{ab}$.

In this case, $a = 81$ and $b = 121$.

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

3. Final Answer

a) i) $x = -4$

a) ii) The common ratio $= \frac{1}{2}$

a) iii) Sum to infinity $= -8$

b) The geometric mean of 81 and 121 is

9

9.