The problem has two parts. Part (a) states that the first three terms of a geometric progression (GP) are $x$, $x+2$, and $x+3$ respectively. We are asked to find: (i) the value of $x$, (ii) the common ratio of the GP, (iii) the sum to infinity of the GP. Part (b) asks to find the geometric mean of 81 and 121.

1. Problem Description

The problem has two parts. Part (a) states that the first three terms of a geometric progression (GP) are $x$, $x+2$, and $x+3$ respectively. We are asked to find:

(i) the value of $x$,

(ii) the common ratio of the GP,

(iii) the sum to infinity of the GP.

Part (b) asks to find the geometric mean of 81 and

1

2

1.

2. Solution Steps

(a) For a geometric progression with consecutive terms $a, b, c$, we have $b/a = c/b$, which implies $b^2 = ac$.

Applying this to the given terms $x, x+2, x+3$, we have:

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

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

$4x - 3x = -4$

$x = -4$

(i) The value of $x$ is $-4$.

(ii) The terms are $x = -4$, $x+2 = -4+2 = -2$, and $x+3 = -4+3 = -1$.

The common ratio $r$ is given by the ratio of consecutive terms.

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

Also, $r = \frac{-1}{-2} = \frac{1}{2}$

The common ratio is $1/2$.

(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, provided $|r| < 1$.

In this case, $a = -4$ and $r = 1/2$. Since $|1/2| < 1$, the sum to infinity exists.

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

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

Here, $a = 81$ and $b = 121$.

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

3. Final Answer

(a)

(i) $x = -4$

(ii) The common ratio is $\frac{1}{2}$

(iii) Sum to infinity is $-8$

(b) The geometric mean of 81 and 121 is

9

9.