# Question 8 Exercise 4.4

Solutions of Question 8 of Exercise 4.4 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

## Question 8(i)

Find the geometric mean of $3.14$ and $2.71$

### Solution

Here $a=3.14$ and $b=2.71$

then $$G= \pm \sqrt{(3.14)(2.71)}= \pm 2.94$$
Thus $$G=2.94 \quad \text{or} \quad -2.94$$

## Question 8(ii)

Find the geometric mean of $-6$ and $-216$

### Solution

Here $a=-6$ and $b=-216$ then

\begin{align}G&= \pm \sqrt{(-6)(-216)}= \pm \sqrt{1296} \\
\Rightarrow G&= \pm 36\end{align}
Thus

$$G=36 \quad \text{or} \quad -36$$

## Question 8(iii)

Find the geometric mean of $x+y$ and $x-y$

### Solution

Here $a=x+y$ and $b=x-y$

then $$G= \pm \sqrt{(x+y)(x-y)}= \pm \sqrt{x^2-y^2}$$

## Question 8(iv)

Find the geometric mean of $\sqrt{2}+3$ and $\sqrt{2}-3$

### Solution

Here $a=\sqrt{2}+3$ and \begin{align}b&=\sqrt{2}-3 \text { then } \\ G&= \pm \sqrt{(\sqrt{2}+3)(\sqrt{2}-3)} \\ \Rightarrow G&= \pm \sqrt{4-9}= \pm \sqrt{5}i\end{align} It does not exists.

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