# Question 12 & 13 Exercise 4.2

Solutions of Question 12 & 13 of Exercise 4.2 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 12

A man earned dollars 3500 the first year he worked. If he received a raise of dollars 750 at the end of each year for 20 years, what was his salary during his twenty first year of work?

### Solution

Suppose $a_1$ represents salary of worker at first year. Then $$a_1=3500.$$ Increase in salary in each year $=d=750$. The given problem is of A.P and we have to find $a_{21}$.

As, we have \begin{align} a_{21}&=a_1+20d\\ &=3500+20(750) \\ &=18500. \end{align} Hence, the salary of the man during his 21st year of work is dollars 18,500.

## Question 13(i)

Find the arithmetic mean between $12$ and $18$.

### Solution

Here $a=12, b=18$.

Let say $A$ be arithmetic means. Then

\begin{align}A&=\dfrac{a+b}{2}\\&=\dfrac{12+18}{2}\\&=\dfrac{30}{2}=15.\end{align}
Hence 15 is A.M between 12 and 18.

## Question 13(ii)

Find the arithmetic mean between $\dfrac{1}{3}$ and $\dfrac{1}{4}$.

### Solution

Here $a=\dfrac{1}{3}, b=\dfrac{1}{4}$,

Let $A$ be arithmetic mean. Then

\begin{align}A&=\dfrac{a+b}{2}\\&=\dfrac{\dfrac{1}{3}+\dfrac{1}{4}}{2}\\&=\dfrac{\dfrac{4+3}{12}}{2}\\&=\dfrac{7}{24}\end{align}

## Question 13(iii)

Find the arithmetic mean between $-6,-216$.

### Solution

Here $a=-6, b=-216$.

Let $A$ be arithmetic mean. Then

$$A=\dfrac{a+b}{2}=\dfrac{-6-216}{2}=-111$$

## Question 13(iv)

Find the arithmetic mean between $(a+b)^2,(a-b)^2$.

### Solution

Here $a^{\prime}=(a+b)^2$, $b^{\prime}=(a-b)^2$.

Let $A$ be arithmetic mean. Then

\begin{align}
A&=\dfrac{a^{\prime}+b^{\prime}}{2}\\
&=\dfrac{(a+b)^2+(a-b)^2}{2} \\
\Rightarrow A&=\dfrac{a^2+b^2+2 a b+a^2+b^2-2 a b}{2} \\
& =\dfrac{2a^2+2b^2}{2}\\
&=a^2+b^2.\end{align}

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