# Question 14 Exercise 4.2

Solutions of Question 14 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 14(i)

Insert three arithmetic means between 6 and 41.

### Solution

Let $A_1, A_2, A_3$ be three arithmetic means between 6 and 41. Then $6, A_1, A_2, A_3, 41$ are in A.P. We have $$a_1=6 \text{ and } a_6=41.$$ Now \begin{align}& a_5=11\\ \Rightarrow &a_1+4 d=41 \\ \Rightarrow &6+4 d=41 \\ \Rightarrow &d=\dfrac{41-6}{4}\\ &=\dfrac{35}{4}.\end{align} Now \begin{align} A_1&=a+d=6+\dfrac{35}{4} \\ &=\dfrac{59}{4}=14\dfrac{3}{4},\end{align} \begin{align} A_2&=a+2 d\\ &=6+2 \cdot \dfrac{35}{4} \\ &=\dfrac{47}{2}=23\dfrac{1}{2}\end{align} and \begin{align}A_3&=a+3 d=6+3 \cdot \dfrac{35}{4}\\ &=\dfrac{129}{4}=32\dfrac{1}{4}.\end{align} Hence three arithmetic means between 6 and 41 are $$14\dfrac{3}{4},23\dfrac{1}{2},32\dfrac{1}{4}.$$

## Question 14(ii)

Insert four arithmetic means between 17 and 32.

### Solution

Let $A_1, A_2, A_3, A_4$ be four arithmetic means between 17 and 32.

then $17, A_1, A_2, A_3, A_4, 32$ are in A.P, where
$$a_1=17 \text{ and } a_6=32.$$
Now
\begin{align} & a_6=a_1+5 d \\
\Rightarrow &17+5 d=32 \\
\Rightarrow &d=\dfrac{32-17}{5}\\
&=\dfrac{15}{5}=3 \end{align}
Now
\begin{align} A_1&=a_1+d=17+3=20,\\
A_2&=a_1+2 d=17+2(3)=23,\\
A_3&=a_1+3 d=17+3(3)=26 \end{align}
and
\begin{align}A_4&=17+4(3)=29. \end{align}
Hence four arithmetic means between 17 and 32 are
$$20,23,26,29.$$

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