Question 12, Exercise 4.6
Solutions of Question 12 of Exercise 4.6 of Unit 04: Sequence and Series. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.
Question 12
Find four H.Ms. between $\dfrac{1}{3}$ and $\dfrac{1}{11}$.
Solution.
Let $H_1, H_2, H_3, H_4$ be four $H.Ms$ between $\dfrac{1}{3}$ and $\dfrac{1}{11}$.
Then $$\dfrac{1}{3},H_1, H_2, H_3, H_4, \dfrac{1}{11} \text{ are in H.P.}$$
$$\quad 3,\dfrac{1}{H_1},\dfrac{1}{H_2}, \dfrac{1}{H_3}, \dfrac{1}{H_4},11 \text{ are in A.P.}$$
Here $a_1=3$, $a_n=11$ and $n=6$.
Since general term of A.P is given as
$$a_n=a_1+(n-1)d.$$
Thus
\begin{align*}\\
&11 =3+(6-1)d\\
\implies &5d = 11-3\\
\implies &5d=8\\
\implies & d=\frac{8}{5}\end{align*}
Now
\begin{align*}
&\frac{1}{H_1}=a_1+d=3+\frac{8}{5}=\frac{23}{5}\\
\implies &H_1=\frac{5}{23}
\end{align*}
\begin{align*}
&\frac{1}{H_2}=a_1+2d=3+2\times\frac{8}{5}=\frac{31}{5}\\
\implies &H_2=\frac{5}{31}
\end{align*}
\begin{align*}
&\frac{1}{H_3}=a_1+3d=3+3\times\frac{8}{5}=\frac{39}{5}\\
\implies &H_3=\frac{5}{39}
\end{align*}
\begin{align*}
&\frac{1}{H_4}=a_1+4d=3+4\times\frac{8}{5}=\frac{47}{5}\\
\implies &H_4=\frac{5}{47}
\end{align*}
Hence $\dfrac{5}{23},\dfrac{5}{31},\dfrac{5}{39},\dfrac{5}{47}$ are four H.Ms. 
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