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.
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. GOOD
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