Question 9 & 10, Exercise 4.6
Solutions of Question 9 & 10 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 9
Find the 8th term in H.P. $\frac{1}{7}, \frac{1}{6},-1,-\frac{1}{3}, \ldots$
Solution.
$$\frac{1}{7}, \frac{1}{6}, -1, -\frac{1}{3}, \ldots \text{ is in H.P.}$$ $$7, 6, -1, -3, \ldots \text{ is in A.P.}$$ Here $a_1 = 7$, $d = 6 - 7 = -1$, $a_8=?$
The general term of the A.P. is given as $$ a_n = a_1 + (n-1)d. $$ Thus, \begin{align*} a_8 &= 7 + (7)(-1) \\ &= 7 - 7 \\ &= 0 \quad \text{is in A.P.} \end{align*} Hence, $a_8 = \dfrac{1}{0} $ is in A.P.
Question 10
Find H.M. between 9 and 11 . Also find $A, H, G$ and show that $A H=G^{2}$.
Solution.
Here $a = 9, b = 11$
\begin{align*}
H &= \frac{2 a b}{a+b} = \frac{2(9)(11)}{9+11} \\
&= \frac{198}{21} =\frac{66}{7}
\end{align*}
Now
\begin{align*}
A & = \frac{a+b}{2} = \frac{9+11}{2}\\ &= \frac{21}{2}\\
G &= \sqrt{a b} = \sqrt{9 \times 11}\\& = \sqrt{99}.
\end{align*}
We have to prove
\begin{align*}
A H = G^{2}
\end{align*}
Now
\begin{align*}
L.H.S. &=A \times H = \frac{21}{2}\times \frac{66}{7} = 99\\
R.H.S. &=G^{2} = (\sqrt{99})^{2} = 99\end{align*}
Thus $A H = G^2$ is verified. 
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