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.

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.

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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