2 Marks Questions

Question 12 Marks

Insert 5 G.M.s between $\frac{1}{3}$ and 9 and verify that their product is the 5 th power of the G.M. between $\frac{1}{3}$ and 9 .

Answer

Let $G_1, G_2, G_3, G_4$ and $G_5$, be five geometric means between $\frac{1}{3}$ and 9 . So, $\frac{1}{3}, G_1, G_2, G_3, G_4, G_5, 9$ are in GP., whose first term $=\frac{1}{3}$ and 7 th term $=9$
Let the common ratio of this sequence be $r$.
$\therefore \quad 9=a_7=\frac{1}{3} r^6 \Rightarrow r^6=27=(\sqrt{3})^6$
$\begin{aligned} & \therefore & r & =\sqrt{3} \\ \text { Now, } & & G _1 & =\frac{1}{3} r=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}},\end{aligned}$
$
\begin{array}{l}G_2=G_1 \cdot r=\frac{1}{\sqrt{3}} \cdot \sqrt{3}=1 \\
G_3=G_2 \cdot r=1 \cdot \sqrt{3}=\sqrt{3} \\
G_4=G_3 \cdot r=\sqrt{3} \cdot \sqrt{3}=3 \\
G_5=G_4 \cdot r=3 \cdot \sqrt{3}=3 \sqrt{3}
\end{array}$
Product of geometric means $=G_1, G_2, G_3, G_4, G_5$
$
\begin{array}{l}
=\frac{1}{\sqrt{3}} \times 1 \times \sqrt{3} \times 3 \times 3 \sqrt{3} \\
=9 \sqrt{3}.....(i)
\end{array}
$
The 5 th power of GM. between $\frac{1}{3}$ and 9
$
=\left[\sqrt{\frac{1}{3} \times 9}\right]^5=(\sqrt{3})^5
$
$
=(\sqrt{3})^4 \cdot \sqrt{3}=9 \sqrt{3}
$
From equations (i) and (ii), it is clear that given statement is true.

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Question 22 Marks

Find four numbers forming a geometric progression in which the third term is greater than the first term by 9 and the second term is greater than the 4th by 18.

Answer

Let the four terms of the geometric progression be $a$, $a r, a r^2$ and $a r^3$.
According to question,
$a r^2-a=9$
$\Rightarrow \quad a\left(r^2-1\right)=9$..........(i)
and $\quad a r-a r^3=18$
$\Rightarrow \quad a r\left(1-r^2\right)=18$.........(ii)
On dividing by equation (i) in equation (ii)
$\frac{a r\left(1-r^2\right)}{a\left(r^2-1\right)}=\frac{18}{9}$
$\Rightarrow \quad-r=2 \quad \Rightarrow r=-2$
Putting value of $r$ in equation (i)
$a(4-1)=9 \quad \Rightarrow a=3$
$\begin{array}{l}\therefore \text { Required numbers are } 3,3(-2), 3(-2)^2, 3(-2)^3 \\ \quad=3-6 \cdot 12-24 \text {}\end{array}$

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