Answer :

The given $G.P.$ is $\dfrac{5}{2}, \dfrac{5}{4}, \dfrac{5}{8}, \ldots$

Here, $a=$ First term $=\dfrac{5}{2}$

$r=\text{ Common ratio }=\dfrac{\dfrac{5}{4}}{\dfrac{5}{2}}=\dfrac{1}{2} $

$a _ {20}=a r^{20-1}=\dfrac{5}{2}\left(\dfrac{1}{2}\right)^{19}=\dfrac{5}{\left(2\right)\left(2\right)^{19}}=\dfrac{5}{\left(2\right)^{20}} $

$a_n=a r^{n-1}=\dfrac{5}{2}\left(\dfrac{1}{2}\right)^{n-1}=\dfrac{5}{\left(2\right)\left(2\right)^{n-1}}=\dfrac{5}{\left(2\right)^{n}}$

Answer :

Common ratio, $r=2$

Let $a$ be the first term of the $G.P.$

$ \begin{aligned} & a r^{8.1} \ \Rightarrow a r^{7}=192 \\ \\ & a\left(2\right)^{7}=192 \\ \\ & a\left(2\right)^{7}=\left(2\right)^{6}\left(3\right) \\ \\ & \Rightarrow a=\dfrac{\left(2\right)^{6} \times 3}{\left(2\right)^{7}}=\dfrac{3}{2} \\ \\ & \therefore \ \ a _ {12}=a r^{12-1}=\left(\dfrac{3}{2}\right)\left(2\right)^{11}=\left(3\right)\left(2\right)^{10}=3072 \end{aligned} $

Answer :

Let $a$ be the first term and $r$ be the common ratio of the $G.P.$

According to the given condition,

$a_5=a r^{5 - 1}=a r^{4}=p\qquad \ldots(1)$

$a_8=a r^{8 - 1}=a r^{7}=q \qquad \ldots(2)$

$a_{11}=a r^{11 -1}=a r^{10}=s \qquad \ldots(3)$

Dividing equation $\left(2\right)$ by $\left(1\right),$

we obtain $\dfrac{a r^{7}}{a r^{4}}=\dfrac{q}{p}$

$r^{3}=\dfrac{q}{p}\qquad \ldots(4)$

Dividing equation $\left(3\right)$ by $\left(2\right),$

we obtain $\dfrac{a r^{10}}{a r^{7}}=\dfrac{s}{q}$

$\Rightarrow r^{3}=\dfrac{s}{q}\qquad \ldots(5)$

Equating the values of $r^{3}$ obtained in $\left(4\right)$ and $\left(5\right),$

we obtain $\dfrac{q}{p}=\dfrac{s}{q}$

$\Rightarrow q^{2}=p s$

Thus, the given result is proved.

Answer :

Let $a$ be the first term and $r$ be the common ratio of the $G.P.$

$\therefore \ \ a=-3$

It is known that, $a_n=a r^{n-1}$

$\therefore \ \ a_4=a r^{3}=\left(-3\right) r^{3}$

$\quad \ \ a_2=a r^{1}=\left(-3\right) r$

According to the given condition,

$\left(-3\right) r^{3}=[\left(-3\right) r]^{2}$

$\Rightarrow-3 r^{3}=9 r^{2}$

$\Rightarrow r=-3$

Now, $7^{\text{th }}$ term of series is

$a r^{7-1}=a r^{6}=\left(-3\right)\left(-3\right)^{6}=-\left(3\right)^{7}=-2187$

Thus, the seventh term of the $G.P.$ is $-2187 .$

Answer :

$\left(a\right) \ $ The given sequence is $2,2 \sqrt{2}, 4, \ldots$

Here, $a=2$ and $r=\dfrac{2 \sqrt{2}}{2}=\sqrt{2}$

Let the $n^{\text{th }}$ term of the given sequence be $128 .$

$a_n=a r^{n-1}$

$\Rightarrow\left(2\right)\left(\sqrt{2}\right)^{n-1}=128$

$\Rightarrow\left(2\right)\left(2\right)^{\frac{n-1}{2}}=\left(2\right)^{7}$

$\Rightarrow\left(2\right)^{\frac{n-1}{2}+1}=\left(2\right)^{7}$

On comparing both sides.

$\therefore \ \ \dfrac{n-1}{2}+1=7$

$\Rightarrow \dfrac{n-1}{2}=6$

$\Rightarrow n-1=12$

$\Rightarrow n=13$

Thus, the $13^{\text{th }}$ term of the given sequence is $128 .$

$\left(b\right) \ $ The given sequence is $\sqrt{3}, 3,3 \sqrt{3}, \ldots$

Here,

$ a=\sqrt{3} \text{ and } r=\dfrac{3}{\sqrt{3}}=\sqrt{3} $

Let the $n^{\text{th }}$ term of the given sequence be $729 .$

$\qquad a_n=a r^{n-1}$

$\therefore \ \ a r^{n-1}=729$

$\Rightarrow\left(\sqrt{3}\right)\left(\sqrt{3}\right)^{n-1}=729$

$\Rightarrow\left(3\right)^{\frac{1}{2}}\left(3\right)^{\frac{n-1}{2}}=\left(3\right)^{6}$

$\Rightarrow\left(3\right)^{\frac{1}{2}+\frac{n-1}{2}}=\left(3\right)^{6}$

On comparing both sides.

$\therefore \ \ \dfrac{1}{2}+\dfrac{n-1}{2}=6$

$\Rightarrow \dfrac{1+n-1}{2}=6$

$\Rightarrow n=12$

Thus, the $12^{\text{th }}$ term of the given sequence is $729 .$

$\left(c\right) \ $ The given sequence is $\dfrac{1}{3}, \dfrac{1}{9}, \dfrac{1}{27}, \ldots$

Here, $\quad a=\dfrac{1}{3}$ and $r=\dfrac{1}{9} \div \dfrac{1}{3}=\dfrac{1}{3}$

Let the $n^{\text{th }}$ term of the given sequence be $\dfrac{1}{19683}$

$a_n=a r^{n-1}$

$\therefore \ \ a r^{n-1}=\dfrac{1}{19683}$

$\Rightarrow\left(\dfrac{1}{3}\right)\left(\dfrac{1}{3}\right)^{n-1}=\dfrac{1}{19683}$

$\Rightarrow\left(\dfrac{1}{3}\right)^{n}=\left(\dfrac{1}{3}\right)^{9}$

On comparing both sides.

$\Rightarrow n=9$

Thus, the $9^{\text{th }}$ term of the given sequence is $\dfrac{1}{19683}$

Answer :

The given numbers are $\dfrac{-2}{7}, x, \dfrac{-7}{2}$

$ \text { Common ratio }=\dfrac{x}{-2 / 7}=\dfrac{-7 x}{2} $

$ \text { Also, common ratio }=\dfrac{-7 / 2}{x}=\dfrac{-7}{2 x} $

$ \begin{aligned} & \therefore \ \ \dfrac{-7 x}{2}=\dfrac{-7}{2 x} \\ & \Rightarrow x^2=\dfrac{-2 \times 7}{-2 \times 7}=1 \\ & \Rightarrow x=\sqrt{1} \\ & \Rightarrow x= \pm 1 \end{aligned} $

Thus, for $x= \pm 1$, the given numbers will be in G.P.

Answer :

The given $G.P.$ is $0.15,0.015,0.00015, \ldots$

Here, $a=0.15$ and

$ r=\dfrac{0.015}{0.15}=0.1 $

$S_n=\dfrac{a\left(1-r^{n}\right)}{1-r}$

$\therefore \ \ S _ {20}=\dfrac{0.15[1-\left(0.1\right)^{20}]}{1-0.1}$

$ \qquad\quad=\dfrac{0.15}{0.9}[1-\left(0.1\right)^{20}] $

$ \qquad\quad=\dfrac{15}{90}[1-\left(0.1\right)^{20}] $

$\qquad\quad=\dfrac{1}{6}[1-\left(0.1\right)^{20}]$

Answer :

The given G.P. is $\sqrt{7}, \sqrt{21}, 3 \sqrt{7}, \ldots$

Here, $a=\sqrt{7}$

$ \ r=\dfrac{\sqrt{21}}{\sqrt{7}}=\sqrt{3}$

$S _ {n}=\dfrac{a\left(1-r^{n}\right)}{1-r}$

$\therefore \ \ S_n=\dfrac{\sqrt{7}[1-\left(\sqrt{3}\right)^{n}]}{1-\sqrt{3}}$

$\qquad \quad =\dfrac{\sqrt{7}[1-\left(\sqrt{3}\right)^{n}]}{1-\sqrt{3}} \times \dfrac{1+\sqrt{3}}{1+\sqrt{3}}$

$($ By rationalizing $)$

$\qquad \quad =\dfrac{\sqrt{7}\left(1+\sqrt{3}\right)[1-\left(\sqrt{3}\right)^{n}]}{1-3}$

$\qquad \quad =\dfrac{-\sqrt{7}\left(1+\sqrt{3}\right)}{2}\big[1-\left(3\right)^{\frac{n}{2}}\big]$

$\qquad \quad =\dfrac{\sqrt{7}\left(1+\sqrt{3}\right)}{2}\big[\left(3\right)^{\frac{n}{2}}-1\big]$

Answer :

The given $G.P.$ is $1,-a, a^{2},-a^{3}, \ldots n$ terms

Here, first term $=a_1=1$

Common ratio $=r=- a$

$ \begin{aligned} & S_n=\dfrac{a_1\left(1-r^{n}\right)}{1-r} \\ \\ & \therefore \ \ S_n=\dfrac{1[1-\left(-a\right)^{n}]}{1-\left(-a\right)}=\dfrac{[1-\left(-a\right)^{n}]}{1+a} \end{aligned} $

Answer :

The given $G.P.$ is $x^{3}, x^{5}, x^{7}, \ldots$

Here,

$a=x^{3}$ and $r=x^{2}$

$S_n=\dfrac{a\left(1-r^{n}\right)}{1-r}=\dfrac{x^{3}[1-\left(x^{2}\right)^{n}]}{1-x^{2}}=\dfrac{x^{3}\left(1-x^{2 n}\right)}{1-x^{2}}$

Answer :

$\sum _ {k=1}^{11}\left(2+3^{k}\right)=\sum _ {k=1}^{11}\left(2\right)+\sum _ {k=1}^{11} 3^{k}=2\left(11\right)+\sum _ {k=1}^{11} 3^{k}=22+\sum _ {k=1}^{11} 3^{k}\qquad \ldots(1)$

$\sum _ {k=1}^{11} 3^{k}=3^{1}+3^{2}+3^{3}+\ldots+3^{11}$

The terms of this sequence $3,3^{2}, 3^{3}, \ldots$ forms a $G.P.$

$\quad S_n=\dfrac{a\left(r^{n}-1\right)}{r-1}$

$\Rightarrow S _ {11}=\dfrac{3[\left(3\right)^{11}-1]}{3-1}$

$\Rightarrow S _ {11}=\dfrac{3}{2}\left(3^{11}-1\right)$

$\therefore \ \ \sum _ {k=1}^{11} 3^{k}=\dfrac{3}{2}\left(3^{11}-1\right)$

Substituting this value in equation $\left(1\right),$

we obtain $\sum _ {k=1}^{11}\left(2+3^{k}\right)=22+\dfrac{3}{2}\left(3^{11}-1\right)$

Answer :

Let $\dfrac{a}{r}, a, a r$ be the first three terms of the $G.P.$

$\dfrac{a}{r}+a+a r=\dfrac{39}{10}\qquad \ldots(1)$

$\left(\dfrac{a}{r}\right)\left(a\right)\left(a r\right)=1\qquad \ldots(2)$

From $\left(2\right),$

we obtain $a^{3}=1$

$\Rightarrow a=1$ (Considering real roots only)

Substituting $a=1$ in equation $\left(1\right),$

we obtain $\dfrac{1}{r}+1+r=\dfrac{39}{10}$

$\Rightarrow 1+r+r^{2}=\dfrac{39}{10} r$

$\Rightarrow 10+10 r+10 r^{2}-39 r=0$

$\Rightarrow 10 r^{2}-29 r+10=0$

$\Rightarrow 10 r^{2}-25 r-4 r+10=0$

$\Rightarrow 5 r\left(2 r-5\right)-2\left(2 r-5\right)=0$

$\Rightarrow\left(5 r-2\right)\left(2 r-5\right)=0$

$\Rightarrow r=\dfrac{2}{5}$ or $\dfrac{5}{2}$

Thus, the three terms of $G.P.$ are $\dfrac{5}{2}, 1$, and $\dfrac{2}{5}$.

Answer :

The given $G.P.$ is $3,3^{2}, 3^{3}, \ldots$

Let $n$ terms of this $G.P.$ be required to obtain the sum as $120 .$

$S_n=\dfrac{a\left(r^{n}-1\right)}{r-1}$

Here, $a=3$ and $r=3$

$\therefore \ \ S_n=120=\dfrac{3\left(3^{n}-1\right)}{3-1}$

$\Rightarrow 120=\dfrac{3\left(3^{n}-1\right)}{2}$

$\Rightarrow \dfrac{120 \times 2}{3}=3^{n}-1$

$\Rightarrow 3^{n}-1=80$

$\Rightarrow 3^{n}=81$

$\Rightarrow 3^{n}=3^{4}$

$\therefore \ \ n=4$

Thus, four terms of the given $G.P.$ are required to obtain the sum as $120 .$

Answer :

Let the $G.P.$ be $a, a r, a r^{2}, a r^{3}, \ldots$

According to the given condition,

$a+a r+a r^{2}=16$ and $a r^{3}+a r^{4}+a r^{5}=128$

$\Rightarrow a\left(1+r+r^{2}\right)=16\qquad \ldots(1)$

$\Rightarrow ar^{3}\left(1+r+r^{2}\right)=128\qquad \ldots(2)$

Dividing equation $\left(2\right)$ by $\left(1\right),$

we obtain $\dfrac{a r^{3}\left(1+r+r^{2}\right)}{a\left(1+r+r^{2}\right)}=\dfrac{128}{16}$

$\Rightarrow r^{3}=8$

$\therefore \ \ r=2$

Substituting $r=2$ in $\left(1\right),$

we obtain $a\left(1+2+4\right)=16$

$\Rightarrow a\left(7\right)=16$

$\Rightarrow a=\dfrac{16}{7}$

$\quad S_n=\dfrac{a\left(r^{n}-1\right)}{r-1}$

$\Rightarrow S_n=\dfrac{16}{7} \dfrac{\left(2^{n}-1\right)}{(2-1)}=\dfrac{16}{7}\left(2^{n}-1\right)$

Answer :

$a=729$

$a_7=64$

Let $r$ be the common ratio of the $G.P.$

It is known that, $a_n = ar^{n-1}$

$a_7=a r ^{7-1}=\left(729\right) r^{6}$

$\Rightarrow 64=729 r^{6}$

$\Rightarrow r^{6}=\dfrac{64}{729}$

$\Rightarrow r^{6}=\left(\dfrac{2}{3}\right)^{6}$

$\Rightarrow r=\dfrac{2}{3}$

Also, it is known that, $S_n=\dfrac{a\left(1-r^{n}\right)}{1-r}$

$ \begin{aligned} \therefore \ \ S_7 & =\dfrac{729\bigg[1-\left(\dfrac{2}{3}\right)^{7}\bigg]}{1-\dfrac{2}{3}} \\ \\ & =3 \times 729\bigg[1-\left(\dfrac{2}{3}\right)^{7}\bigg] \\ \\ & =\left(3\right)^{7}\bigg[\dfrac{\left(3\right)^{7}-\left(2\right)^{7}}{\left(3\right)^{7}}\bigg] \\ \\ & =\left(3\right)^{7}-\left(2\right)^{7} \\ \\ & =2187-128 \\ \\ & =2059 \end{aligned} $

Answer :

Then $G.P.$ is $a, a r, a r^2,\ldots$

Given, $T_1+T_2=-4\qquad \ldots(1)$

and $T_5=4 T_3\qquad \ldots(2)$

Now, $(1)$ implies,

$ a+a r=-4 \Rightarrow a(1+r)=-4\qquad \ldots(3) $

and $(2)$ implies,

$ \begin{aligned} & a r^4=4 a r^2 \\ \\ & \Rightarrow r^4-4 r^2=0 \qquad [\because \ \ a \neq 0] \\ & \Rightarrow r^2\left(r^2-4\right)=0 \\ \\ & \Rightarrow r^2=4 \qquad [\because \ \ r \neq 0] \\ & \Rightarrow r= \pm 2 \Rightarrow r=2 \text { or } r=-2 \end{aligned} $

Substitute $r=2$ in $(3),$ we get

$ a(1+2)=-4 \Rightarrow a=\frac{-4}{3} $

Substitute $r=-2$, in $(3),$ we get

$ a(1-2)=-4 \Rightarrow a=4 $

Case (i): If $a=-\frac{4}{3}, r=2$, then required $G.P.$ is

$ \begin{aligned} & \frac{-4}{3}, \frac{-4}{3} \times 2, \frac{-4}{3} \times 2^{2}, \ldots \\ & \text { i.e., } \frac{-4}{3}, \frac{-8}{3}, \frac{-16}{3}, \cdots \end{aligned} $

Case (ii): If $a=4, r=-2$, then required $G.P. \ $ is $ \ 4,4 \times(-2), 4 \times(-2)^2, \cdots \cdots$

ie., $4,-8,16, \cdots \cdots$

Answer :

Let $a$ be the first term and $r$ be the common ratio of the $G.P.$

According to the given condition,

$a_4=a r^{3}=x\qquad \ldots(1)$

$a _ {10}=a r^{9}=y\qquad \ldots(2)$

$a _ {16}=a r^{15}=z\qquad \ldots(3)$

Dividing $\left(2\right)$ by $\left(1\right),$

we obtain $\dfrac{y}{x}=\dfrac{a r^{9}}{a r^{3}} $

$\Rightarrow \dfrac{y}{x}=r^{6}$

Dividing $\left(3\right)$ by $\left(2\right),$

we obtain $\dfrac{z}{y}=\dfrac{a r^{15}}{a r^{9}} $

$\Rightarrow \dfrac{z}{y}=r^{6}$

$\therefore \ \ \dfrac{y}{x}=\dfrac{z}{y}$

Thus, $x, y, z$ are in $G. P.$

Answer :

The given sequence is $8,88,888,8888 \ldots$

This sequence is not a G.P. However, it can be changed to G.P. by writing the terms as

$S_n=8+88+888+8888+$ to $n$ terms

$\quad \ =\dfrac{8}{9}\bigg[9+99+999+9999+….$ to $n$ terms $\bigg]$

$\quad \ =\dfrac{8}{9}\bigg[\left(10-1\right)+\left(10^{2}-1\right)+\left(10^{3}-1\right)+\left(10^{4}-1\right)+\ldots \ldots $ to $n$ terms $\bigg]$

$\quad \ =\dfrac{8}{9}\bigg[(10+10^{2}+\ldots . . n..$ terms $)-(1+1+1+\ldots . n$ terms $.)\bigg]$

$\quad \ =\dfrac{8}{9}\bigg[\dfrac{10\left(10^{n}-1\right)}{10-1}-n\bigg]$

$\quad \ =\dfrac{8}{9}\bigg[\dfrac{10\left(10^{n}-1\right)}{9}-n\bigg]$

$\quad \ =\dfrac{80}{81}\left(10^{n}-1\right)-\dfrac{8}{9} n$

Answer :

Required sum $=2 \times 128+4 \times 32+8 \times 8+16 \times 2+32 \times \dfrac{1}{2}$

$\hspace{1.9cm}=64\bigg[4+2+1+\dfrac{1}{2}+\dfrac{1}{2^{2}}\bigg]$

Here, $4,2,1, \dfrac{1}{2}, \dfrac{1}{2^{2}} \ $ is a $G.P.$

First term, $a=4$

Common ratio, $r=\dfrac{1}{2}$

It is known that, $S_n=\dfrac{a\left(1-r^{n}\right)}{1-r}$

$\therefore \ \ S_5=\dfrac{4\bigg[1-\left(\dfrac{1}{2}\right)^{5}\bigg]}{1-\dfrac{1}{2}}=\dfrac{4\bigg[1-\dfrac{1}{32}\bigg]}{\dfrac{1}{2}}=8\left(\dfrac{32-1}{32}\right)=\dfrac{31}{4}$

$\therefore \ \ $ Required sum $=$ $ 64\left(\dfrac{31}{4}\right)=\left(16\right)\left(31\right)=496 $

Answer :

It has to be proved that the sequence, $a A, \ a r A R, \ a r^{2} A R^{2}, \ldots a r^{n -1} A R^{n - 1}$, forms a $G.P.$

$ \begin{aligned} & \dfrac{\text{ Second term }}{\text{ First term }}=\dfrac{a r A R}{a A}=r R \\ \\ & \dfrac{\text{ Third term }}{\text{ Second term }}=\dfrac{a r^{2} A R^{2}}{a r A R}=r R \end{aligned} $

Thus, the above sequence forms a $G.P.$ and the common ratio is $r R$.

Answer :

Let $a$ be the first term and $r$ be the common ratio of the $G.P.$

$a_1=a, a_2=a r, a_3=a r^{2}, a_4=a r^{3}$

By the given condition,

$a_3=a_1+9$

$\Rightarrow a r^{2}=a+9\qquad \ldots(1)$

$a_2=a_4+18$

$\Rightarrow a r=a r^{3}+18\qquad \ldots(2)$

From $\left(1\right)$ and $\left(2\right),$

we obtain

$a\left(r^{2}- 1\right)=9\qquad \ldots(3)$

$ar\left(1-{2}\right)=18\qquad \ldots(4)$

Dividing $\left(4\right)$ by $\left(3\right),$

we obtain $\dfrac{a r\left(1-r^{2}\right)}{a\left(r^{2}-1\right)}=\dfrac{18}{9}$

$\Rightarrow-r=2$

$\Rightarrow r=-2$

Substituting the value of $r$ in $\left(1\right),$

we obtain $4 a=a+9$

$\Rightarrow 3 a=9$

$\therefore \ \ a=3$

Thus, the first four numbers of the $G.P.$ are $3,-6,12,-24$

Answer :

Let $A$ be the first term and $R$ be the common ratio of the $G.P.$

According to the given information,

$A R^{p-1}=a$

$A R^{q-1}=b$

$A R^{r-1}=c$

$a^{q-r} b^{r-p} C^{p-q}$ $=A^{q-r} \times R^{\left(p-1\right)\left(q-r\right)} \times A^{r-p} \times R^{\left(q-1\right)\left(r-p\right)} \times A^{p-q} \times R^{\left(r-1\right)\left(p-q\right)}$

$\hspace{2cm}=A ^{(q-r+r-p+p-q)} \times R^{\left( p q-p r-q+r\right)+\left(r q-r+p-p q\right)+\left(p r-p-q r+q\right)}$

$\hspace{2cm}=A^{0} \times R^{0}=1$

Thus, the given result is proved.

Answer :

The first term of the $G.P$ is $a$ and the last term is $b$.

Therefore, the $G.P.$ is $a, a r, a r^{2}, a r^{3}, \ldots a r^{n-1}$, where $r$ is the common ratio.

$b=ar^{n-1}$

$P=$ Product of $n$ terms

$\quad=\left(a\right)\left(a r\right)\left(a r^{2}\right) \ldots\left(a r^{n-1}\right)$

$\quad=\left(a \times a \times \ldots a\right)\left(r \times r^{2} \times \ldots r^{{n-1}}\right)$

$\quad=a^{n} r^{1+2+3\ldots \left(n-1\right) }$

Here, $1,2,\ldots\left(n-1\right)$ is an A.P.

$\therefore \ \ 1+2+\ldots \ldots \ldots+\left(n- 1\right)=\dfrac{n-1}{2}\big[2+\left(n-1-1\right) \times 1\big]=\dfrac{n-1}{2}\big[2+n-2\big]=\dfrac{n\left(n-1\right)}{2}$

$\qquad P=a^{n} r^{\frac{n\left(n-1\right)}{2}}$

$\therefore \ \ P^{2}=a^{2 n} r^{n\left(n-1\right)}$

$\quad\qquad=\big[a^{2} r^{\left(n-1\right)}\big]^{n}$

$\quad\qquad=\big[a \times ar^{n-1}\big]^{n}$

$\quad\qquad=\left(ab\right)^{n}$

Thus, the given result is proved.

Answer :

Sum of first n terms,

$ S_{\mathrm{n}}=\dfrac{\mathrm{a}\left(1-\mathrm{r}^{\mathrm{n}}\right)}{1-\mathrm{r}} $

Sum of terms from $(n+1)$ th to $2 n$th term is,

$ \begin{aligned} S_{2 n}-S_n & =\dfrac{a\left(1-r^{2 n}\right)}{1-r}-\dfrac{a\left(1-r^n\right)}{1-r} \\ \\ & =\dfrac{a\left(r^n-r^{2 n}\right)}{1-r} \end{aligned} $

Ratio,

$\hspace{1.1cm} \begin{aligned} & =\dfrac{\dfrac{a\left(1-r^n\right)}{1-r}}{\dfrac{a\left(r^n-r^{2 n}\right)}{1-r}} =\dfrac{1-r^n}{r^n-r^{2 n}} \\ \\ & =\dfrac{1-r^n}{r^n\left(1-r^n\right)} =\dfrac{1}{r^n} \end{aligned} $

Hence proved.

Answer :

$a, b, c, d$ are in $G.P.$

Therefore,

$b c=a d\qquad \ldots(1)$

$b^{2}=a c\qquad \ldots(2)$

$c^{2}=b d\qquad \ldots(3)$

It has to be proved that, $\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=\left(a b+b c \text{ - } c d\right)^{2}$

$\text{R.H.S.}=\left(a b+b c+c d\right)^{2}$

$\hspace{1.2cm}=\left(a b+a d+c d\right)^{2}\quad[$ Using $\left(1\right)]$

$\hspace{1.2cm}=[a b+d\left(a+c\right)]^{2}$

$\hspace{1cm} \begin{aligned} & =a^{2} b^{2}+2 a b d\left(a+c\right)+d^{2}\left(a+c\right)^{2} \\ \\ & =a^{2} b^{2}+2 a^{2} b d+2 a c b d+d^{2}\left(a^{2}+2 a c+c^{2}\right) \\ \\ & =a^{2} b^{2}+2 a^{2} c^{2}+2 b^{2} c^{2}+d^{2} a^{2}+2 d^{2} b^{2}+d^{2} c^{2}\quad[\text{ Using (1) and (2)}] \\ \\ & =a^{2} b^{2}+a^{2} c^{2}+a^{2} c^{2}+b^{2} c^{2}+b^{2} c^{2}+d^{2} a^{2}+d^{2} b^{2}+d^{2} b^{2}+d^{2} c^{2} \\ \\ & =a^{2} b^{2}+a^{2} c^{2}+a^{2} d^{2}+b^{2} \times b^{2}+b^{2} c^{2}+b^{2} d^{2}+c^{2} b^{2}+c^{2} \times c^{2}+c^{2} d^{2} \end{aligned} $

$[$ Using $\left(2\right)$ and $\left(3\right)$ and rearranging terms $]$

$\hspace{1.1cm}=a^{2}\left(b^{2}+c^{2}+d^{2}\right)+b^{2}\left(b^{2}+c^{2}+d^{2}\right)+c^{2}\left(b^{2}+c^{2}+d^{2}\right)$

$\hspace{1.1cm}=\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)$

$\hspace{1.1cm}=\text{L.H.S.}$

$\therefore \ \ \text{L.H.S.} =\text{R.H.S.}$

$\therefore \ \ \left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=\left(a b+b c+c d\right)^{2}$

Answer :

Let $G_1$ and $G_2$ be two numbers between $3$ and $81$ such that the series, $3, G_1, G_2, 81,$ forms a $G.P.$

Let $a$ be the first term and $r$ be the common ratio of the $G.P.$

$\therefore \ \ 81=\left(3\right)\left(r\right)^{3}$

$\Rightarrow r^{3}=27$

$\therefore \ \ r=3$ (Taking real roots only)

For $r=3$,

$G_1=a r=\left(3\right)\left(3\right)=9$

$G_2=a r^{2}=\left(3\right)\left(3\right)^{2}=27$

Thus, the required two numbers are $9$ and $27.$

Answer :

G. M. of $a$ and $b$ is $\sqrt{a b}$.

By the given condition, $\dfrac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}=\sqrt{a b}$

Squaring both sides,

we obtain $ \dfrac{\left(a^{n+1}+b^{n+1}\right)^{2}}{\left(a^{n}+b^{n}\right)^{2}}=a b $

$\Rightarrow a^{2 n+2}+2 a^{n+1} b^{n+1}+b^{2 n+2}=\left(a b\right)\left(a^{2 n}+2 a^{n} b^{n}+b^{2 n}\right)$

$\Rightarrow a^{2 n+2}+2 a^{n+1} b^{n+1}+b^{2 n+2}=a^{2 n+1} b+2 a^{n+1} b^{n+1}+a b^{2 n+1}$

$\Rightarrow a^{2 n+2}+b^{2 n+2}=a^{2 n+1} b+a b^{2 n+1}$

$\Rightarrow a^{2 n+2}-a^{2 n+1} b=a b^{2 n+1}-b^{2 n+2}$

$\Rightarrow a^{2 n+1}\left(a-b\right)=b^{2 n+1}\left(a-b\right)$

$\Rightarrow\left(\dfrac{a}{b}\right)^{2 n+1}=1=\left(\dfrac{a}{b}\right)^{0}$

On comparing both sides.

$\Rightarrow 2 n+1=0$

$\Rightarrow n=\dfrac{-1}{2}$

Answer :

Let the two numbers be $a$ and $b$.

G.M. $=\sqrt{a b}$

According to the given condition,

$a+b=6 \sqrt{a b}\qquad \ldots(1)$

$\Rightarrow\left(a+b\right)^{2}=36\left(a b\right)$

Also,

$\left(a-b\right)^{2}=\left(a+b\right)^{2}-4 a b=36 a b-4 a b=32 a b$

$\Rightarrow a-b=\sqrt{32} \sqrt{a b}$

$\Rightarrow a-b=4 \sqrt{2} \sqrt{a b}\qquad \ldots(2)$

Adding $\left(1\right)$ and $\left(2\right),$

we obtain $2 a=\left(6+4 \sqrt{2}\right) \sqrt{a b}$

$\Rightarrow a=\left(3+2 \sqrt{2}\right) \sqrt{a b}$

Substituting the value of $a$ in $\left(1\right),$

we obtain $b=6 \sqrt{a b}-\left(3+2 \sqrt{2}\right) \sqrt{a b}$

$\Rightarrow b=\left(3-2 \sqrt{2}\right) \sqrt{a b}$

$\dfrac{a}{b}=\dfrac{\left(3+2 \sqrt{2}\right) \sqrt{a b}}{\left(3-2 \sqrt{2}\right) \sqrt{a b}}=\dfrac{3+2 \sqrt{2}}{3-2 \sqrt{2}}$

Thus, the required ratio is $\left(3+2 \sqrt{2}\right):\left(3-2 \sqrt{2}\right)$.

Answer :

It is given that $A$ and $G$ are $A.M.$ and $G.M.$ between two positive numbers. Let these two positive numbers be $a$ and $b$.

$\therefore \ \ AM=A=\dfrac{a+b}{2}\qquad \ldots(1)$

$GM=G=\sqrt{ab}\qquad \ldots(2)$

From $\left(1\right)$ and $\left(2\right),$

we obtain $a+b=2 A\qquad \ldots(3)$

$a b=G^{2}\qquad \ldots(4)$

Substituting the value of $a$ and $b$ from $\left(3\right)$ and $\left(4\right)$ in the identity $\left(a - b\right)^{2}=\left(a+b\right)^{2}- 4 a b$,

we obtain $\left(a- b\right)^{2}=4 A^{2}- 4 G^{2}=4\left(A^{2} -G^{2}\right)$

$\left(a - b\right)^{2}=4\left(A+G\right)\left(A- G\right)$

$\left(a-b\right)=2 \sqrt{\left(A+G\right)\left(A-G\right)}\qquad \ldots(5)$

From $\left(3\right)$ and $\left(5\right),$ we obtain

$ \begin{aligned} & 2 a=2 A+2 \sqrt{\left(A+G\right)\left(A-G\right)} \\ \\ & \Rightarrow a=A+\sqrt{\left(A+G\right)\left(A-G\right)} \end{aligned} $

Substituting the value of $a$ in $\left(3\right),$

we obtain $b=2 A-A-\sqrt{\left(A+G\right)\left(A-G\right)}=A-\sqrt{\left(A+G\right)\left(A-G\right)}$

Thus, the two numbers are $ A \pm \sqrt{\left(A+G\right)\left(A-G\right)} $

Answer :

It is given that the number of bacteria doubles every hour.

Therefore, the number of bacteria after every hour will form a $G.P.$

with first term $(a=30)$ and common ratio $(r=2)$

$ \therefore \ a_3=a^2=(30)(2)^2=120 $

Therefore, the number of bacteria at the end of $2nd$ hour will be $480 $.

$a_5=a r^4=(30)(2)^4=480 \ $ and $ \ a_{n+1}=a r^n=(30)(2)^n$

Thus the number of bacteria at the end of $n$th hour will be $(30)(2)^n$

Answer :

The principal is Rs. $500 $

So, the interest for this principal for one year is $500\left(\dfrac{10}{100}\right)=50$

Thus, the principal for the $2^{\text {nd }}$ year $=$ Principal for $1^{\text {st }}$ year + Interest

$ \hspace{4.6cm}=500+500\left(\dfrac{10}{100}\right)=500\left(1+\dfrac{10}{100}\right) $

Now, the interest for the second year $=\left(500\left(1+\dfrac{10}{100}\right)\right)\left(1+\dfrac{10}{100}\right)$

So, the principal for the third year $=500\left(1+\dfrac{10}{100}\right)+500\left(1+\dfrac{10}{100}\right) \dfrac{10}{100}$

$ \hspace{4.4cm}=500\left(1+\dfrac{10}{100}\right)^2 $

Continuing in this way we see that the principal for the $\mathrm{n}^{\text {th }}$ year $ =500\left(1+\dfrac{10}{100}\right)^{\mathrm{n}-1} $

The amount at the end of $(n-1)^{\text {th }}$ year $=$ Principal for the $n^{\text {th }}$ year.

Thus, the amount in the account at the end of $\mathrm{n}^{\text {th }}$ year.

$ =500\left(1+\dfrac{10}{100}\right)^{\mathrm{n}-1}$ $=500\left(1+\dfrac{10}{100}\right)^{\mathrm{n}-1}\left(\dfrac{10}{100}\right)$ $=500\left(\dfrac{11}{10}\right)^{\mathrm{n}} $

The amount in the account at the end of $10^{\text {th }}$ year

$ =\text { Rs. } 500\left(1+\dfrac{10}{100}\right)^{10}$

$=\operatorname{Rs} .500\left(\dfrac{11}{10}\right)^{10} $

Answer :

Let the root of the quadratic equation be $a$ and $b$.

According to the given condition,