Answer :

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

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

$ \begin{gathered} r=\text{ Common ratio }= \quad \dfrac{\dfrac{5}{4}}{\dfrac{5}{2}}=\dfrac{1}{2}\\ a _{20}=a r^{20-1}=\dfrac{5}{2}(\dfrac{1}{2})^{19}=\dfrac{5}{(2)(2)^{19}}=\dfrac{5}{(2)^{20}} \\ a_n=a r^{n-1}=\dfrac{5}{2}(\dfrac{1}{2})^{n-1}=\dfrac{5}{(2)(2)^{n-1}}=\dfrac{5}{(2)^{n}} \end{gathered} $

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(2)^{7}=192 \\ & a(2)^{7}=(2)^{6}(3) \\ & \Rightarrow a=\dfrac{(2)^{6} \times 3}{(2)^{7}}=\dfrac{3}{2} \\ & \therefore a _{12}=a r^{12-1}=(\dfrac{3}{2})(2)^{11}=(3)(2)^{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,

$a_5=a r^{5 - 1}=a r^{4}=p$.

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

$a_{11}=a r^{11 -1}=a r^{10}=s \ldots$

Dividing equation (2) by (1), we obtain

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

$r^{3}=\dfrac{q}{p}$

Dividing equation (3) by (2), we obtain

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

$\Rightarrow r^{3}=\dfrac{s}{q}$

Equating the values of $r^{3}$ obtained in (4) and (5), 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, $a_n=a r^{n-1}$

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

$a_2=a r^{1}=(-3) r$

According to the given condition,

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

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

$\Rightarrow r=-3$

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

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

Answer :

(a) 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 .

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

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

$\Rightarrow(2)(2)^{\dfrac{n-1}{2}}=(2)^{7}$

$\Rightarrow(2)^{\dfrac{n-1}{2}+1}=(2)^{7}$

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

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

$\Rightarrow n-1=12$

$\Rightarrow n=13$

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

(b) 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 . $a_n=a r^{n-1}$

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

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

$\Rightarrow(3)^{\dfrac{1}{2}}(3)^{\dfrac{n-1}{2}}=(3)^{6}$

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

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

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

$\Rightarrow n=12$

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

(c) 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 $\dfrac{1}{19683}$.

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

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

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

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

$\Rightarrow n=9$

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

Answer :

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

Common ratio $ =\dfrac{x}{(\frac{-2}{7})}=\dfrac{-7 x}{2}$

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

$\therefore \dfrac{-7 x}{2}=\dfrac{-7}{2 x}$

$\Rightarrow x^{2}=\dfrac{-2 \times 7}{-2 \times 7}=1$

$\Rightarrow x=\sqrt{1}$

$\Rightarrow x= \pm 1$

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 $

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

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

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

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

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

Answer :

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

Here, $a=\sqrt{7}$

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

$S_{12}=\dfrac{a(1-r^{n})}{1-r}$

$\therefore S_{12}=\dfrac{\sqrt{7}[1-(\sqrt{3})^{n}]}{1-\sqrt{3}}$

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

(By rationalizing)

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

$=\dfrac{-\sqrt{7}(1+\sqrt{3})}{2}[1-(3)^{\dfrac{n}{2}}]$

$=\dfrac{\sqrt{7}(1+\sqrt{3})}{2}[(3)^{\dfrac{n}{2}}-1]$

Answer :

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

Here, first term $=a_1=1$

Common ratio $=r=- a$

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

Answer :

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

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

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

Answer :

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

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

The terms of this sequence $3,3^{2}, 3^{3}, \ldots$ forms a G.P. $S_n=\dfrac{a(r^{n}-1)}{r-1}$

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

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

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

Substituting this value in equation (1), we obtain

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

Answer :

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

$\dfrac{a}{r}+a+a r=\dfrac{39}{10}$

$(\dfrac{a}{r})(a)(a r)=1$

From (2), we obtain

$a^{3}=1$

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

Substituting $a=1$ in equation (1), we obtain $\dfrac{1}{r}+1+r=\dfrac{39}{10}$

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

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

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

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

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

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

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

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

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

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

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

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

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

$\Rightarrow 3^{n}=81$

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

$\therefore n=4$

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

$\Rightarrow a(1+r+r^{2})=16$

$ar^{3}(1+r+r^{2})=128$

Dividing equation (2) by (1), we obtain

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

$\Rightarrow r^{3}=8$

$\therefore r=2$

Substituting $r=2$ in (1), we obtain

$a(1+2+4)=16$

$\Rightarrow a(7)=16$

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

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

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

Answer :

$a=729$

$a_7=64$

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

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

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

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

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

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

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

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

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

Answer :

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

According to the given conditions,

$S_2=-4=\dfrac{a(1-r^{2})}{1-r}$

$a_5=4 \times a_3$

$a r^{4}=4 a r^{2}$ $\Rightarrow r^{2}=4$

$\therefore r= \pm 2$

From (1), we obtain

$-4=\dfrac{a[1-(2)^{2}]}{1-2}$ for $r=2$

$\Rightarrow-4=\dfrac{a(1-4)}{-1}$

$\Rightarrow-4=a(3)$

$\Rightarrow a=\dfrac{-4}{3}$

Also, $-4=\dfrac{a[1-(-2)^{2}]}{1-(-2)}$ for $r=-2$

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

$\Rightarrow-4=\dfrac{a(-3)}{3}$

$\Rightarrow a=4$

Thus, the required G.P. is

$\dfrac{-4}{3}, \dfrac{-8}{3}, \dfrac{-16}{3}, \ldots$ or

4, -8, 16, -32, 64, …

Answer :

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

According to the given condition,

$a_4=a r^{3}=x$

$a _{10}=a r^{9}=y$..

$a _{16}=a r^{15}=z$

Dividing (2) by (1), we obtain

$\dfrac{y}{x}=\dfrac{a r^{9}}{a r^{3}} \Rightarrow \dfrac{y}{x}=r^{6}$

Dividing (3) by (2), 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

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

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

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

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

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

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

Answer :

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

$=64[4+2+1+\dfrac{1}{2}+\dfrac{1}{2^{2}}]$

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

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

$\therefore$ Required sum $=$

$ 64(\dfrac{31}{4})=(16)(31)=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,

$a_3=a_1+9$

$\Rightarrow a r^{2}=a+9$

$a_2=a_4+18$

$\Rightarrow a r=a r^{3}+18$.

From (1) and (2), we obtain

$a(r^{2}- 1)=9$

$ar(1-{2})=18$

Dividing (4) by (3), we obtain

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

$\Rightarrow-r=2$

$\Rightarrow r=-2$

Substituting the value of $r$ in (1), 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,

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

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

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

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

$=A q^{-r+r-p+p-q} \times R^{(p r-p r-q+r+(r q-r+p-p q)+(p r-p-q r+q)}$

$=A^{0} \times R^{0}$

$=1$

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

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

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

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

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

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

$P=a^{n} r^{\dfrac{n(n-1)}{2}}$

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

$=[a^{2} r^{(n-1)}]^{n}$

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

$=(ab)^{n}$

Thus, the given result is proved.

Answer :

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

Sum of first $n$ terms $=\dfrac{a(1-r^{n})}{(1-r)}$

Since there are $n$ terms from $(n+1)^{\text{th }}$ to $(2 n)^{\text{th }}$ term,

Sum of terms from $(n+1)^{\text{th }}$ to $(2 n)^{\text{th }}$ term

$ =\dfrac{a _{n+1}(1-r^{n})}{(1-r)} $

$a_{n+1}=a r^{n+1-1}=a r^{n}$

Thus, required ratio $=\dfrac{a(1-r^{n})}{(1-r)} \times \dfrac{(1-r)}{a^{n}(1-r^{n})}=\dfrac{1}{r^{n}}$

Thus, the ratio of the sum of first $n$ terms of a G.P. to the sum of terms from $(n+1)^{\text{th }}$ to $(2 n)^{\text{th }}$ term is $\dfrac{1}{r^{n}}$.

Answer :

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

Therefore,

$b c=a d$.

$b^{2}=a c$.

$c^{2}=b d$.

It has to be proved that,

$(a^{2}+b^{2}+c^{2})(b^{2}+c^{2}+d^{2})=(a b+b c \text{ - } c d)^{2}$

R.H.S.

$=(a b+b c+c d)^{2}$

$=(a b+a d+c d)^{2}[$ Using (1)]

$=[a b+d(a+c)]^{2}$

$ \begin{aligned} & =a^{2} b^{2}+2 a b d(a+c)+d^{2}(a+c)^{2} \\ & =a^{2} b^{2}+2 a^{2} b d+2 a c b d+d^{2}(a^{2}+2 a c+c^{2}) \\ & =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}[\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 (2) and (3) and rearranging terms]

$=a^{2}(b^{2}+c^{2}+d^{2})+b^{2}(b^{2}+c^{2}+d^{2})+c^{2}(b^{2}+c^{2}+d^{2})$

$=(a^{2}+b^{2}+c^{2})(b^{2}+c^{2}+d^{2})$

$=$ L.H.S.

$\therefore$ L.H.S. $=$ R.H.S.

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

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.

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

$\Rightarrow r^{3}=27$

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

For $r=3$,

$G_1=a r=(3)(3)=9$

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

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{(a^{n+1}+b^{n+1})^{2}}{(a^{n}+b^{n})^{2}}=a b $

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

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

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

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

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

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

$\Rightarrow 2 n+1=0$

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

Answer :

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

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

According to the given condition,

$a+b=6 \sqrt{a b}$

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

Also,

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

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

$=4 \sqrt{2} \sqrt{a b}$

Adding (1) and (2), we obtain

$2 a=(6+4 \sqrt{2}) \sqrt{a b}$

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

Substituting the value of $a$ in (1), we obtain

$b=6 \sqrt{a b}-(3+2 \sqrt{2}) \sqrt{a b}$

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

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

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

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 $b$.

$\therefore AM=A=\dfrac{a+b}{2}$

$GM=G=\sqrt{ab}$

From (1) and (2), we obtain

$a+b=2 A$.

$a b=G^{2}$.

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

$(a- b)^{2}=4 A^{2}- 4 G^{2}=4(A^{2} -G^{2})$

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

$(a-b)=2 \sqrt{(A+G)(A-G)}$

From (3) and (5), we obtain

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

Substituting the value of $a$ in (3), we obtain $b=2 A-A-\sqrt{(A+G)(A-G)}=A-\sqrt{(A+G)(A-G)}$

Thus, the two numbers are

$ A \pm \sqrt{(A+G)(A-G)} $

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.

Here, $a=30$ and $r=2$

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

Therefore, the number of bacteria at the end of $2^{\text{nd }}$ hour will be 120 .

$a_5=a r^{4}=(30)(2)^{4}=480$

The number of bacteria at the end of $4^{\text{th }}$ hour will be 480 .

$a _{n+1}=a r^{n}=(30) 2^{n}$

Thus, number of bacteria at the end of $n^{\text{th }}$ hour will be $30(2)^{n}$.

Answer :

The amount deposited in the bank is Rs 500 .

At the end of first year, amount $=Rs 500(1+\dfrac{1}{10})=Rs 500$ (1.1)

At the end of $2^{\text{nd }}$ year, amount $=$ Rs 500 (1.1) (1.1)

At the end of $3^{\text{rd }}$ year, amount = Rs 500 (1.1) (1.1) (1.1) and so on

$\therefore$ Amount at the end of 10 years $=$ Rs 500 (1.1) (1.1) $\ldots$ (10 times)

$=Rs 500(1.1)^{10}$

Answer :

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

According to the given condition,