Ex 9.3 Class 11 Maths Ncert Solutions

The given G.P. is \( \frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \ldots \)
Here, \( {a}= \) first term \( =\frac{5}{2} \)
\( r= \) common ratio \( =\frac{\frac{5}{4}}{\frac{5}{2}}=\frac{1}{2} \)
\( {a}_{20}=ar^{20-1}=\frac{5}{2}\left(\frac{1}{2}\right)^{19}=\frac{5}{2(2)^{19}}=\frac{5}{(2)^{20}} \)
\( a_{n}=ar^{n-1}=\frac{5}{2}\left(\frac{1}{2}\right)^{n-1}=\frac{5}{2(2)^{n-1}}=\frac{5}{(2)^{n}} \)

Common ratio, \( r=0 \)
Let a be the first term of the G.P.
\( {a}_{8}={ar}^{8-1}={ar}^{7}=192 {a}(2)^{7}=(2)^{6}(3) \)
\( ={a}=\frac{(2)^{6} \times 3}{(2)^{7}}=\frac{3}{2} \)
\( {a}_{12}=ar^{12-1}=\left(\frac{3}{2}\right)(2)^{11}=(3)(2)^{10}=3072 \)

Let a be the first term and \( r \) be the common ratio of the G.P. according to the given condition,
\(a_{5}=a r^{5-1}=a r^{4}=p \ldots(1)\)
\(a_{8}=a r^{8-1}=a r^{7}=q \ldots(2)\)
\(a_{11}=a r^{11-1}=a r^{10}={s} \ldots(3)\)
dividing eq. (2) by (1), we obtain
\( \frac{a r^{7}}{a r^{4}}=\frac{q}{p} \)
\({r}^{3}=\frac{q}{p} \ldots(4)\)
dividing eq. (3) by (2), we obtain
\(\frac{a r^{10}}{a r^{7}}=\frac{s}{q} \ldots(5)\)
Equation the value of \( r^{3} \) obtained in (4) and (5), we obtain
\(\frac{q}{p}=\frac{s}{q}\)
\(={q}^{2}=p {s}\)
Thus, the given result is proved.

Let, a be the first term and \( r \) be the common ratio of the G.P.
\(a=-3\)
It is known that, \(a_{n} = ar ^{n-1}\)
\(a_{4}=a r^{3}=(-3) r^{3}\)
\(a_{2}=a r^{1}=(-3) r\)
According to the given condition, \((-3) {r}^{3}=[(-3) {r}]^{2} \)
\(=-3 r^{3}=9 r^{2}=r=-3 a^{7}=a r^{7-1}=a\)
\(r^{6}=(-3)(-3)^{6}=(-3)^{7}=-2187\)
thus, the \( 7^{\text {th}} \) term of the G.P. is \(-2187 \).

the given sequence is \( 2,2 \sqrt{2}, 4, \ldots \) is \(128 \) ?
Here, \( {a}=2 \) and \( {r}=\frac{(2 \sqrt{2}) }{ 2}=\sqrt{2} \)
Let, the \(n^{th}\) term of the given sequence be \(128 \).
\({a}_{{n}}={a} {r}^{{n}-1}\)
\(=(2)(\sqrt{2})^{n-1}=128\)
\(=(2)(2)^{\frac{n-1}{2}}=(2)^{7}\)
\(=(2)^{\frac{n-1}{2}+1}=(2)^{7}\)
\(=\frac{n-1}{2}+1=7\)
\(=\frac{n-1}{2}=6\)
\(={n}-1=12\)
\(={n}=13\)
Thus, the \( 13^{\text {th}} \) term of the given sequence is \(128 \).

the given sequence is \( \sqrt{3}, 3,3 \sqrt{3}, \ldots \)
\( {a}=\sqrt{3} \) and \( {r}=\frac{3}{\sqrt{3}}=\sqrt{3} \)
Let, the \(n^{th}\) term of the given sequence be \(729 \).
\({a}_{{n}}={ar}^{{n}-1}\)
\(={a} {r}^{{n}-1}=729\)
\(=(\sqrt{3})(\sqrt{3})^{n-1}=729\)
\(=(3)^{\frac{1}{2}}(3)^{\frac{n-1}{2}}=(3)^6\)
\(=(3)^{\frac{1}{2}+\frac{n-1}{2}}=(3)^6\)
\(=\frac{1}{2}+\frac{n-1}{2}=6\)
\(=\frac{1+n-1}{2}=6\)
\(={n}=12\)
Thus, the \( 12^{\text {th}} \) term of the given sequence is \(729 \).

the given sequence is \( \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots \)
Here, \( {a}=\frac{1}{3} \) and \( {r}=\frac{1}{9} \div \frac{1}{3}=\frac{1}{3} \)
Let, the \(n^{th}\) term of the given sequence be \( \frac{1}{19683} \).
\(a_{n}=a r^{n-1}\)
\(=ar^{n-1}=\frac{1}{19683}\)
\(=\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)^{n-1}=\frac{1}{19683}\)
\(=\left(\frac{1}{3}\right)^{n}=\left(\frac{1}{3}\right)^{9}\)
\(=n=9\)
Thus, the \( 9^{\text {th}} \) term of the given sequence is \( \frac{1}{19683} \).

The given numbers are \( \frac{-2}{7}, x, \frac{-7}{2} \)
Common ratio \( =\frac{x}{\frac{-2}{7}}=\frac{-7 x}{2} \)
Also, common ratio \( =\frac{\frac{-7}{2}}{x}=\frac{-7}{2 x} \)
\(=\frac{-7 x}{2}=\frac{-7}{2 x}\)
\(=x^{2}=\frac{-2 \times 7}{-2 \times 7}=1\)
\(=x=\sqrt{1}\)
\(=x= \pm 1\)
Thus, for \( {x}= \pm 1 \), the given numbers will be in G.P.

The given G.P. is \( 0.15,0.015,0.00015, \ldots \)
Here \( {a}=0.15 \) and \( {r}=\frac{0.015}{0.15}=0.1 \)
\({S}_{{n}}=\frac{a\left(1-r^{n}\right)}{1-r}\)
\({S}_{20}=\frac{0.15\left[1-(0.1)^{20}\right]}{1-0.1}\)
\(=\frac{0.15}{0.9}[1-(0.1) ^{20}]\)
\(=\frac{15}{90}[1-(0.1) ^{20}]\)
\(=\frac{1}{6}[1-(0.1) ^{20}]\)

The given G.P. is \( \sqrt{7}, \sqrt{21}, 3 \sqrt{7}, \ldots \)
Here, \( a=\sqrt{7} \) and \( r=\frac{\sqrt{21}}{7}=\sqrt{3} \)
\({S}_{{n}}=\frac{a\left(1-r^{n}\right)}{1-r}\)
\({S}_{{n}}=\frac{\sqrt{7}\left[1-(\sqrt{3})^{n}\right]}{1-\sqrt{3}}\)
\({S}_{{n}}=\frac{\sqrt{7}\left[1-(\sqrt{3})^{n}\right]}{1-\sqrt{3}} \times \frac{1+\sqrt{3}}{1+\sqrt{3}}\)
\({S}_{{n}}=\frac{\sqrt{7}(\sqrt{3}+1)\left[1-(\sqrt{3})^{n}\right]}{1-3}\)
\({S}_{{n}}=\frac{-\sqrt{7}(\sqrt{3}+1)\left[1-(\sqrt{3})^{n}\right]}{2}\)

The given G.P. is \( 1,-a, a^{2},-a^{3}, \ldots \)
Here, first term \( =a_{1}=1 \)
Common ratio \( ={r}=-{a} \)
\({S}_{{n}}=\frac{a_{1}\left(1-r^{n}\right)}{1-r}\)
\({S}_{{n}}=\frac{1\left[1-(-a)^{n}\right]}{1-(-a)}=\frac{\left[1-(-a)^{n}\right]}{1+a}\)

The given G.P. is \( x^{3}, x^{5}, x^{7}, \ldots \)
Here, \( a=x^{3} \) and \( r=x^{2} \)
\({S}_{{n}}=\frac{a\left(1-r^{n}\right)}{1-r}=\frac{x^{3}\left[1-\left(x^{2}\right)^{n}\right]}{1-x^{2}}=\frac{x^{3}\left(1-x^{2 n}\right)}{1-x^{2}}\)

\(\sum_{k=1}^{11}\left(2+3^{k}\right)=\sum_{k=1}^{11}(2)+\sum_{k=1}^{11} 3^{k}=2(11)\) \(+\sum_{k=1}^{11} 3^{k}=22+\sum_{k=1}^{11} 3^{k} \ldots(1)\)
\(\sum_{k=1}^{11} 3^{k}=31+32+33+\ldots+311\)
The terms of the sequence \( 3,3^{2}, 3^{3} \ldots \) Forms a G.P.
\({S}_{{n}}=\frac{a\left(r^{n}-1\right)}{r-1}\)
\({S}_{{n}}=\frac{3\left[(3)^{11}-1\right]}{3-1}\)
\({S}_{{n}}=\frac{3}{2}\left(3^{11}-1\right)\)
\(=\sum_{k=1}^{11} 3^{k}=\frac{3}{2}\left(3^{11}-1\right)\)
Substituting this value in eq. (1), we obtain
\(\sum_{k=1}^{11}\left(2+3^{k}\right)=22+\frac{3}{2}\left(3^{11}-1\right)\)

Let, \( \frac{a}{r} , a, ar\) be the first three terms of the G.P.
\(\frac{a}{r}+{a}+{ar}=\frac{39}{10} \ldots(1)\)
\(\left(\frac{a}{r}\right)(a)(a r)=1\ldots(2)\)
From (2), we obtain \( a^{3} =1 \)
\( ={a}=1 \) (considering real roots only)
Substituting \( {a}=1 \) in eq. (1), we obtain
\(\frac{1}{r}+1+r=\frac{39}{10}\)
\(=1+r+r^{2}=\frac{39}{10} r\)
\(=10+10 r+10 r^{2}-39 r=0\)
\(=10 r^{2}-29 r+10=0\)
\(=10 r^{2}-25 r-4 r+10=0\)
\(=5 r(2 r-5)-2(2 r-5)=0\)
\(=(2 r-5)(5 r-2)=0\)
\(=r=\frac{2}{5} \text { or } \frac{5}{2}\)
Thus, the three terms if G.P. are \( \frac{5}{2}, 1 \) and \( \frac{2}{5} \)

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 \(120 \).
\({S}_{{n}}=\frac{a\left(1-r^{n}\right)}{1-r}\)
Here, \( {a}=3 \) and \( {r}=3 \)
\(S_{n}=120=\frac{3\left(3^{n}-1\right)}{3-1}\)
\(=120=\frac{3\left(3^{n}-1\right)}{2}\)
\(=\frac{120 \times 2}{3}=3 n-1\)
\(=3 n-1=80\)
\(=3 n=81\)
\(=3 n=34\)
\(=n=4\)
Thus, four terms of the given G.P. are required to obtain the sum as \(120 \).

Let, the G.P. be \( a , a {r}, {ar}^{2}, {ar}^{3}, \ldots \) According to the given condition,
\(a+a r+a r^{2}=16\text{ and }a r^{3}+a r^{4}+a r^{5}=128\)
\(=a\left(1+r+r^{2}\right)=16 \ldots (1) \)
\(=a r^{3}\left(1+r+r^{2}\right)=128 \ldots(2)\)
Dividing eq. (2) by (1), we obtain
\(\frac{a r^{3}\left(1+r+r^{2}\right)}{a\left(1+r+r^{2}\right)}=\frac{128}{16}\)
\(=r^{3}=8\)
\(={r}=2\)
Substituting \( {r}=2 \) in (1), we obtain \( a (1+2+4)=16 \)
\(={a}(7)=16\)
\(={a}=\frac{16}{7}\)
\({S}_{{n}}=\frac{a\left(r^{n}-1\right)}{r-1}\)
\({S}_{{n}}=\frac{16}{7} \frac{\left(2^{n}-1\right)}{2-1}=\frac{16}{7}\left(2^{n}-1\right)\)

\(A=729 a^{7}=64\)
Let, \( r \) be the common ratio of the G.P. it is known that,
\(=a_{n}=a r n^{-1}\)
\(=a_{7}=a r^{7-1}=(729) r^{6}\)
\(=64=729 r^{6}\)
\(=r^{6}=\frac{64}{729}\)
\(=r^{6}=\left(\frac{2}{3}\right)^{6}\)
\(=r=\frac{2}{3}\)
Also, it is known that,
\({S}_{{n}}=\frac{a\left(1-r^{n}\right)}{1-r}\)
\({S}_{7}=\frac{729\left[1-\left(\frac{2}{3}\right)^{7}\right]}{1-\frac{2}{3}}\)
\(=3 \times 729\left[1-\left(\frac{2}{3}\right)^{7}\right]\)
\(=(3)^{7}\left[\frac{\left[(3)^{7}-(2)^{7}\right.}{(3)^{7}}\right]\)
\(=(3)^{7}-(2)^{7}\)
\(=2187-128\)
\(=2059\)

Let, a be the first term and \( r \) be the common ratio of the G.P.
According to the given condition,
\({S}_{2}=-4=\frac{a\left(1-r^{2}\right)}{1-r} \ldots(1)\)
\(={a}_{5}=4 \times {a}^{3}\)
\(={a} {r}^{4}=4 {ar}^{2}={r}^{2}=4\)
\(={r}= \pm 2\)
From (1), we obtain
\(-4=\frac{a\left[1-(2)^{2}\right]}{1-2} \text { for } {r}=2\)
\(=-4=\frac{a(1-4)}{-1}\)
\(=-4={a}(3)\)
\(={a}=\frac{-4}{3}\)
Also, \( -4=\frac{a\left[1-(-2)^{2}\right]}{1-(-2)} \) for \( {r}=-2 \)
\(=-4=\frac{a(1-4)}{1+2}\)
\(=-4=\frac{a(-3)}{3}\)
\(={a}=4\)
Thus, the required G.P. is \( \frac{-4}{3}, \frac{-8}{3}, \frac{-16}{3}, \ldots \) or \( 4,-8,16,-32, \ldots \)

Let, a be the first term and \( r \) be the common ratio of the G.P.
According to the given condition,
\(=a_{4}=a r^{3}=x \ldots(1)\)
\(=a_{10}=a r^{9}=y \ldots(2)\)
\(=a_{16}=a r^{15}=z \ldots(3)\)
Dividing (2) by (1), we obtain
\(\frac{y}{x}=\frac{a r^{9}}{a r^{3}}=\frac{y}{x}={r}^{6}\)
Dividing (3) by (2), we obtain
\(\frac{z}{y}=\frac{a r^{15}}{a r^{9}}=\frac{z}{y}={r}^{6}\)
\(=\frac{y}{x}=\frac{z}{y}\)
Thus, \( x, y, z \) is in G.P.

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
\({S}_{{n}}=8+88+888+8888+\ldots .+{n} \text { terms }\)
\(=\frac{8}{9}[9+99+999+9999+\ldots \text { to n terms }]\)
\(=\frac{8}{9}\left[(10-1)+\left(10^{2}-1\right)+\left(10^{3}-1\right)+\left(10^{4}-1\right)+\ldots \text { to n terms}\right]\)
\(=\frac{8}{9}\left[\left(10+10^{2}+\ldots . {n} \text { terms}\right)-(1+1+1+{n} \text { terms})\right]\)
\(=\frac{8}{9}\left(\frac{10\left(10^{n}-1\right)}{10-1}-n\right)\)
\(=\frac{8}{9}\left[\frac{10\left(10^{n}-1\right)}{9}-n\right]\)
\(=\frac{80}{81}\left(10^{n}-1\right)-\frac{8}{9} n\)

Required sum \( =2 \times 128+4 \times 32+8 \times 8+16 \times 2+32 \times \frac{1}{2} \)
\(=64\left[4+2+1+\frac{1}{2}+\frac{1}{2^{2}}\right]\)
Here, \( 4,2,1, \frac{1}{2}, \frac{1}{2^{2}} \) is a G.P.
First term, \( {a}=4 \)
Common ratio \( {r}=\frac{1}{2} \)
It is known that,
\({S}_{{n}}=\frac{a\left(1-r^{n}\right)}{1-r}\)
\({S}_{5}=\frac{4\left[1-\left(\frac{1}{2}\right)^{5}\right]}{1-\frac{1}{2}}=\frac{4\left[1-\frac{1}{32}\right]}{\frac{1}{2}}=8\left(\frac{32-1}{32}\right)=\frac{31}{4}\)
Required sum \( =64 \times \frac{31}{4}=(16)(31)=496 \)

It has to be proved that the sequence: \( a {A} , ar {AR}, ar^{2} {AR}^{2}, \ldots a {r}^{{n}-1} {AR}^{{n}-1} \), forms a G.P.
\(\frac{\text { second term }}{\text { first term }}=\frac{a r A R}{a A}={Rr}\)
\(\frac{\text { third term }}{\text {second term }}=\frac{a r^{2} A R^{2}}{a r A R}={Rr}\)
Thus, the above sequence forms a G.P. and the common ratio is \( r {R} \).

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
\(=a_{3}=a_{1}+9=a r^{2}=a+9 \ldots(1)\)
\(=a_{2}=a_{4}+18=a r=a r^{3}+18 \ldots(2)\)
From eq. (1) and (2), we obtain
\(=a\left(r^{2}-1\right)=9 \ldots(3)\)
\(=ar\left(1-r^{2}\right)=18 \ldots(4)\)
Dividing (4) by (3), we obtain
\(\frac{a r\left(1-r^{2}\right)}{a\left(r^{2}-1\right)}=\frac{18}{9}\)
\(=-r=2\)
\(=r=-2\)
Substituting the value of \( r \) in (1), we obtain
\(4 a=a+9\)
\(=3 a=9\)
\(=a=3\)
Thus, the first four numbers of the G.P. are \( 3,3(-2), 3(-2)^{2} \) and \( 3(-2)^{3} \) i.e., \( 3,-6,12 \), and \(-24\) .

Let, A be the first term and R be the common ratio of the G.P.
According to the given information,
\({AR}^{{p}-1}={a}\)
\({AR}^{{q}-1}={b}\)
\({AR}^{{r}-1}={c}\)
\(=a^{q-r} \cdot b^{r-p} \cdot c^{p-q}\)
\(=A^{q-r} \times R^{(p-1)(q-r)} \times A^{r-p} \times R^{(q-1)(r-p)} \times A^{p-q} \times R^{(r-1)(p-q)}\)
\(=A^{q-r-r-p+p-q} \times R^{(p r-p r-q+r)+(r q-r+p-p q)+(p r-p-q r+q)}\)
\(=A^{0} \times R^{0}\)
\(=1\)
Thus, the given result is proved.

The first terms of the A.P. is a and the last term is \( b \).
Therefore, the G.P. is \(a, a {r}, {ar}^{2}, {ar}^{3} \ldots {ar}^{{n}-1} \), where \(r\) is common ratio
\(={b}={ar}^{{n}-1} \quad\ldots\ldots(1)\)
\(={p}=\text { product of } {n} \text { terms }\)
\(=({a})(a {r})\left({ar}^{2}\right) \ldots\left({ar}^{{n}-1}\right)\)
\(=({a} \times {a} \times {a})\left({r} \times {r}^{2} \times \ldots{r}^{{n}-1}\right)\)
\(=an. r^{1+2+\ldots({n}-1)}\quad \ldots\ldots(2)\)
Here, \( 1,2, \ldots({n}-1) \) is an A.P.
\(=1+2+\ldots+({n}-1)\)
\(=\frac{n-1}{2}[2+(n-1-1) \times 1]\)
\(=\frac{n-1}{2}[2+n-2]\)
\(=\frac{n(n-1)}{2}\)
\({P}={a}^{{n}} r^{\frac{n(n-1)}{2}}\)
\({P}^{2}={a}^{2 {n}} r^{n(n-1)}\)
\(=\left[a^{2} r^{(n-1)}\right]^{n}\)
\(=\left[a \times a r^{n-1}\right]^{n}\)
\(=({ab})^{{n}} \text { [using eq. (1)] }\)
Thus, the given result id proved.

Let, a be the first term and \( r \) be the common ratio of the G.P.
Sum of first n terms \( =\frac{a\left(1-r^{n}\right)}{1-r} \)
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
\({S}_{{n}}=\frac{a_{n+1}\left(1-r^{n}\right)}{1-r}\)
Thus, required ratio \( =\frac{a\left(1-r^{n}\right)}{1-r} \times \frac{1-r}{a r^{n}\left(1-r^{n}\right)}=\frac{1}{r^{n}} \)
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 \( (2 n)^{\text {th}} \) terms is \( \frac{1}{r^{n}} \)

\( {a}, {b}, {c} \) and \(d\) are in G.P. therefore,
\(=b {c}={ad} \ldots(1)\)
\(={b} ^{2}={ac} \ldots(2)\)
\(={c} ^{2}=b {d} \ldots(3)\)
It has to be proved that,
\(\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c-c d)^{2}\)
R.H.S.
\(=(a b+b c+c d)^{2}\)
\(=(a b+a d+c d)^{2}\quad[\text {using (1)}]\)
\(=[a b+d(a+c)]^{2}\)
\(=a^{2} b^{2}+2 a b d(a+c)+d^{2}(a+c)^{2}\)
\(=a^{2} b^{2}+2 a^{2} b d+2 a c b d+d^{2}\left(a^{2}+2 a c+c^{2}\right)\)
\(=a^{2} b^{2}+2 a^{2} c^{2}+2 b^{2} c^{2}+d^{2} a^{2}+2 d^{2} b^{2}+d^{2} c^{2}\quad[\text {using (1) and (2)}]\)
\(=a^{2} b^{2}+a^{2} c^{2}+a^{2} c^{2}+b^{2} c^{2}+b^{2} c^{2}+d^{2} a^{2}+d^{2} b^{2}+d^{2} b^{2}+d^{2} c^{2}\)
\(=a^{2} b^{2}+a^{2} c^{2}+a^{2} d^{2}+b^{2} \times b^{2}+b^{2} c^{2}+b^{2} d^{2}+c^{2} d^{2}+c^{2} \times c^{2}+c^{2} d^{2}\)
[using (2) and (3) and rearranging terms]
\(=a^{2}\left(b^{2}+c^{2}+d^{2}\right)+b^{2}\left(b^{2}+c^{2}+d^{2}\right)+c^{2}\left(b^{2}+c^{2}+d^{2}\right)\)
\(=\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=\text { L.H.S. }\)
L.H.S. \( = \) R.H.S.
\(=\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c-c d)^{2}\)

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.
\(81=(3)(r)^{3}\)
\(=r^{3}=27\)
\(=r=3 \text { (taken real roots only) }\)
For \( r=3 \)
\({G}_{1}={ar}=(3)(3)=9\)
\({G}_{2}={a} {r}^{2}=(3)(3)^{2}=27\)
Thus, the required two numbers are \(9\) and \(27 \).

A.M. is a and b is \( \sqrt{a b} \)
By the given condition: \( \frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}=\sqrt{a b} \)
Squaring both sides, we obtain
\(=\frac{\left(a^{n+1}+b^{n+1}\right)^{2}}{\left(a^{n}+b^{n}\right)^{2}}=a b\)
\(={a}^{2 {n}+2}+2 {a}^{{n}+1} b^{{n}+1}+{b}^{2 {n}+2}=({ab})\left({a}^{2 {n}}+2 {a}^{{n}} b^{{n}}+b^{2 {n}}\right)\)
\(={a}^{2 {n}+2}+2 {a}^{{n}+1} {b}^{{n}+1}+{b}^{2 {n}+2}={a}^{2 {n}+1} {b}+2 {a}^{{n}+1} b^{{n}+1}+{ab}^{2 {n}+1}\)
\(={a}^{2 {n}+2}+{b}^{2 {n}+2}={a}^{2 {n}+1} {b}+{a} b^{2 {n}+1}\)
\(={a}^{2 {n}+2}-{a}^{2 {n}+1} {b}={ab}^{2 {n}+1}-{b}^{2 {n}+2}\)
\(={a}^{2 {n}+1}({a}-{b})={b}^{2 {n}+1}({a}-{b})\)
\(=\left(\frac{a}{b}\right)^{2 n+1}=1=\left(\frac{a}{b}\right)^{0}\)
\(=2 {n}+1=0\)
\(={n}=\frac{-1}{2}\)

Let, the two numbers be \(a\) and \(b \).
\(\text {G.M. }=\sqrt{a b}\)
According to the given condition,
\(=a+b=6 \sqrt{a b} \ldots(1)\)
\(=(a+b)^{2}=36(a b)\)
Also, \( ({a}-{b})^{2}=({a}+{b})^{2}-4 {ab}=36 {ab}-4 {ab}=32 {ab} \)
\(=a-b=\sqrt{32} \sqrt{a b}\)
\(=4 \sqrt{2} \sqrt{a b} \ldots (2) \)
Adding (1) and (2), we obtain
\(2 a=(6+4 \sqrt{2}) \sqrt{a b}\)
\(=a=(3+2 \sqrt{2}) \sqrt{a b}\)
Substituting the value of a in (1), we obtain
\(={b}=6 \sqrt{a b}-(3+2 \sqrt{2}) \sqrt{a b}\)
\(={b}=(3-2 \sqrt{2}) \sqrt{a b}\)
\(=\frac{a}{b}=\frac{(3+2 \sqrt{2}) \sqrt{a b}}{(3-2 \sqrt{2}) \sqrt{a b}}=\frac{3+2 \sqrt{2}}{3-2 \sqrt{2}}\)
Thus, the required ratio is \( (3+2 \sqrt{2}):(3-2 \sqrt{2}) \)

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 \(b \).
\(=\text {A.M.}={A}=\frac{a+b}{2} \ldots(1)\)
\(\text {G.M.}={G}=\sqrt{a b} \ldots(2)\)
From (1) and (2), we obtain
\(=a+b=2 A \ldots(3)\)
\(=a b=G^{2} \ldots (4)\)
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
\( (a-b)^{2}=4 A^{2}-4 G^{2}=4\left(A^{2}-G^{2}\right) \)
\( ({a}-{b})^{2}=4({A}+{G})({A}-{G}) \)
\( ({a}-{b})=2 \sqrt{(A+G)(A-G)}\ldots (5) \)
From (3) and (5), we obtain
\(2 {a}=2 {A}+2 \sqrt{(A+G)(A-G)}\)
\(={a}={A}+\sqrt{(A+G)(A-G)}\)
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 \( {A} \pm \sqrt{(A+G)(A-G)} \)

It is given that the number of bacteria doubles every hour. Therefore, the numbers of bacteria after every hour will form a G.P.
Here, \( {a}=30 \) and \( {r}=2 \)
\(={a}_{3}=ar^{2}=(30)(2)^{2}=120\)
Therefore, the numbers of bacteria at the end of \( 2^{\text {nd}} \) hour will be \(120 \).
\(={a}_{5}={ar}^{4}=(30)(2)^{4}=480\)
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, numbers of bacteria at the end of \( {n}^{\text {th}} \) hour will be \( 30(2)^{{n}} \)

The amount deposited in the bank is ₹ \(500 \).
At the end of first year, amount \( =₹ 500\left(1+\frac{1}{10}\right)= \) rs \(500(1.1)\)
At the end of \( 2^{\text {nd}} \) year, amount \( =₹ 500(1.1)(1.1) \)
At the end of \( 3^{\text {rd}} \) year, amount \( =₹ 500(1.1)(1.1)(1.1) \) and so on.
Amount at the end of 10 years \( =₹ 500(1.1)(1.1) \ldots\) \((10 \text{ times})= \text{Rs }500 (1.1)^{10} \)

Let, the root of the quadratic equation be a and b.
According to the given condition,
\(\text {A.M.}=\frac{a+b}{2}=8={a}+{b}=16 \ldots(1)\)
\(\text {G.M.}=\sqrt{a b}=5={ab}=25 \ldots (2)\)
The quadratic equation is given by,
\( x^{2}-x \) (sum of roots) \(+\) (product of roots) \( =0 \)
\(x^{2}-x(a+b)+(a b)=0\)
\( x^{2}-16 x+25=0\quad \) [using (1) and (2)]
thus, the required quadratic equation is \( x^{2}-16 x+25=0 \)