Chapter 8: Sequences and Series

Ans: According to the given conditions in the question,

\[f\left( x+y \right)=f\left( x \right)\times f\left( y \right)\] for all \[x,y,\in N\]

\[f\left( 1 \right)=3\]

Let \[x=y=1\].

Then,

\[f\left( 1+1 \right)=f\left( 1+2 \right)=f\left( 1 \right)f\left( 2 \right)=3\times 3=9\]

We can also write 

\[f\left( 1+1+1 \right)=f\left( 3 \right)=f\left( 1+2 \right)=f\left( 1 \right)f\left( 2 \right)=3\times 9=27\]

\[f\left( 4 \right)=f\left( 1+4 \right)=f\left( 1 \right)f\left( 3 \right)=3\times 27=81\]

Both the first term and common ratio of \[f\left( 1 \right),f\left( 2 \right),f\left( 3 \right),...,\]that is \[3,9,27,...,\] that forms s G.P. is equal to \[3\]

We know that, \[{{S}_{n}}=\frac{a\left( {{r}^{n}}-1 \right)}{r-1}\]

Given that, \[\sum\limits_{k=1}^{n}{f}\left( x \right)=120\] 

Then,

\[120=\frac{3\left( {{3}^{n}}-1 \right)}{3-1}\]

\[\Rightarrow 120=\frac{3}{2}\left( {{3}^{n}}-1 \right)\]

\[\Rightarrow {{3}^{n}}-1=80\] 

\[\Rightarrow {{3}^{n}}=80={{3}^{4}}\]

\[\Rightarrow {{3}^{n}}-1=80\]

\[n=4\]

Therefore, \[4\] is the value of \[n\].

Ans: Let \[315\] be the sum of \[n\] terms of the G.P.

We know that, \[{{S}_{n}}=\frac{a\left( {{r}^{n}}-1 \right)}{r-1}\]

The first term \[a\] of the A.P. is \[5\] and the common difference \[r\] is \[2\].

Substitute the values of \[a\] and \[r\] in the equation

\[315=\frac{5\left( {{2}^{n}}-1 \right)}{2-1}\]

\[\Rightarrow {{2}^{n}}-1=63\]

\[\Rightarrow {{2}^{n}}=63={{\left( 2 \right)}^{2}}\]

\[\Rightarrow n=6\]

Therefore, the \[{{6}^{th}}\] term is the last term of the G.P.

 \[{{6}^{th}}\]term \[=a{{r}^{6-1}}=\left( 5 \right){{\left( 2 \right)}^{5}}=\left( 5 \right)\left( 32 \right)=160\]

Therefore, \[160\] is the last term of the G.P  and the number of terms is \[6\]. 

Ans: Let the first term of the G.P. be \[a\] and the common ratio be \[r\] .

Then, \[a=1\]

\[{{a}_{3}}=a{{r}^{2}}={{r}^{2}}\]

\[{{a}_{5}}=a{{r}^{4}}={{r}^{4}}\]

Therefore,

\[{{r}^{2}}+{{r}^{4}}=90\]

\[\Rightarrow {{r}^{4}}+{{r}^{2}}-90=0\]

\[\Rightarrow {{r}^{2}}=\frac{-1+\sqrt{1+360}}{2}\]

\[=\frac{-1+\sqrt{361}}{2}\]

\[=-10\] or \[9\]

\[\Rightarrow r=\pm 3\]

Therefore, \[\pm 3\] is the common ratio of the G.P.

Ans: Let \[a,ar\] and \[a{{r}^{2}}\] be the three numbers in G.P.

According to the conditions given in the question,

\[a+ar+a{{r}^{2}}=56\]

\[\Rightarrow a\left( 1+r+{{r}^{2}} \right)=56\]            …(1)

An A.P. is formed by

\[a-1,ar-7,a{{r}^{2}}-21\]

Therefore,

\[\left( ar-7 \right)-\left( a-1 \right)=\left( a{{r}^{2}}-21 \right)-\left( ar-7 \right)b\]

\[\Rightarrow ar-a-6=a{{r}^{2}}-ar-14\]

\[\Rightarrow a{{r}^{2}}-2ar+a=8\]

\[\Rightarrow a{{r}^{2}}-ar-ar+a=8\]

\[\Rightarrow a\left( {{r}^{2}}+1-2r \right)=8\]

\[\Rightarrow a{{\left( {{r}^{2}}-1 \right)}^{2}}=8\]                   …(2)

Equating (1) and (2), we get

\[\Rightarrow 7\left( {{r}^{2}}-2r+1 \right)=1+r+{{r}^{2}}\]

\[\Rightarrow 7{{r}^{2}}-14r+7-1-r-{{r}^{2}}\]

\[\Rightarrow 6{{r}^{2}}-15r+6=0\]

\[\Rightarrow 6{{r}^{2}}-12r-3r+6=0\]

\[\Rightarrow 6\left( r-2 \right)-3\left( r-2 \right)=0\]

\[\Rightarrow \left( 6r-3 \right)\left( r-2 \right)=0\]

Then,\[8,16\] and \[32\] are the three numbers when \[r=2\]  and \[32,16\] and \[8\] are the numbers when \[r=\frac{1}{2}\].  

Therefore, \[8,16\] and \[32\] are the three required numbers in either case. 

Ans:  Let \[{{T}_{1}},{{T}_{2}},{{T}_{3}},{{T}_{4}},...{{T}_{2n}}\] be the G.P.

\[2n\] is the number of terms.

According to the conditions given in the question,

\[{{T}_{1}}+{{T}_{2}}+{{T}_{3}}+...+{{T}_{2n}}=5\left[ {{T}_{1}}+{{T}_{3}}+...+{{T}_{2n-1}} \right]\]

\[\Rightarrow {{T}_{1}}+{{T}_{2}}+{{T}_{3}}+...+{{T}_{2n}}-5\left[ {{T}_{1}}+{{T}_{3}}+...+{{T}_{2n-1}} \right]=0\]

\[\Rightarrow {{T}_{2}}+{{T}_{4}}+...+{{T}_{2n}}=4\left[ {{T}_{1}}+{{T}_{3}}+...+{{T}_{2n-1}} \right]\]

Let \[a,ar,a{{r}^{2}},a{{r}^{3}}\] be the G.P.

Therefore,

\[\frac{ar\left( {{r}^{n}}-1 \right)}{r-1}=\frac{4\times a\left( {{r}^{n}}-1 \right)}{r-1}\]

\[\Rightarrow ar=4a\]

\[\Rightarrow r=4\]

Therefore, \[4\] is the common ratio of the G.P.

Ans: Given,

\[\frac{a+bx}{a-bx}=\frac{b+cx}{b-cx}\]

\[\Rightarrow \left( a+bx \right)\left( b-cx \right)=\left( b+cx \right)\left( a-bx \right)\]

\[\Rightarrow ab-acx+{{b}^{2}}x-bc{{x}^{2}}=ab-{{b}^{2}}x+-acx-bc{{x}^{2}}\]

\[\Rightarrow 2{{b}^{2}}x=2acx\]

\[\Rightarrow {{b}^{2}}=ac\]

\[\Rightarrow \frac{b}{a}=\frac{c}{b}\]

It is also given that,

\[\frac{b+cx}{b-cx}=\frac{c+dx}{c-dx}\]

\[\Rightarrow \left( b+cx \right)\left( c-dx \right)=\left( b-cx \right)\left( c+dx \right)\]

\[\Rightarrow bc-bdx+{{c}^{2}}x-cd{{x}^{2}}=bc+bdx-{{c}^{2}}x-cd{{x}^{2}}\]

\[\Rightarrow 2{{c}^{2}}x=2bdx\]

\[\Rightarrow {{c}^{2}}=bd\]

\[\Rightarrow \frac{c}{d}=\frac{d}{c}\]

Equating both results, we get

\[\frac{b}{a}=\frac{c}{b}=\frac{d}{b}\]

Therefore, it is proved that \[a,b,c\] and \[d\] are in G.P.

Ans: Let \[a,ar,a{{r}^{2}},a{{r}^{3}}...a{{r}^{n-1}}\] be the G.P.

According to the conditions given in the question,

\[S=\frac{a\left( {{r}^{n}}-1 \right)}{r-1}\] 

\[P={{a}^{n}}\times {{r}^{1+2+...+n-1}}\] 

Since the sum of first \[n\] natural numbers is \[n\frac{\left( n+1 \right)}{2}\]

\[\Rightarrow P={{a}^{n}}{{r}^{\frac{n\left( n-1 \right)}{2}}}\] 

\[R=\frac{1}{a}+\frac{1}{ar}+...+\frac{1}{a{{r}^{n-1}}}\]

\[=\frac{{{r}^{n-1}}+{{r}^{n-2}}+...r+1}{a{{r}^{n-1}}}\]

Since \[1,r,...{{r}^{n-1}}\]forms a G.P.,

\[\Rightarrow R=\frac{1\left( {{r}^{n}}-1 \right)}{\left( r-1 \right)}\times \frac{1}{a{{r}^{n-1}}}\]            

\[=\frac{{{r}^{n}}-1}{a{{r}^{n-1}}\left( r-1 \right)}\]

Then,

\[{{P}^{2}}{{R}^{n}}={{a}^{2n}}{{r}^{n\left( n-1 \right)}}\frac{{{\left( {{r}^{n}}-1 \right)}^{n}}}{{{a}^{n}}{{r}^{n\left( n-1 \right)}}{{\left( r-1 \right)}^{n}}}\]

\[=\frac{{{a}^{n}}{{\left( {{r}^{n}}-1 \right)}^{n}}}{{{\left( r-1 \right)}^{n}}}\]

\[={{\left[ \frac{a\left( {{r}^{n}}-1 \right)}{\left( r-1 \right)} \right]}^{n}}\]

\[={{S}^{n}}\]

Therefore, \[{{P}^{2}}{{R}^{n}}={{S}^{n}}\].

Ans:  Given: 

\[a,b,c\] and \[d\] are in G.P.

Therefore,

\[{{b}^{2}}=ac\]

\[{{c}^{2}}=bd\]

\[ad=bc\]

To prove: 

\[\left( {{a}^{n}}+{{b}^{n}} \right),\left( {{b}^{n}}+{{c}^{n}} \right),\left( {{c}^{n}}+{{d}^{n}} \right)\] are in G.P.

That is, \[{{\left( {{b}^{n}}+{{c}^{n}} \right)}^{2}}=\left( {{a}^{n}}+{{b}^{n}} \right),\left( {{c}^{n}}+{{d}^{n}} \right)\]

Then, 

L.H.S \[={{\left( {{b}^{n}}+{{c}^{n}} \right)}^{2}}\]

\[={{b}^{2n}}+2{{b}^{n}}{{c}^{n}}+{{c}^{2n}}\]

\[={{\left( {{b}^{2}} \right)}^{n}}+2{{b}^{n}}{{c}^{n}}+{{\left( {{c}^{2}} \right)}^{n}}\]

\[={{\left( ac \right)}^{n}}+2{{b}^{n}}{{c}^{n}}+{{\left( bd \right)}^{n}}\]

\[={{a}^{n}}{{c}^{n}}+{{b}^{n}}{{c}^{n}}+{{b}^{n}}{{c}^{n}}+{{b}^{n}}{{d}^{n}}\]

\[={{a}^{n}}{{c}^{n}}+{{b}^{n}}{{c}^{n}}+{{a}^{n}}{{d}^{n}}+{{b}^{n}}{{d}^{n}}\]

\[={{c}^{n}}\left( {{a}^{n}}+{{b}^{n}} \right)+{{d}^{n}}\left( {{a}^{n}}+{{b}^{n}} \right)\]

\[=\left( {{a}^{n}}+{{b}^{n}} \right)\left( {{a}^{n}}+{{d}^{n}} \right)\]

\[=\]R.H.S

Therefore,

\[{{\left( {{b}^{n}}+{{c}^{n}} \right)}^{2}}=\left( {{a}^{n}}+{{b}^{n}} \right)\left( {{c}^{n}}+{{d}^{n}} \right)\]

Therefore, \[\left( {{b}^{n}}+{{c}^{n}} \right),\left( {{b}^{n}}+{{c}^{n}} \right)\] and \[\left( {{c}^{n}}+{{d}^{n}} \right)\] are in G.P.

Ans: Given: \[a\] and \[b\] are the roots of \[{{x}^{2}}-3x+p=0\].

Therefore,

\[a+b=3\] and \[ab=p\]    …(1)      

We also know that \[c\] and \[d\] are the roots of \[{{x}^{2}}-12x+q=0\].

Therefore,

\[c+d=12\] and \[cd=q\]   …(2)      

Also, \[a,b,c,d\] are in G.P.

Let us take \[a=x,b=xr,c=x{{r}^{2}}\] and \[d=x{{r}^{3}}\].

We get from (1) and (2) that,

\[x+xr=3\]

\[\Rightarrow x\left( 1+r \right)=3\]

Also,

\[x{{r}^{2}}+x{{r}^{3}}=12\]

\[\Rightarrow x{{r}^{2}}+\left( 1+r \right)=12\]

Divide both the equations obtained.

\[\frac{x{{r}^{2}}\left( 1+r \right)}{x\left( 1+r \right)}=\frac{12}{3}\]

\[\Rightarrow {{r}^{2}}=4\]

\[\Rightarrow r=\pm 2\]

\[x=\frac{3}{1+2}=\frac{3}{3}=1\], when \[r=2\] and

\[x=\frac{3}{1-2}=\frac{3}{-1}=-3\], when \[r=-2\].

Case I:

\[ab={{x}^{2}}r=2\], \[cd={{x}^{2}}{{r}^{5}}=32\] when \[r=2\] and \[x=1\] .

Therefore, 

\[\frac{q+p}{q-p}=\frac{32+2}{32-2}=\frac{34}{30}=\frac{17}{15}\]

\[\Rightarrow \left( q+p \right):\left( q-p \right)=17:15\]

Case II:

\[ab={{x}^{2}}r=18\], \[cd={{x}^{2}}{{r}^{5}}=-288\] when \[r=-2\] and \[x=-3\] .

Therefore,

\[\frac{q+p}{q-p}=\frac{-288-18}{-288+18}=\frac{-306}{-270}=\frac{17}{15}\]

\[\Rightarrow \left( q+p \right):\left( q-p \right)=17:15\]

Therefore, it is proved that \[\left( q+p \right):\left( q-p \right)=17:15\]as we obtain the same for both cases. 

Ans: Let \[a\] and \[b\] be the two numbers.

The arithmetic mean, A.M \[=\frac{a+b}{2}\] and the geometric mean, G.M \[=\sqrt{ab}\]

According to the conditions given in the question,

\[\frac{a+b}{2\sqrt{ab}}=\frac{m}{n}\]

\[\Rightarrow \frac{{{\left( a+b \right)}^{2}}}{4\left( ab \right)}=\frac{{{m}^{2}}}{{{n}^{2}}}\]

\[\Rightarrow \left( a+b \right)=\frac{4ab{{m}^{2}}}{{{n}^{2}}}\]

\[\Rightarrow \left( a+b \right)=\frac{2\sqrt{ab}m}{n}\]                         …(1)

Using the above equation in the identity \[{{\left( a-b \right)}^{2}}={{\left( a+b \right)}^{2}}-4ab\] , we obtain

\[{{\left( a-b \right)}^{2}}=\frac{4ab{{m}^{2}}}{{{n}^{2}}}-4ab=\frac{4ab\left( {{m}^{2}}-{{n}^{2}} \right)}{{{n}^{2}}}\]

\[\Rightarrow \left( a-b \right)=\frac{2\sqrt{ab}\sqrt{{{m}^{2}}-{{n}^{2}}}}{n}\]             …(2)

Add equation (1) and (2)

\[2a=\frac{2\sqrt{ab}}{n}\left( m+\sqrt{{{m}^{2}}-{{n}^{2}}} \right)\]

\[\Rightarrow a=\frac{\sqrt{ab}}{n}\left( m+\sqrt{{{m}^{2}}-{{n}^{2}}} \right)\]

Substitute in (1) the value of \[a\].

\[b=\frac{2\sqrt{ab}}{n}m-\frac{\sqrt{ab}}{n}\left( m+\sqrt{{{m}^{2}}-{{n}^{2}}} \right)\]

\[=\frac{\sqrt{ab}}{n}m-\frac{\sqrt{ab}}{n}\sqrt{{{m}^{2}}-{{n}^{2}}}\]

\[=\frac{\sqrt{ab}}{n}\left( m-\sqrt{{{m}^{2}}-{{n}^{2}}} \right)\]

Therefore,

\[a:b=\frac{a}{b}=\frac{\frac{\sqrt{ab}}{n}\left( m+\sqrt{{{m}^{2}}-{{n}^{2}}} \right)}{\frac{\sqrt{ab}}{n}\left( m-\sqrt{{{m}^{2}}-{{n}^{2}}} \right)}=\frac{\left( m+\sqrt{{{m}^{2}}-{{n}^{2}}} \right)}{\left( m-\sqrt{{{m}^{2}}-{{n}^{2}}} \right)}\]

Therefore, it is proved that \[a:b=\left( m+\sqrt{{{m}^{2}}-{{n}^{2}}} \right):\left( m-\sqrt{{{m}^{2}}-{{n}^{2}}} \right)\].