Chapter 8: Sequences and Series

Ans:

Suppose that the first term and the common ratio of the geometric progression respectively are $a$ and $r$.

Then the first four terms of the sequence are $a,\,ar,\,a{{r}^{2}},\,a{{r}^{3}}$.

Now, according to the condition provided,

$a{{r}^{2}}=a+9$                                                    …… (i)

$ar=a{{r}^{3}}+18$                                                 …… (ii)

Therefore, rewriting the equations, give

$a\left( {{r}^{2}}-1 \right)=9$                                                …… (iii)

$ar\left( 1-{{r}^{2}} \right)=18$                                              …… (iv)

Now, divide the equation (iv) by the equation (iii). Then, it gives

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

$\Rightarrow r=-2$                                                   …… (v)

Now, using the equation (v) into the equation (i) gives

$\begin{align}   & 4a=a+9 \\  & \Rightarrow 3a=9 \\  & \Rightarrow a=3 \\ \end{align}$

Hence, the first four numbers that forms a geometric progression are $3,\,3\left( -2 \right),\,3{{\left( -2 \right)}^{2}},$ and $3{{\left( -2 \right)}^{3}}$, that is $3,\,-6,\,\,12,\,$ and $-24$.

Ans:

First suppose that the first term and the common ratio of the geometric progression are respectively $A$ and $R$.

Then, by the condition provided to us,

$A{{R}^{p-1}}=a$,                                       …… (i)

$A{{R}^{q-1}}=b$,                                       …… (ii)

and $A{{R}^{r-1}}=c$.                                 …… (iii)

Therefore, we have

${{a}^{q-r}}{{b}^{r-p}}{{c}^{p-q}}$

$={{A}^{q-r}}\times {{R}^{\left( p-1 \right)\left( q-r \right)}}\times {{A}^{r-p}}\times {{R}^{\left( q-1 \right)\left( r-p \right)}}\times {{A}^{p-q}}\times {{R}^{\left( r-1 \right)\left( p-q \right)}}$, using the equations (i), (ii) and (iii).

$\begin{align}   & ={{A}^{q-r+r-p+p-q}}\times {{R}^{\left( pr-pr-q+r \right)\left( rq-r+p-pq \right)+\left( pr-p-qr+q \right)}} \\  & ={{A}^{0}}\times {{R}^{0}} \\  & =1. \\ \end{align}$

Hence, it has been proved that, ${{a}^{q-r}}{{b}^{r-p}}{{c}^{p-q}}=1$.

Ans:

First suppose that, $r$ is the common ratio of the geometric progression.

Since, $a$ is the first of the G.P., so it takes the form $a,\,ar,\,a{{r}^{2}},a{{r}^{3}},...,a{{r}^{n-1}}$.

Therefore, by the given condition,

$b=a{{r}^{n-1}}$                                              …… (i)

Also, 

$P=$ Product of $n$ terms

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

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

$={{a}^{n}}{{r}^{1+2+\cdot \cdot \cdot +\left( n-1 \right)}}$                                      …… (ii)

Now, notice that the numbers $1,2,...,\left( n-1 \right)$ on the power of $r$ are in A.P.

So, the sum of the $n-1$ terms, 

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

$\begin{align}   & =\dfrac{n-1}{2}\left[ 2+n-2 \right] \\  & =\dfrac{n\left( n-1 \right)}{2}. \\ \end{align}$

Thus, from the equation (ii) we have

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

$\begin{align}   & \Rightarrow {{P}^{2}}={{a}^{2n}}{{r}^{n\left( n-1 \right)}} \\  & ={{\left[ {{a}^{2}}{{r}^{\left( n-1 \right)}} \right]}^{n}} \\  & ={{\left( a\times a{{r}^{n-1}} \right)}^{n}} \\ \end{align}$

$={{\left( ab \right)}^{n}}$, applying the result of equation (i).

Hence, it has been proved that $P={{\left( ab \right)}^{n}}$.

Ans:

Suppose that $a$ and $r$ respectively are the first term and the common ratio of the geometric progression.

Therefore, by the formula, sum of the first $n$ terms of a G.P.,

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

Now, note that from ${{\left( \text{n+1} \right)}^{\text{th}}}$ to ${{\left( \text{2n} \right)}^{\text{th}}}$, there are $n$ terms.

So, the sum of the terms between these two terms,

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

Thus, the ${{\left( n+1 \right)}^{th}}$ term, ${{a}_{n+1}}=a{{r}^{n+1-1}}=a{{r}^{n}}$.

So, the ratio of the sums,

$\begin{align}   & \dfrac{{{S}_{n}}}{{{{{S}'}}_{n}}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}\times \dfrac{\left( 1-r \right)}{a{{r}^{n}}\left( 1-{{r}^{n}} \right)} \\  & =\dfrac{1}{{{r}^{n}}}. \\ \end{align}$

Hence, it has been shown that the ratio of the sum of the sum of first $\text{n}$ terms of a G.P. to the sum of terms from ${{\left( \text{n+1} \right)}^{\text{th}}}$ to ${{\left( \text{2n} \right)}^{\text{th}}}$ term is $\dfrac{\text{1}}{{{\text{r}}^{\text{n}}}}$. 

Ans:

It is given that the numbers $a,b,c$ and $d$ are in geometric progression (G.P.). So, we have

$\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}$.

That is,

$bc=ad$                                    …… (i)

${{b}^{2}}=ac$                                     …… (ii) 

${{c}^{2}}=bd$                                     …… (iii)

Now, Right-Hand-Side of the statement that needs to be proved,

$={{\left( ab+bc+cd \right)}^{2}}$

$={{\left( ab+ad+cd \right)}^{2}}$, applying the result of the equation (i)

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

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

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

\[={{a}^{2}}{{b}^{2}}+2{{a}^{2}}{{c}^{2}}+2{{b}^{2}}{{c}^{2}}+{{d}^{2}}{{a}^{2}}+2{{d}^{2}}{{b}^{2}}+{{d}^{2}}{{c}^{2}}\], applying the results of equations (i) and (ii).

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

$={{a}^{2}}{{b}^{2}}+{{a}^{2}}{{c}^{2}}+{{a}^{2}}{{d}^{2}}+{{b}^{2}}\cdot {{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{b}^{2}}{{d}^{2}}+{{c}^{2}}{{b}^{2}}+{{c}^{2}}\times {{c}^{2}}+{{c}^{2}}{{d}^{2}}$, applying the results of equations (ii) and (iii) and rewriting the terms.

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

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

$=$ Left-Hand-Side of the proof statement.

Hence, it has been proved that

$\left( {{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{+}{{\text{c}}^{\text{2}}} \right)\left( {{\text{b}}^{\text{2}}}\text{+}{{\text{c}}^{\text{2}}}\text{+}{{\text{d}}^{\text{2}}} \right)\text{=}{{\left( \text{ab+bc+cd} \right)}^{\text{2}}}$.

Ans:

Suppose the two numbers that needed to be inserted between $3$ and $31$ are ${{G}_{1}}$ and ${{G}_{2}}$, so that it forms a G.P. sequence $3,\,{{G}_{1}},\,{{G}_{2}},\,81$.

Also, assume that the first term and the common ratio respectively are $a$ and $r$.

Then, according to the given condition we have

$3{{\left( r \right)}^{4-1}}=81$

$\Rightarrow {{r}^{3}}=27$

$\Rightarrow r=3$, neglecting the imaginary roots.

Therefore, the terms 

${{G}_{1}}=ar=3\times 3=9$ and 

${{G}_{2}}=a{{r}^{2}}=3{{\left( 3 \right)}^{2}}=27$.

Hence, the numbers that need to be inserted between $3$ and $81$ are $9$ and $27$.

Ans:

It is known that the geometric mean between $a$ and $b$ is $\sqrt{ab}$.

Therefore, according to the condition provided,

$\dfrac{{{a}^{n+1}}+{{b}^{n+1}}}{{{a}^{n}}+{{b}^{n}}}=\sqrt{ab}$

$\Rightarrow \dfrac{{{\left( {{a}^{n+1}}+{{b}^{n+1}} \right)}^{2}}}{{{\left( {{a}^{n}}+{{b}^{n}} \right)}^{2}}}=ab$, squaring both sides of the equation.

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

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

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

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

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

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

$\Rightarrow 2n+1=0$

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

Hence, for $n=-\dfrac{1}{2}$, $\dfrac{{{\text{a}}^{\text{n-1}}}\text{+}{{\text{b}}^{\text{n-1}}}}{{{\text{a}}^{\text{n}}}\text{+}{{\text{b}}^{\text{n}}}}$ may be the geometric mean between $\text{a}$ and $b$.

Ans:

Suppose that $a$ and $b$ are the two numbers.

Then, their geometric mean $=\sqrt{ab}$.

So, by the condition provided to us,

$a+b=6\sqrt{ab}$                                                  …… (i)

Squaring both sides of the equation (i), we have

${{\left( a+b \right)}^{2}}=36ab$                                             …… (ii)

Now,

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

$=36ab-4ab$, using the equation (ii).

$=32ab$.

$\Rightarrow a-b=\sqrt{32}\sqrt{ab}=4\sqrt{2}\sqrt{ab}$                      …… (iii)

Therefore, adding the equation (i) and (iii), gives

$2a=\left( 6+4\sqrt{2} \right)\sqrt{ab}$

$\Rightarrow a=\left( 3+2\sqrt{2} \right)\sqrt{ab}$                                        …… (iv)

Now, using the equation (iv) into the equation (i), gives

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

$\Rightarrow b=\left( 3-2\sqrt{2} \right)\sqrt{ab}$                                         ….. (v)

Thus, dividing the equation (iv) by (v) we have,

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

Hence, the ratio of the numbers is $\left( 3+2\sqrt{2} \right):\left( 3-2\sqrt{2} \right)$.

Ans:

Suppose that $a$ and $b$ are the two positive numbers.

Then, according to the definition, A.M and G.M of the numbers are

$A=\dfrac{a+b}{2}$                                 …… (i)

$G=\sqrt{ab}$                                  …… (ii)

Rewriting the equations (i) and (ii), gives

$a+b=2A$                               …… (iii)

$ab={{G}^{2}}$                                   …… (iv)

Now, it is known that,

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

$={{\left( 2A \right)}^{2}}-4{{G}^{2}}$, using the equations (iii) and (iv).

$\begin{align}   & =4{{A}^{2}}-4{{G}^{2}} \\  & =4\left( {{A}^{2}}-{{G}^{2}} \right) \\ \end{align}$

Therefore,

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

$\Rightarrow a-b=2\sqrt{\left( A+G \right)\left( A-G \right)}$       …… (v)

Adding the equations (iii) and (v) we have

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

$\Rightarrow a=A+\sqrt{\left( A+G \right)\left( A-G \right)}$        …… (vi)

Thus, the equations (iii) and (vi) gives

$\begin{align}   & b=2A-A-\sqrt{\left( A+G \right)\left( A-G \right)} \\  & =A-\sqrt{\left( A+G \right)\left( A-G \right)}. \\ \end{align}$

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

Ans:
Since, the number of bacteria is increasing by two times in each hour, so the problem makes a geometric progression.

It is given that, originally there were $30$ bacteria.

So, the first term of the G.P. is $a=30$ and the common ratio is $r=2$.

Thus, the number of bacteria will present at the end of ${{2}^{nd}}$ hour is given by  ${{a}_{3}}=a{{r}^{3-1}}=\left( 30 \right){{\left( 2 \right)}^{2}}=120$.

The number of bacteria will present at the end of ${{4}^{th}}$ hour is given by

${{a}_{5}}=a{{r}^{5-1}}=\left( 30 \right){{\left( 2 \right)}^{4}}=480$.

Also, the number of bacteria will present at the end of ${{n}^{th}}$ hour is given by ${{a}_{n+1}}=a{{r}^{n}}=\left( 30 \right){{2}^{n}}$.