Chapter 8: Sequences and Series

Ans:

The G.P. is provided as $\dfrac{\text{5}}{\text{2}}\text{,}\dfrac{\text{5}}{\text{4}}\text{,}\dfrac{\text{5}}{\text{8}}\text{,}...$.

Clearly, $a=\dfrac{5}{2}$ is the first term and $r=\dfrac{\dfrac{5}{4}}{\dfrac{5}{2}}=\dfrac{1}{2}$ is the common ratio of the G.P.

Therefore, the ${{20}^{th}}$ term of the G.P. is

$\begin{align}   & {{a}_{20}}=a{{r}^{20-1}} \\  & =\dfrac{5}{2}{{\left( \dfrac{1}{2} \right)}^{19}} \\   & =\dfrac{5}{2\times {{2}^{19}}} \\   & =\dfrac{5}{{{2}^{20}}}. \\ \end{align}$ 

Also, the ${{n}^{th}}$ term of the G.P. is 

$\begin{align}   & {{a}_{n}}=a{{r}^{n-1}} \\  & =\dfrac{5}{2}{{\left( \dfrac{1}{2} \right)}^{n-1}} \\  & =\dfrac{5}{2\cdot {{2}^{n-1}}} \\  & =\dfrac{5}{{{2}^{n}}}. \\ \end{align}$

Hence, the ${{20}^{th}}$ and ${{n}^{th}}$ terms of the G.P. are respectively $\dfrac{5}{{{2}^{20}}}$ and $\dfrac{5}{{{2}^{n}}}$.

Ans:

The G.P. described in the problem has the common ratio $r=2$.

Suppose that the first term of the G.P. is $a$.

Then, the ${{8}^{th}}$ term of the G.P. is given by

${{a}_{8}}=a{{r}^{8-1}}$$=a{{r}^{7}}$

$\Rightarrow a{{r}^{7}}=192$

$\Rightarrow a{{\left( 2 \right)}^{7}}=192$

$\Rightarrow a{{\left( 2 \right)}^{7}}={{\left( 2 \right)}^{6}}\times 3$

$\Rightarrow a=\dfrac{{{2}^{6}}\times 3}{{{2}^{7}}}=\dfrac{3}{2}$.

Thus, the ${{12}^{th}}$ term,

$\begin{align}   & {{a}_{12}}=a{{r}^{12-1}} \\  & =a{{r}^{11}} \\  & =\dfrac{3}{2}{{\left( 2 \right)}^{11}} \\  & =3{{\left( 2 \right)}^{10}} \\  & =3072. \\ \end{align}$

Hence, the ${{12}^{th}}$ term of the G.P. is $3072$.

Ans:

Suppose that $a$ and $r$ are the first term and the common ratio of the given G.P. Then, by the condition provided to us,

the ${{5}^{th}}$ term of G.P.,

${{a}_{5}}=a{{r}^{5-1}}=a{{r}^{4}}=p$                                  …… (i)

The ${{8}^{th}}$ term of the G.P.,

${{a}_{8}}=a{{r}^{8-1}}=a{{r}^{7}}=q$                                  …… (ii)

Also, the ${{11}^{th}}$  term of the G.P.,

${{a}_{11}}=a{{r}^{11-1}}=a{{r}^{10}}=s$                                …… (iii)

Now, divide the equation (ii) by the equation (i). Then, it gives

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

$\Rightarrow {{r}^{3}}=\dfrac{q}{p}$                                               …… (iv)

Again, divide the equation (iii) by the equation (ii). Then, we have

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

$\Rightarrow {{r}^{3}}=\dfrac{s}{q}$                                           …… (v)

Therefore, from the equation (iv) and (v) gives

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

$\Rightarrow {{q}^{2}}=ps$.

Hence, the required result has been shown.

Ans:

Suppose that $r$ is the common ratio of the G.P.

It is given that the first term $a=-3$.

Now, by the formula, ${{n}^{th}}$ term of a G.P.,
${{a}_{n}}=a{{r}^{n-1}}$.

Therefore,

${{a}_{4}}=a{{r}^{3}}=\left( -3 \right){{r}^{3}}$ and ${{a}_{2}}=a{{r}^{1}}=\left( -3 \right)r$.

So, by the condition provided to us,

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

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

$\Rightarrow r=-3$.

Thus, the ${{7}^{th}}$ term of the G.P. is given by

$\begin{align}   & {{a}_{7}}=\left( -3 \right){{\left( -3 \right)}^{7-1}} \\  & ={{\left( -3 \right)}^{7}} \\  & =-2187.  \end{align}$

Hence, the ${{7}^{th}}$ term of the geometric progression is $-2187$.

Ans:

The sequence provided to us is in a G.P. since the first term is $a=2$ and the common ratio is $r=\dfrac{2\sqrt{2}}{2}=\sqrt{2}$.

So, suppose that the ${{n}^{th}}$ term of the G.P. is $128$.

Therefore, we have

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

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

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

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

$\Rightarrow n+1=12$

$\Rightarrow n=11$.

Hence, $128$ is the ${{11}^{th}}$ term of the G.P. (sequence) $\text{2,2}\sqrt{\text{2}}\text{,4,}...$.

Ans:

The sequence provided to us is in a G.P. since the first term is $a=\sqrt{3}$, and the common ratio is $r=\dfrac{3}{\sqrt{3}}=\sqrt{3}$.

Now, suppose that the ${{n}^{th}}$ term of the G.P. is $729$.

Therefore, we have

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

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

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

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

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

$\Rightarrow n=12$.

Hence, $729$ is the ${{n}^{th}}$ term of the geometric progression (sequence) $\sqrt{\text{3}}\text{,3,3}\sqrt{\text{3}}\text{,}...$.

Ans:

The sequence given to us is in a geometric progression (G.P.) since the first term of it is $a=\dfrac{1}{3}$ and the common ratio is

$r=\dfrac{\dfrac{1}{9}}{\dfrac{1}{3}}=\dfrac{1}{3}$.

Now, suppose that $\dfrac{1}{19683}$ is the ${{n}^{th}}$ term of the G.P.

Therefore, we have

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

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

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

$\Rightarrow n=9$.

Hence, $\dfrac{1}{19683}$ is the ${{9}^{th}}$ term of the G.P. (sequence) $\dfrac{\text{1}}{\text{3}}\text{,}\dfrac{\text{1}}{\text{9}}\text{,}\dfrac{\text{1}}{\text{27}}\text{,}...$.

Ans:

The numbers provided to us are $-\dfrac{2}{7},\,x,\,-\dfrac{7}{2}$.

Now, let ${{a}_{1}}=-\dfrac{2}{7}$, ${{a}_{2}}=x$ and ${{a}_{3}}=-\dfrac{7}{2}$.

Then, $\dfrac{{{a}_{2}}}{{{a}_{1}}}=\dfrac{x}{-\dfrac{2}{7}}=\dfrac{-7x}{2}$ and $\dfrac{{{a}_{3}}}{{{a}_{2}}}=\dfrac{-\dfrac{7}{2}}{x}=-\dfrac{7}{2x}$.

Now, if the numbers are in G.P., then

$r=\dfrac{{{a}_{2}}}{{{a}_{1}}}=\dfrac{{{a}_{3}}}{{{a}_{2}}}$

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

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

$\Rightarrow x=\sqrt{1}$

$\Rightarrow x=\pm 1$.

Hence, the numbers $-\dfrac{2}{7},\,x,\,-\dfrac{7}{2}$  are in G.P. if $x=\pm 1$.

Ans:

It is provided that the numbers $\text{0}\text{.15,}\,\,\text{0}\text{.015,}\,\,\text{0}\text{.0015,}\,...$are in geometric progression.

Note that, the first term of the sequence is $a=0.15$ and the common ratio is $r=\dfrac{0.015}{0.15}=0.1$

Now, by the formula, sum of $n$ terms in G.P., we have

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

Therefore,

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

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

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

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

Ans:

It is provided to us that the numbers $\sqrt{7},\sqrt{21},3\sqrt{7},...$ are in geometric progression.

Note that, the first term of the G.P. is $a=\sqrt{7}$ and the common ratio is $r=\dfrac{\sqrt{21}}{\sqrt{7}}=\sqrt{3}$.

Now, by the formula, sum of the first $n$ terms in G.P.,

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

$\begin{align}   & =\dfrac{\sqrt{7}\left[ 1-{{\left( \sqrt{3} \right)}^{n}} \right]}{1-\sqrt{3}} \\  & =\dfrac{\sqrt{7}\left[ 1-{{\left( \sqrt{3} \right)}^{n}} \right]}{1-\sqrt{3}}\cdot \dfrac{1+\sqrt{3}}{1+\sqrt{3}} \\  & =\dfrac{\sqrt{7}\left[ 1-{{\left( \sqrt{3} \right)}^{n}} \right]\left( 1+\sqrt{3} \right)}{1-3} \\  & =\dfrac{-\sqrt{7}\left( \sqrt{3}+1 \right)\left[ 1-{{\left( \sqrt{3} \right)}^{n}} \right]}{2} \\ \end{align}$

Hence, the sum of the first $n$ terms is given by

${{S}_{n}}=\dfrac{\sqrt{7}\left( 1+\sqrt{3} \right)}{2}\left[ {{\left( 3 \right)}^{\dfrac{n}{2}}}-1 \right]$.

Ans:

The numbers $1,-a,{{a}^{2}},-{{a}^{3}},...$ provided to us, is in G.P. with the first term $a=1$ and the common ratio $r=\dfrac{-a}{1}=-a$.

Now, by the formula, sum of the first $n$ terms in G.P.,

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

$=\dfrac{1\left[ 1-{{\left( -a \right)}^{n}} \right]}{1-\left( -a \right)}$

Hence, the sum of $n$ terms of the sequence is given by

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

Ans:

The numbers ${{x}^{3}},\,{{x}^{5}},\,{{x}^{7}},...$ are in G.P. with its first term $a={{x}^{3}}$ and the common ratio $r=\dfrac{{{x}^{5}}}{{{x}^{3}}}={{x}^{2}}$.

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

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

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

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

Hence, the sum of the first $n$ terms of the sequence ${{\text{x}}^{\text{3}}}\text{,}{{\text{x}}^{\text{5}}}\text{,}{{\text{x}}^{\text{7}}}\text{,}...$ is given by  ${{S}_{n}}=\dfrac{{{x}^{3}}\left( 1-{{x}^{2n}} \right)}{1-{{x}^{2}}}$.