Chapter 8: Sequences and Series

Ans:

The given sum can be written as

$\sum\limits_{\text{k=1}}^{\text{11}}{\left( \text{2+}{{\text{3}}^{\text{k}}} \right)}=\sum\limits_{k=1}^{11}{2}+\sum\limits_{\text{k=1}}^{\text{11}}{{{\text{3}}^{\text{k}}}}=22+\sum\limits_{\text{k=1}}^{\text{11}}{{{\text{3}}^{\text{k}}}}$                    …… (i)

Again, $\sum\limits_{\mathbf{k=1}}^{\mathbf{11}}{{{3}^{k}}}={{3}^{1}}+{{3}^{2}}+{{3}^{3}}+\cdot \cdot \cdot +{{3}^{11}}$, which is in a geometric progression with the first term $a=3$ and the common ratio $r=3$.

Therefore, we have

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

$\Rightarrow {{S}_{n}}=\dfrac{3\left[ {{3}^{11}}-1 \right]}{3-1}$

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

Thus,

$\sum\limits_{k=1}^{11}{{{3}^{k}}}=\dfrac{3}{2}\left( {{3}^{11}}-1 \right)$.                                                 …… (ii)

Hence, from the equation (i) and (ii), it gives

$\sum\limits_{\text{k=1}}^{\text{11}}{\left( \text{2+}{{\text{3}}^{\text{k}}} \right)}=22+\dfrac{3}{2}\left( {{3}^{11}}-1 \right)$.

Ans:

Suppose that the first three terms of the geometric progression are $\dfrac{a}{r},\,\,a,\,\,ar$.

Now, by the given conditions,

$\dfrac{a}{r}+a+ar=\dfrac{39}{10}$                                        …… (i)

$\left( \dfrac{a}{r} \right)\cdot \left( a \right)\cdot \left( ar \right)=1$                                    …… (ii)

Simplifying the equation (ii) gives

${{a}^{3}}=1$

$\Rightarrow a=1$, neglecting the imaginary roots.

Thus, replacing $a$ by $1$ in the equation (i) gives

$\dfrac{1}{r}+1+r=\dfrac{39}{10}$

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

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

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

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

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

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

Hence, $\dfrac{5}{2},$ $1$, and $\dfrac{2}{5}$ are the three terms of the geometric progression.

Ans:

It is provided that the terms $\text{3,}\,{{\text{3}}^{\text{2}}}\text{,}\,{{\text{3}}^{\text{3}}}\text{,}...$ are geometric progression with the first term $a=3$ and the common ratio $r=3$.

Suppose that to get the sum of the G.P. as $120$, $n$ terms are needed to be considered.

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

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

Therefore,

$\begin{align}  & \dfrac{3\left( {{3}^{n}}-1 \right)}{3-1}=120 \\  & \Rightarrow {{3}^{n}}-1=120\times \dfrac{2}{3} \\  & \Rightarrow {{3}^{n}}-1=80 \\  & \Rightarrow {{3}^{n}}=81 \\  & \Rightarrow {{3}^{n}}={{3}^{4}} \\ \end{align}$

$\Rightarrow n=4$.

Hence, to get the sum of $\text{3,}\,{{\text{3}}^{\text{2}}}\text{,}\,{{\text{3}}^{\text{3}}}\text{,}...$ as $120$, $4$ terms are needed to be considered.

Ans:

Suppose that the first six terms of the given geometric progression are

$a,\,ar,\,a{{r}^{2}},\,a{{r}^{3}},\,a{{r}^{4}},\,a{{r}^{5}}$.

Then, by the condition provided to us,

$a+ar+a{{r}^{2}}=16$              

 $\Rightarrow a\left( 1+r+{{r}^{3}} \right)=16$                                           …… (i)

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

$\Rightarrow a{{r}^{3}}\left( 1+r+{{r}^{2}} \right)=128$                                       …… (ii)

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

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

$\begin{align}   & \Rightarrow {{r}^{3}}=8 \\  & \Rightarrow {{r}^{3}}={{2}^{3}} \\ \end{align}$

$\Rightarrow r=2$                                                            …… (iii)

Therefore, the equation (i) and (iii) together implies

$a\left( 1+2+{{2}^{2}} \right)=16$

$\begin{align}   & \Rightarrow a\times 7=16 \\  & \Rightarrow a=\dfrac{16}{7} \\ \end{align}$

Hence, the sum of $n$ terms of the G.P. is given by

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

$=\dfrac{16}{7}\dfrac{\left( {{2}^{n}}-1 \right)}{2-1}$

$=\dfrac{16}{7}\left( {{2}^{n}}-1 \right)$.

Ans:

The first term of the G.P. is $a=729$.

Suppose that $r$ is the common ratio of the geometric progression.

Therefore, by the formula, ${{n}^{th}}$ term of the sequence in G.P.,

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

So, ${{a}_{7}}=\left( 729 \right){{r}^{7-1}}=729\cdot {{r}^{6}}$

$\begin{align}   & \Rightarrow 64=729{{r}^{6}} \\  & \Rightarrow {{r}^{6}}={{\left( \dfrac{2}{3} \right)}^{6}} \\  & \Rightarrow r=\dfrac{2}{3}. \\ \end{align}$

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

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

Therefore, for $n=7$, we have

${{S}_{7}}=\dfrac{729\left[ 1-{{\left( \dfrac{2}{3} \right)}^{7}} \right]}{1-\dfrac{2}{3}}$

$={{3}^{7}}\times \dfrac{{{3}^{7}}-{{2}^{7}}}{{{3}^{7}}}$

$\begin{align}   & ={{3}^{7}}-{{2}^{7}} \\  & =2187-128 \\ \end{align}$

$=2059$.

Hence, ${{S}_{7}}=2059$.

Ans:

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

So, by the condition provided to us,

$-4=\dfrac{a\left( 1-{{r}^{2}} \right)}{1-r}$                                                    …… (i)

${{a}_{5}}=4\times {{a}_{3}}$                                                          …… (ii)

Now, we know that, the ${{n}^{th}}$ term of a G.P.,

${{a}_{n}}=a{{r}^{n-1}}$.
Therefore, 

${{a}_{3}}=a{{r}^{3-1}}=a{{r}^{2}}$                                                   …… (iii)

and ${{a}_{5}}=a{{r}^{5-1}}=a{{r}^{4}}$.                                           …… (iv)

Thus, from the equation (ii), (iii), and (iv) we have

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

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

$\Rightarrow r=\pm 2$.

Substituting $r=2$ into the equation (i), it gives

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

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

$\Rightarrow -4=3a$

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

Also, substituting $r=-2$ into the equation (i), gives

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

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

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

$\Rightarrow a=4$.

Hence, the G.P. can be of the form $-\dfrac{4}{3},-\dfrac{8}{3},-\dfrac{16}{3},...$ or $4,-8,-16,-32,...$

Ans:

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

Then, by the condition provided to us,

${{a}_{4}}=a{{r}^{3}}=x$                                             …… (i)

${{a}_{10}}=a{{r}^{9}}=y$                                            …… (ii)

${{a}_{16}}=a{{r}^{15}}=z$                                           …… (iii)

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

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

$\Rightarrow \dfrac{y}{x}={{r}^{6}}$                                                  ……(iv)

Again, divide the equation (iii) by the equation (ii). Then, it gives

$\dfrac{z}{y}=\dfrac{a{{r}^{15}}}{a{{r}^{9}}}$

$\Rightarrow \dfrac{z}{y}={{r}^{6}}$                                                  ..,…. (v)

Equating the equations (iv) and (v), we have

$\dfrac{y}{x}=\dfrac{z}{y}$.

Hence, it has been proved that $x,y,z$ are in G.P.

Ans:

The provided sequence $\text{8,}\,\text{88,}\,\text{888,}\,\text{8888,}...$ is not in a geometric progression (G.P.), but by rearranging and rewriting the terms, it can be converted into a G.P.

The sum of the $n$ terms of the given sequence is given by

${{S}_{n}}=8+88+888+8888+\cdot \cdot \cdot \,\text{to n terms}$

$=\dfrac{8}{9}\left[ 9+99+999+9999+\cdot \cdot \cdot \,\text{to n terms} \right]$

$=\dfrac{8}{9}\left[ \left( 10-1 \right)+\left( {{10}^{2}}-1 \right)+\left( {{10}^{3}}-1 \right)+\left( {{10}^{4}}-1 \right)+\cdot \cdot \cdot \,\,\text{to n terms} \right]$

$=\dfrac{8}{9}\left[ \left( 10+{{10}^{2}}+\cdot \cdot \cdot +\,\,\text{n terms} \right)-\left( 1+1+1+\cdot \cdot \cdot \,\text{n terms} \right) \right]$

$\begin{align}   & =\dfrac{8}{9}\left[ \dfrac{10\left( {{10}^{n}}-1 \right)}{10-1}-n \right] \\  & =\dfrac{80}{81}\left( {{10}^{n}}-1 \right)-\dfrac{8}{9}n. \\  \end{align}$

Hence, the sum of $n$ terms of the sequence $\text{8,}\,\text{88,}\,\text{888,}\,\text{8888,}...$ is $\dfrac{80}{81}\left( {{10}^{n}}-1 \right)-\dfrac{8}{9}n$.

Ans:

The sum of the product of the corresponding terms $\text{2,}\,\text{4,}\,\text{8,}\,\text{16,}\,\text{32}$ and $\text{128,}\,\text{32,}\,\text{8,}\,\text{2,}\,\dfrac{\text{1}}{\text{2}}$, is given by

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

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

Now, note that, the terms $4,\,2,\,1,\,\dfrac{1}{2},\,\dfrac{1}{{{2}^{2}}}$ are in geometric progression, with the first term $a=4$ and the common ratio $r=\dfrac{2}{4}=\dfrac{1}{2}$.

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

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

Therefore,

$\begin{align}  & {{S}_{5}}=\dfrac{4\left[ 1-{{\left( \dfrac{1}{2} \right)}^{5}} \right]}{1-\dfrac{1}{2}} \\  & =\dfrac{4\left[ 1-\dfrac{1}{32} \right]}{\dfrac{1}{2}} \\  & =8\left( \dfrac{32-1}{32} \right) \\  & =\dfrac{31}{4}. \\ \end{align}$

Hence, the sum of the products of the corresponding terms of the given sequences is $\dfrac{31}{4}$.

Ans:

The product of the corresponding terms of the sequences $\text{a,}\,\text{ar,}\,\text{a}{{\text{r}}^{\text{2}}}\text{,}...\text{,}\,\text{a}{{\text{r}}^{\text{n-1}}}$ and $\text{A,}\,\text{AR,}\,\text{A}{{\text{R}}^{\text{2}}}\text{,}...\text{,}\,\text{A}{{\text{R}}^{\text{n-1}}}$ is $aA,\,arAR,\,a{{r}^{2}}A{{R}^{2}},...,a{{r}^{n-1}}A{{R}^{n-1}}$.

Now, we are to be proved that the product is in G.P.

So. $\dfrac{\text{Second term }}{\text{First term}}=\dfrac{arAR}{aA}=rR$                                  …… (i)

Also,

$\dfrac{\text{Third term}}{\text{Second term}}=\dfrac{a{{r}^{2}}A{{R}^{2}}}{arAR}=rR$                                      …… (ii)

Thus, by equating the equations (i) and (ii), we have

$\dfrac{\text{Second term }}{\text{First term}}=\dfrac{\text{Third term}}{\text{Second term}}$$=rR$, the common ratio.

Hence, it has been proved that the product of the corresponding terms of the sequences $\text{a,}\,\text{ar,}\,\text{a}{{\text{r}}^{\text{2}}}\text{,}...\text{,}\,\text{a}{{\text{r}}^{\text{n-1}}}$ and $\text{A,}\,\text{AR,}\,\text{A}{{\text{R}}^{\text{2}}}\text{,}...\text{,}\,\text{A}{{\text{R}}^{\text{n-1}}}$ is in G.P. with the common ratio $rR$.