Question

The ${{9}^{th}}$ term of the series $27+9+5\dfrac{2}{5}+3\dfrac{6}{7}+.......$ will be
1)$1\dfrac{10}{17}$
2)$\dfrac{10}{17}$
3)$\dfrac{16}{27}$
4)$\dfrac{17}{27}$

Answer 1

Hint: In this question we will first learn some basic & important concepts of sequences and series. Then we will try to change our series in simpler form so that we can observe and find a general formula for the ${{n}^{th}}$term. After that we will substitute $n=9$ in the general formula that we have calculated to get our required answer.

Complete step-by-step solution:
To understand series we should have basic knowledge of sequence
Sequence is a list of numbers which are arranged in a sequential way.
Let us observe some examples of sequences
Example 1: $2,4,6,8,.........$
Example 2: $2,4,8,16,....$
In example 1 observe that if we subtract any two successive terms we will obtain $2$
In example 2 observe that if we divide any two successive terms we will obtain $2$
First example is of Arithmetic sequence and the second example is of Geometric sequence.
Now we will learn some basic concepts of Arithmetic Sequence and Geometric Sequence.
Arithmetic Sequence or Arithmetic Progression
In arithmetic progression (A.P) we add or subtract a particular number to create progression. That particular number is called common difference.
First term of a progression is denoted by $a$ and we represent the terms of arithmetic progression by ${{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}},{{a}_{5}},{{a}_{6,}}........$
Common difference is denoted as $d$ . We will find the value of $d$ by subtracting two successive terms of progression.
$d={{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}={{a}_{n}}-{{a}_{n-1}}$ .
${{n}^{th}}$ Term of arithmetic progression is denoted as ${{a}_{n}}$ .
The formula for ${{n}^{th}}$ term of arithmetic progression is ${{a}_{n}}=a+\left( n-1 \right)d$ where $a$is the first term and $d$ is common difference of arithmetic progression.
$2,4,6,8,.........$In this example $a=2\And d=2$
Geometric Sequence or Geometric Progression:
In geometric progression (G.P) we multiply or divide a particular number to create progression. That particular number is called the common ratio.
First term of a progression is denoted by $a$ and we represent the terms of geometric progression by ${{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}},{{a}_{5}},{{a}_{6,}}........$
Common ratio is denoted as $r$ . We find the value of $r$ by dividing two successive terms of progression.
$r=\dfrac{{{a}_{2}}}{{{a}_{1}}}=\dfrac{{{a}_{3}}}{{{a}_{2}}}=\dfrac{{{a}_{n}}}{{{a}_{n-1}}}$ .
${{n}^{th}}$ Term is denoted as ${{a}_{n}}$ .
The formula for ${{n}^{th}}$ term of geometric progression is
 ${{a}_{n}}=a{{r}^{n-1}}$
Where $a$ is first term and $r$ is the common ratio of geometric progression.
$2,4,8,16,....$ In this example $a=2\And r=2$
Now we will learn about series
Series is the sum or difference of terms of sequence.
Series can be finite as well as infinite
Example1: $1+2+3+4+5$
Example2: $1+2+3+4+5+...........$
Example 1 is a series of finite terms and example 2 is a series of infinite terms.
We denote series as,
$\sum{{{a}_{i}}=}$ $\sum{{{a}_{i}}={{a}_{1}}+{{a}_{2}}+{{a}_{3}}+......}$
Where ${{a}_{1}},{{a}_{2}},{{a}_{3}},......$ are terms of sequence.
Now we will proceed to our question
We have to find the ${{9}^{th}}$ term of the series
$27+9+5\dfrac{2}{5}+3\dfrac{6}{7}+.......$
We will simplify the terms as
$27+9+\dfrac{27}{5}+\dfrac{27}{7}+.......$
We can also write it as
$\dfrac{27}{1}+\dfrac{27}{3}+\dfrac{27}{5}+\dfrac{27}{7}+.......$
We can see that numerator is $27$ for all terms and denominators are $1,3,5,7...$ for ${{1}^{st,}}{{2}^{nd}},{{3}^{rd}},{{4}^{th}}....$ terms respectively. This is a sequence with $a=1\And d=2$where $a$ is first term and $d$ is common difference
Now we will make general formula of ${{n}^{th}}$ term of sequence
${{a}_{n}}=a+\left( n-1 \right)d$
Substituting$a=1\And d=2$, we will get
\[\begin{align}
  & {{a}_{n}}=1+\left( n-1 \right)2 \\
 & \Rightarrow {{a}_{n}}=1+2n-2 \\
 & \Rightarrow {{a}_{n}}=2n-1 \\
\end{align}\]
$\therefore $ Our general term of denominator sequence is ${{a}_{n}}=2n-1$
So our general term for series as,
Let us denote general term of series as ${{T}_{n}}$
${{T}_{n}}=\dfrac{27}{2n-1}$
We have to find ${{9}^{th}}$ term so we will substitute $n=9$
$\begin{align}
  & {{T}_{9}}=\dfrac{27}{18-1} \\
 & \Rightarrow {{T}_{9}}=\dfrac{27}{17} \\
\end{align}$
We can write it as our given terms of series
${{T}_{9}}=1\dfrac{10}{17}$
$\therefore {{T}_{9}}=1\dfrac{10}{17}$
Hence answer is option 1.

Note: Series can have finite and infinite terms also. We can generalize series by observing their terms. If a series has only positive terms, we say it as a series of positive terms. If a series has alternating $+\And -$ sign between its terms we say it as an alternating series. Series can be convergent and divergent both.