Question

The 5th term of the series \[\dfrac{{10}}{9},\dfrac{1}{3}\sqrt {\dfrac{{20}}{3}} ,\dfrac{2}{3},...\] is
1. \[1\]
2. \[\dfrac{1}{3}\]
3. \[\dfrac{2}{5}\]
4. \[\dfrac{{\sqrt 2 }}{3}\]

Answer 1

Hint: First of all, do not interpret this question for some other progression, it is a question of geometric progression involving its basic properties. Now as it is clear that the question is of geometric progression, we should be aware of its formulas like,
\[{a_n} = a{r^{n - 1}}\]where (\[a = \]first term and \[r = \]the common ratio);
 are sufficient to solve the question. Also, common ratio can be found out by the division of consecutive terms, i.e., \[r = \dfrac{{{a_n}}}{{{a_{n - 1}}}}\].

Complete step-by-step answer:
We have, \[\dfrac{{10}}{9},\dfrac{1}{3}\sqrt {\dfrac{{20}}{3}} ,\dfrac{2}{3},...\]
The first term of the progression is given by a or a1
\[ \Rightarrow {a_1} = \dfrac{{10}}{9}\]
Here, the value of r can be calculated by dividing the second term with the first term,
Or \[r = \dfrac{{{a_2}}}{{{a_1}}}\]
Here, \[{a_2} = \dfrac{1}{3}\sqrt {\dfrac{{20}}{3}} \]and \[{a_1} = \dfrac{{10}}{9}\]
\[ \Rightarrow r = \dfrac{1}{3}\sqrt {\dfrac{{20}}{3}} \times \dfrac{9}{{10}}\]
\[ \Rightarrow r = \sqrt {\dfrac{3}{5}} \]
Now, the nth term of the geometric progression is given by
\[ \Rightarrow {a_n} = a{r^{n - 1}}\]
For the 5th term n=5,
\[ \Rightarrow {a_5} = a{r^{5 - 1}}\]
\[ \Rightarrow {a_5} = a{r^4}\]
By putting \[a = \dfrac{{10}}{9}\]and \[r = \sqrt {\dfrac{3}{5}} \]we get,
\[ \Rightarrow {a_5} = \dfrac{{10}}{9}{(\sqrt {\dfrac{3}{5}} )^4}\]
Carefully solving the above expression, keeping in mind the powers of the irrationals
\[ \Rightarrow {a_5} = \dfrac{{10}}{9} \times \dfrac{3}{5} \times \dfrac{3}{5}\]
Now, equating the value of the 5th term to the correct answer
\[ \Rightarrow {a_5} = \dfrac{2}{5}\]
Thus, option(3) is the correct answer.
So, the correct answer is “Option 3”.

Note: This question can be overwhelming at first, but things seem doable after applying some little geometric progression concepts. One should be comfortable with geometric progression series and how to manipulate them. Then, take care of the calculation mistakes and be sure of the final answer. It is easy to misinterpret this question and consider it to be of different progression concepts, but once you take a good look at it, an interesting geometric progression seems to pop out.