Question

The nth term of the series $ 3,\sqrt 3 ,1,.... $ is $ \dfrac{1}{{243}} $ , then n is
A. 12
B. 13
C. 14
D. 15

Answer 1

Hint: The given series is clearly in Geometric progression. The first term of the given series is 3. So next we have to find the common ratio of the series. Using the first term, obtained a common ratio we have to find the value of n using the below mentioned formula when the nth term is $ \dfrac{1}{{243}} $ .
Formulas used:
The nth term of a G.P is $ a{r^{n - 1}} $ , where a is the first term and r is the common ratio of the series.

Complete step by step solution:
We are given that the nth term of the series $ 3,\sqrt 3 ,1,.... $ is $ \dfrac{1}{{243}} $ .
We have to find the value of n.
In a geometric progression, every term starting from the second term is obtained by multiplying a fixed number to its previous term. This fixed number is called common ratio and it is denoted by r.
Common ratio of a G.P can be calculated by the formula $ \dfrac{{{T_{n + 1}}}}{{{T_n}}} $ , where $ {T_{n + 1}} $ is the next term of $ {T_n} $
Here when n is 1, r is equal to $ \dfrac{{{T_2}}}{{{T_1}}} $ . Second term is $ \sqrt 3 $ and the first term is 3.
 $ r = \dfrac{{\sqrt 3 }}{3} = \dfrac{1}{{\sqrt 3 }} $
Common ratio of the series is $ \dfrac{1}{{\sqrt 3 }} $
We now have the first term, common ratio and the nth term. We have to find the value of n.
 $ \dfrac{1}{{243}} = a{r^{n - 1}};a = 3,r = \dfrac{1}{{\sqrt 3 }} $
 $ \Rightarrow \dfrac{1}{{243}} = 3 \times {\left( {\dfrac{1}{{\sqrt 3 }}} \right)^{n - 1}} $
 $ \Rightarrow \dfrac{1}{{{{\left( {\sqrt 3 } \right)}^{n - 1}}}} = \dfrac{1}{{243 \times 3}} $
 $ \Rightarrow \dfrac{1}{{{3^{\left( {\dfrac{1}{2}} \right) \times \left( {n - 1} \right)}}}} = \dfrac{1}{{{3^6}}} $
 $ \Rightarrow {3^{\left( {\dfrac{{n - 1}}{2}} \right)}} = {3^6} $
Either side bases are equal, so the powers must be equated.
 $ \Rightarrow \dfrac{{n - 1}}{2} = 6 $
 $ \Rightarrow n - 1 = 12 $
 $ \therefore n = 13 $
Therefore, $ \dfrac{1}{{243}} $ is the 13th term of the given series.
So, the correct answer is “Option B”.

Note: Do not confuse a G.P with an A.P. A.P is obtained by adding a fixed number to the previous terms whereas G.P is obtained by multiplying a fixed number to the previous terms. r can be calculated by dividing any two consecutive terms over each other like third by second or fourth by third or fifth by fourth terms.