Question

Which term of the G.P. 18, $ - 12$, 8, … is $\dfrac{{512}}{{729}}$?
A) \[{7^{{\text{th}}}}\]
B) \[{9^{{\text{th}}}}\]
C) \[{11^{{\text{th}}}}\]
D) \[{13^{{\text{th}}}}\]

Answer 1

Hint: Firstly, know about the G.P that is geometric progression which means that the sequence of numbers with a common ratio between any two consecutive numbers. Use the formula of geometric progression sequence for the \[{n^{{\text{th}}}}\] terms that is \[{t_n} = a{r^{n - 1}}\] where, a is the first or initial term of the G.P. series or sequence, \[{t_n}\] is the $n^{th}$ term of G.P. series, and r is the common ratio of successive numbers. Calculate the value of n.

Complete step-by-step answer:
The given G.P. series is 18, $ - 12$, 8, … and the \[{n^{{\text{th}}}}\] term is $\dfrac{{512}}{{729}}$.
Now, we know about the formula of geometric progression sequence for the \[{n^{{\text{th}}}}\] terms that is \[{t_n} = a{r^{n - 1}}\].
Now, we calculate the value of r by using the formula \[r = \dfrac{{{a_2}}}{{{a_1}}}\] where \[{a_1} = 18\] and \[{a_2} = - 12\] from the given G.P. series.
Substitute the values \[{a_1} = 18\] and \[{a_2} = - 12\] in the expression \[r = \dfrac{{{a_2}}}{{{a_1}}}\].
\[ \Rightarrow r = \dfrac{{ - 12}}{{18}}\]
Now, in both denominators and numerators, we divide the above equation by 6.
\[ \Rightarrow r = \dfrac{{ - 2}}{3}\]
Now, calculate the value of $n$. Substitute the value of $a = 18,r = - \dfrac{2}{3},$ and ${t_n} = \dfrac{{512}}{{729}}$ in the expression \[{t_n} = a{r^{n - 1}}\].
\[ \Rightarrow \dfrac{{512}}{{729}} = 18{\left( { - \dfrac{2}{3}} \right)^{n - 1}}\]
Now, we simplify the above equation and get the value of n:
\[ \Rightarrow \dfrac{{512}}{{729}} = 18{\left( { - \dfrac{2}{3}} \right)^n}{\left( { - \dfrac{2}{3}} \right)^{ - 1}}\]
\[ \Rightarrow \dfrac{{512}}{{729}} = 18{\left( { - \dfrac{2}{3}} \right)^n}\left( { - \dfrac{3}{2}} \right)\]
\[ \Rightarrow \dfrac{{512}}{{729}} = - 27{\left( { - \dfrac{2}{3}} \right)^n}\]
\[ \Rightarrow - \dfrac{{512}}{{729 \times 27}} = {\left( { - \dfrac{2}{3}} \right)^n}\]
On the further simplification, the following is obtained:
\[ \Rightarrow - \dfrac{{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}}{{\left( {3 \times 3 \times 3 \times 3 \times 3 \times 3} \right) \times \left( {3 \times 3 \times 3} \right)}} = {\left( { - \dfrac{2}{3}} \right)^n}\]
\[ \Rightarrow - \dfrac{{{2^9}}}{{{3^9}}} = {\left( { - \dfrac{2}{3}} \right)^n}\]
\[ \Rightarrow {\left( { - \dfrac{2}{3}} \right)^9} = {\left( { - \dfrac{2}{3}} \right)^n}\]
Now, we compare the values and get the value of n:
\[ \Rightarrow n = 9\]

Hence, the value of n is 9.

Note: If any successive term is generated by multiplying each preceding term with a constant value in a sequence of terms, then the sequence is called a geometric progression. (GP), while the common ratio is called the constant value. For instance, 3, 9, 27, 81, 243, ... is a GP, where 3 is the common ratio.