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

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.