Question

# The ${5^{th}}$ and ${8^{th}}$ terms of a GP are 1458 and 54, respectively. The common ratio of the GP is${\text{A}}{\text{. }}\dfrac{1}{3} \\ {\text{B}}{\text{. 3}} \\ {\text{C}}{\text{. 9}} \\ {\text{D}}{\text{. }}\dfrac{1}{9} \\ {\text{E}}{\text{. }}\dfrac{1}{8} \\$

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Hint: In order to solve this question, First write what is given to us. It will give us a clear picture of what our approach should be to solve this question. Also then use the formula for ${n^{th}}$ term in GP. Thus we will get our required answer.

Now we have given that,
The ${5^{th}}$ and ${8^{th}}$ terms of a GP are 1458 and 54.
And we have to find the common ratio of GP.
Let us assume $a$ and $r$ be the first term and the common ratio of the GP respectively
Here given,the fifth term of the GP, ${T_5}=1458$
And the eight term of the GP, ${T_8} = 54$
Now, we know that ${n^{th}}$ term in the GP series, ${T_n} = a{r^{n - 1}}$
Where $a =$ first term of the GP series,
$r =$ common ratio of the GP series and
Or ${T_5} = a{r^4} = 1458$ and
${T_8} = a{r^7} = 54$
Now,
$\dfrac{{{T_8}}}{{{T_5}}} = \dfrac{{54}}{{1458}}$
Or $\dfrac{{a{r^7}}}{{a{r^4}}} = \dfrac{{54}}{{1458}}$
Or ${r^3} = \dfrac{1}{{27}}$
Or $r = \dfrac{1}{{\sqrt[3]{{27}}}}$
Or $r = \dfrac{1}{3}$
Thus the value of the common ratio of GP is $\dfrac{1}{3}$.

Note: Whenever we face such types of questions the key concept is that we should write what is given to us then find the easiest approach to find the solution of the question. Like in this question we simply write the formula of ${n^{th}}$ term in a GP, then divide them to get a ratio from where we get the value of $r$ . Thus we get our required answer.