Question

If the sum of the three terms of a G.P is 19 and the product is 216, then the common ratio of the series is
A. $-\dfrac{3}{2}$
B. $\dfrac{3}{2}$
C. $2$
D. $3$

Answer 1

Hint: In this question, we are to find the common ratio of the terms in the given geometric series. By adding and multiplying the first three general terms of the geometric series and equating them with the given values respectively, we are able to find the required common ratio.

Formula used: The geometric sequence is written as
$a,ar,a{{r}^{2}},a{{r}^{3}},...,a{{r}^{n}}$
Where $a$ is the first term and $r$ is the common ratio.
The common ratio is calculated by
$r=\dfrac{{{a}_{n}}}{{{a}_{n-1}}}$

Complete step by step solution: Consider the first three terms as $a,ar,a{{r}^{2}}$
It is given that, the sum of these three terms is 19.
So, we can write,
$\begin{align}
& a+ar+a{{r}^{2}}=19 \\
& \Rightarrow a(1+r+{{r}^{2}})=19\text{ }...\text{(1)} \\
\end{align}$
And also given that, the product of these terms is 216.
So, we can write,
$\begin{align}
& a\times ar\times a{{r}^{2}}=216 \\
& \Rightarrow \text{ }{{a}^{3}}{{r}^{3}}=6\times 6\times 6 \\
& \Rightarrow \text{ (}ar{{)}^{3}}={{6}^{3}} \\
& \Rightarrow \text{ }ar=6 \\
\end{align}$
Then,
$a=\dfrac{6}{r}$
On substituting in equation (1), we get
$\begin{align}
& a(1+r+{{r}^{2}})=19 \\
& \dfrac{6}{r}(1+r+{{r}^{2}})=19 \\
& 6+6r+6{{r}^{2}}=19r \\
& \Rightarrow 6{{r}^{2}}-13r+6=0\text{ }...\text{(2)} \\
\end{align}$
On simplifying equation (2),
$\begin{align}
& 6{{r}^{2}}-9r-4r+6=0 \\
& \Rightarrow 3r(2r-3)-2(2r-3)=0 \\
& \Rightarrow (2r-3)(3r-2)=0 \\
& \therefore r=\dfrac{3}{2};r=\dfrac{2}{3} \\
\end{align}$

Thus, Option (B) is the correct value of the common ratio for the given series.

Note: Here the given series is a geometric series. So, using the general terms, the sum and the product are equated with the given values and by simplifying the obtained quadratic equation we get the required common ratio.