Question

If $x$ is added to each of numbers \[3,9,21\]. So, that the resulting numbers may be in G.P, then the value of $x$ will be
A. $3$
B. $\dfrac{1}{2}$
C. $2$
D. $\dfrac{1}{3}$

Answer 1

Hint: In this question, we have to find the value of the given variable which is when added to the given numbers, the numbers form a geometric series. So, using the common ratio formula the required value is calculated.

Formula Used: The geometric sequence is written as
$a,ar,a{{r}^{2}},a{{r}^{3}},...,a{{r}^{n}}$
$a$ 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: The given numbers are \[3,9,21\].
When $x$ is added to the given numbers, we get
$3+x,9+x,21+x$
Since they are in G.P, the common ratios are
\[\begin{align}
  & \dfrac{9+x}{3+x}=\dfrac{21+x}{9+x} \\
 & {{(9+x)}^{2}}=(3+x)(21+x) \\
 & 81+{{x}^{2}}+18x=63+3x+21x+{{x}^{2}} \\
 & 21x-18x=81-63 \\
\end{align}\]
\[\begin{align}
  & \Rightarrow 6x=18 \\
 & \therefore x=3 \\
\end{align}\]
Then, by adding the value of $x$ to the numbers, we get
\[\begin{align}
  & 3+3=6 \\
 & 9+3=12 \\
 & 21+3=24 \\
\end{align}\]
The numbers $6,12,24$ are G.P.

Option ‘A’ is correct

Note: Here the variable $x$ is added to the given numbers. Then, we may think that the resulting series will be an arithmetic series. But the resulting numbers form a geometric series. By using the common ratio, we will get the required value.