Question

$a,b,c$ are the first three terms of a geometric series. If the harmonic mean of $a$ and $b$ is $12$ and that of $b$ and $c$ is $36$, then the first five terms of the series are
A. $9,18,27,36,45$
B. $8,24,72,216,648$
C. $4,22,38,46$
D. None of these

Answer 1

Hint: In this question, we are to find the first five terms of a geometric series. Where we have the harmonic means of the first three terms. By using them we can calculate the required terms of the series. The harmonic mean formula plays a major role here.

Formula Used: A geometric series is written as
$a+ar+a{{r}^{2}}+...$
Here the first term is represented by $a$ and the common ratio is represented by $r$.
Consider three numbers$a,b,c$.
So, the harmonic mean of $a$ and $b$ is $\dfrac{2ab}{a+b}$ and the harmonic mean of $b$ and $c$ is $\dfrac{2bc}{b+c}$.

Complete step by step solution: Given that, $a,b,c$ are the first three terms of a geometric series.
So, we can compare them with the general terms as
$\begin{align}
  & b=ar \\
 & c=a{{r}^{2}} \\
\end{align}$
It is given that the harmonic means of $a$ and $b$; $b$ and $c$ are $12$ and $36$ respectively.
So, we can write them as
$\dfrac{2ab}{a+b}=12\text{ }...(1)$
$\dfrac{2bc}{b+c}=36\text{ }...(2)$
Substituting $a,b,c$ values in the above equations, we get
$\begin{align}
  & \dfrac{2a(ar)}{a+(ar)}=12 \\
 & \Rightarrow \dfrac{a(ar)}{a(1+r)}=6 \\
 & \Rightarrow \dfrac{ar}{1+r}=6 \\
 & \Rightarrow ar=6(1+r)\text{ }...(3) \\
\end{align}$
Similarly,
$\begin{align}
  & \dfrac{2(ar)(a{{r}^{2}})}{(ar)+(a{{r}^{2}})}=36 \\
 & \Rightarrow \dfrac{(ar)(a{{r}^{2}})}{ar(1+r)}=18 \\
 & \Rightarrow \dfrac{a{{r}^{2}}}{(1+r)}=18 \\
 & \Rightarrow a{{r}^{2}}=18(1+r)\text{ }...(4) \\
\end{align}$
On dividing (4) by (3), we get
$\begin{align}
  & \dfrac{a{{r}^{2}}}{ar}=\dfrac{18(1+r)}{6(1+r)} \\
 & \Rightarrow r=3 \\
\end{align}$
Then, substituting the obtained common ratio in (3), we get
$\begin{align}
  & ar=6(1+r) \\
 & \Rightarrow a(3)=6(1+3) \\
 & \Rightarrow 3a=24 \\
 & \therefore a=8 \\
\end{align}$
Thus, with this first term $a=8$ and the common ratio $r=3$, we can calculate the first five terms of the series as
$\begin{align}
  & {{t}_{1}}=a=8 \\
 & {{t}_{2}}=ar=(8)(3)=24 \\
 & {{t}_{3}}=a{{r}^{2}}=(8)(9)=72 \\
 & {{t}_{5}}=a{{r}^{3}}=(8)(27)=216 \\
 & {{t}_{6}}=a{{r}^{4}}=(8)(81)=648 \\
\end{align}$
So, the geometric series is $8,24,72,216,648$

Option ‘B’ is correct

Note: Here, we do not get confused with the question that the series is in geometric progression but the means are the harmonic means. Here we need to remember that, the relation between the first three terms of the geometric series is given by the harmonic mean. But that does not represent that they are in harmonic progression. So, be sure of the type of progression. Then we can easily calculate such types of questions.