Question

If a, b, c are in AP, b, c, d are in GP and c, d, e are in HP, then a, c, e are in?
(a) No particular order
(b) AP
(c) GP
(d) HP

Answer 1

Hint:For a relation between a, b, c using the formula of arithmetic mean given as $2b=a+c$. For the terms b, c, d in GP use the formula of geometric mean given as ${{d}^{2}}=ce$. For the terms c, d, e in HP use the formula of harmonic mean given as $\dfrac{2}{d}=\dfrac{1}{c}+\dfrac{1}{e}$ to form the relation. Eliminate the terms b and d from the three relations and check which mean formula does the terms a, c, e satisfy to get the answer.

Complete step-by-step solution:
Here we have been given that a, b, c are in AP, b, c, d are in GP and c, d, e are in HP. We are asked to find the sequence relation between the terms a, c, e.
Now, we know that the terms arithmetic mean of the three terms a, b, c in AP is given as twice the middle is equal to the sum of first and third term, so we have,
$\Rightarrow 2b=a+c$ …… (1)
The geometric mean of the three terms b, c, d in GP is given as the middle term is equal to the square root of the product of first and third term, so we have,
$\Rightarrow c=\sqrt{bd}$
On squaring both the sides we get,
$\Rightarrow {{c}^{2}}=bd$ …… (2)
We know that the if three terms c, d, e are in HP then their reciprocals $\dfrac{1}{c}$, $\dfrac{1}{d}$, $\dfrac{1}{e}$ respectively will be in AP, so using the formula of arithmetic mean that is known as the harmonic mean in this case we get,
$\Rightarrow \dfrac{2}{d}=\dfrac{1}{c}+\dfrac{1}{e}$ …… (3)
In relation (3) eliminating the term d using the relation (2) we can write,
$\begin{align}
  & \Rightarrow \dfrac{2}{\left( \dfrac{{{c}^{2}}}{b} \right)}=\dfrac{1}{c}+\dfrac{1}{e} \\
 & \Rightarrow \dfrac{2b}{{{c}^{2}}}=\dfrac{1}{c}+\dfrac{1}{e} \\
\end{align}$
Now, in the above relation eliminating the term b using relation (1) we get,
$\begin{align}
  & \Rightarrow \dfrac{a+c}{{{c}^{2}}}=\dfrac{1}{c}+\dfrac{1}{e} \\
 & \Rightarrow \dfrac{a}{{{c}^{2}}}+\dfrac{1}{c}=\dfrac{1}{c}+\dfrac{1}{e} \\
 & \Rightarrow \dfrac{a}{{{c}^{2}}}=\dfrac{1}{e} \\
\end{align}$
By cross multiplication we get,
$\Rightarrow ae={{c}^{2}}$
Clearly we can see that the above relation denotes the geometric mean of the terms a, c, e in the given order. Therefore a, c, e are in GP.
Hence, option (c) is the correct answer.

Note:You must remember the formulas of the three means of different sequences to solve the above question. Do not try to remove any of the terms a, c, e from the given relations because we have to find the relation between these terms and not b or d. You can check the answer by simply assigning some small integral values to the terms a, b, c, d and e.