Question

If a, b, c are in A.P or G.P or H.P then $\dfrac{a-b}{b-c}$ is equal to
A. $\dfrac{b}{a}\text{ or 1 or }\dfrac{b}{c}$
B. $\dfrac{c}{a}\text{ or }\dfrac{c}{b}\text{ or 1}$
C. $1\text{ or }\dfrac{a}{b}\text{ or }\dfrac{a}{c}$
D. $1\text{ or }\dfrac{a}{b}\text{ or }\dfrac{c}{a}$

Answer 1

Hint: To solve this question, take the progressions case by case. We have to assume a common difference as ‘d’ or a common ratio ‘r’ or $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$ in A.P for the progressions A.P or G.P or H.P respectively. We have to substitute the values of a, b, c which we get from our assumptions in the expression $\dfrac{a-b}{b-c}$ to match with the options.

Complete step-by-step solution:
Consider that the progression a, b, c is an A.P. Let the common difference be d and the first term is ‘a’. Then we get
First term = $a = a$
Second term = $b = a + d$
Third term = $c = a + 2d$
Using the above values in $\dfrac{a-b}{b-c}$, we get
$\dfrac{a-\left( a+d \right)}{\left( a+d \right)-\left( a+2d \right)}=\dfrac{a-a-d}{a+d-a-2d}=\dfrac{-d}{-d}=1$
$\therefore $ If a, b, c are in A.P, then $\dfrac{a-b}{b-c}$ is equal to 1.
Consider that the progression a, b, c is a G.P. Let the common ratio be r and the first term is ‘a’. Then we get
First term = a = $a$
Second term = b = $ar$
Third term = c =$a{{r}^{2}}$.
Substituting in $\dfrac{a-b}{b-c}$, we get
$\dfrac{a-b}{b-c}=\dfrac{a-ar}{ar-a{{r}^{2}}}=\dfrac{a\left( 1-r \right)}{a\left( r-{{r}^{2}} \right)}=\dfrac{1-r}{r\left( 1-r \right)}=\dfrac{1}{r}$
Here, $\dfrac{a-b}{b-c}=\dfrac{1}{r}$, there can be multiple answers. For example,
$\begin{align}
  & \dfrac{1}{r}=\dfrac{a}{ar}=\dfrac{a}{b} \\
 & \dfrac{1}{r}=\dfrac{ar}{a{{r}^{2}}}=\dfrac{b}{c} \\
 & \dfrac{a-b}{b-c}=\dfrac{1}{r}=\dfrac{a}{b}=\dfrac{b}{c} \\
\end{align}$
So, we have to match with the options to suit the answer.
Consider that the progression a, b, c is a H.P. Then we know the property of H.P that the reciprocal terms of the terms in H.P will be an A.P. Using this property, we can write that
$\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$ are in A.P. If 3 terms are in A.P, the second term is the arithmetic mean of first and third terms. We can conclude that $\dfrac{1}{b}$is the arithmetic mean of $\dfrac{1}{a}\text{ and }\dfrac{1}{c}$. This means
$\begin{align}
  & \dfrac{1}{b}=\dfrac{\dfrac{1}{a}+\dfrac{1}{c}}{2} \\
 & \dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c} \\
 & \dfrac{2}{b}=\dfrac{a+c}{ac} \\
 & \\
\end{align}$
By cross multiplying, we get
$b=\dfrac{2ac}{a+c}$.
Substituting in the term$\dfrac{a-b}{b-c}$, we get
\[\begin{align}
& \dfrac{a-b}{b-c}=\dfrac{a-\dfrac{2ac}{a+c}}{\dfrac{2ac}{a+c}-c}=\dfrac{\dfrac{{{a}^{2}}+ac-2ac}{a+c}}{\dfrac{2ac-ac-{{c}^{2}}}{a+c}} \\
 & \dfrac{{{a}^{2}}-ac}{ac-{{c}^{2}}}=\dfrac{a\left( a-c \right)}{c\left( a-c \right)}=\dfrac{a}{c} \\
 & \therefore \dfrac{a-b}{b-c}=\dfrac{a}{c} \\
\end{align}\]
Comparing the three answers we got from the three cases with the options given, the answer is
If a, b, c are in A.P or G.P or H.P then $\dfrac{a-b}{b-c}$ is equal to $1\text{ or }\dfrac{a}{b}\text{ or }\dfrac{a}{c}$ respectively that is option C.

Note: This result can be remembered as a property of the progressions. For example,
In an A.P $\dfrac{a-b}{b-c}$= 1 which means that a – b = b – c.
In a G.P $\dfrac{a-b}{b-c}$= $\dfrac{1}{r}$ which means that (a – b) $\times $ r = (b – c).
There is a chance of mistake by students while selecting the options given for the question as there are two answers for G.P. The order of the terms$1\text{ or }\dfrac{a}{b}\text{ or }\dfrac{a}{c}$ in the options does not matter in the question but none of the other options match the answer.