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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{ ...

If a, b, c are in A.P or G.P or H.P then a−bb−c\dfrac{a-b}{b-c} is equal to
A. ba or 1 or bc\dfrac{b}{a}\text{ or 1 or }\dfrac{b}{c}
B. ca or cb or 1\dfrac{c}{a}\text{ or }\dfrac{c}{b}\text{ or 1}
C. 1 or ab or ac1\text{ or }\dfrac{a}{b}\text{ or }\dfrac{a}{c}
D. 1 or ab or ca1\text{ or }\dfrac{a}{b}\text{ or }\dfrac{c}{a}

Explanation

Solution

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 1a,1b,1c\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 a−bb−c\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=aa = a
Second term = b=a+db = a + d
Third term = c=a+2dc = a + 2d
Using the above values in a−bb−c\dfrac{a-b}{b-c}, we get
a−(a+d)(a+d)−(a+2d)=a−a−da+d−a−2d=−d−d=1\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 a−bb−c\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 = aa
Second term = b = arar
Third term = c =ar2a{{r}^{2}}.
Substituting in a−bb−c\dfrac{a-b}{b-c}, we get
a−bb−c=a−arar−ar2=a(1−r)a(r−r2)=1−rr(1−r)=1r\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, a−bb−c=1r\dfrac{a-b}{b-c}=\dfrac{1}{r}, there can be multiple answers. For example,
1r=aar=ab 1r=arar2=bc a−bb−c=1r=ab=bc \begin{aligned} & \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{aligned}
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
1a,1b,1c\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 1b\dfrac{1}{b}is the arithmetic mean of 1a and 1c\dfrac{1}{a}\text{ and }\dfrac{1}{c}. This means
1b=1a+1c2 2b=1a+1c 2b=a+cac  \begin{aligned} & \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{aligned}
By cross multiplying, we get
b=2aca+cb=\dfrac{2ac}{a+c}.
Substituting in the terma−bb−c\dfrac{a-b}{b-c}, we get

& \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{aligned}$$ 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.