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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Sequences and Series

NDA 2018 · Mathematics · Sequences and Series

If a, b, c are in AP or GP HP, then (a-b)/(b-c) is equal to

If a,b,ca, b, c are in AP or GP HP, then a−bb−c\frac{a-b}{b-c} is equal to

Options

  1. Aba\frac{\mathrm{b}}{\mathrm{a}} or 1 or bc\frac{\mathrm{b}}{\mathrm{c}}
  2. Bca\frac{\mathrm{c}}{\mathrm{a}} or cb\frac{\mathrm{c}}{\mathrm{b}} or 1
  3. C1 or ab\frac{\mathrm{a}}{\mathrm{b}} or ac\frac{\mathrm{a}}{\mathrm{c}}
  4. D1 or ab\frac{\mathrm{a}}{\mathrm{b}} or ca\frac{\mathrm{c}}{\mathrm{a}}

Correct answer

C. 1 or ab\frac{\mathrm{a}}{\mathrm{b}} or ac\frac{\mathrm{a}}{\mathrm{c}}

Step-by-step solution

From question, a,b,c∈ A.P. ⇒a−b=b−c⇒a−bb−c=1a,b,c∈ G.P. ⇒ab=bc=a−bb−ca,b,c∈ H.P. ⇒b=2aca+c⇒ab+bc=2ac⇒ab−ac=ac−bc⇒a−bb−c=ac\begin{aligned} & a, b, c \in \text{ A.P. } \Rightarrow a-b=b-c \Rightarrow \frac{a-b}{b-c}=1 \\ & a, b, c \in \text{ G.P. } \Rightarrow \frac{a}{b}=\frac{b}{c}=\frac{a-b}{b-c} \\ & a, b, c \in \text{ H.P. } \Rightarrow b=\frac{2 a c}{a+c} \Rightarrow a b+b c=2 a c \\ \Rightarrow & a b-a c=a c-b c \\ \Rightarrow & \frac{a-b}{b-c}=\frac{a}{c} \end{aligned}
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