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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Sequences and Series

NDA 2024 · Mathematics · Sequences and Series

If a, b, c are in AP ; b, c, d are in GP ; c, d , e are in HP, then...

If a,b,ca, b, c are in AP;b,c,d\mathrm{AP} ; b, c, d are in GP;c,d\mathrm{GP} ; c, d, ee are in HP, then which of the following is/are correct? 1. a,ca, c and ee are in GP 2. 1a,1c,1e\frac{1}{a}, \frac{1}{c}, \frac{1}{e} are in GP Select the correct answer using the code given below:

Options

  1. A1 only
  2. B2 only
  3. CBoth 1 and 2
  4. DNeither 1 nor 2

Correct answer

A. 1 only

Step-by-step solution

∵a,b,c\because a, b, c are in AP ⇒2b=a+c\Rightarrow \quad 2 b=a+c\qquad ....(i) b,c,db, c, d are in GP ⇒c2=bd\Rightarrow \quad c^2=b d\qquad ....(ii) c,d,ec, d, e are in HP ⇒2d=1c+1e\Rightarrow \quad \frac{2}{d}=\frac{1}{c}+\frac{1}{e}\qquad ....(iii) From (i) and (ii) 2c2=(a+c)d⇒2d=a+cc2⇒1c+1e=ac2+1c⇒1e=ac2⇒c2=ae\begin{aligned} & 2 c^2=(a+c) d \\ & \Rightarrow \quad \frac{2}{d}=\frac{a+c}{c^2} \\ & \Rightarrow \quad \frac{1}{c}+\frac{1}{e}=\frac{a}{c^2}+\frac{1}{c} \\ & \Rightarrow \quad \frac{1}{e}=\frac{a}{c^2} \Rightarrow c^2=a e \end{aligned} Hence, a,c,ea, c, e are in GP.
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