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

NDA 2024 · Mathematics · Sequences and Series

Paragraph: The product of 5 consecutive terms of an AP is 229635....

Paragraph: The product of 5 consecutive terms of an AP is 229635. The first, second and fifth terms are in GP.
Question: What is the common difference?

Options

  1. A33
  2. B44
  3. C55
  4. D66

Correct answer

D. 66

Step-by-step solution

Let the required terms of the given AP are- a−2d,a−d,a,a+d,a+2d.\mathrm{a}-2 \mathrm{ d}, \mathrm{a}-\mathrm{d}, \mathrm{a}, \mathrm{a}+\mathrm{d}, \mathrm{a}+2 \mathrm{ d} . According to the question, (a−2d)(a−d)a(a+d)(a+2d)=229635⇒a(a2−d2)(a2−4d2)=229635....(i)∵a−2d,a−d,a+2d are in G.P. ∴(a−d)2=(a−2d)(a+2d)⇒(a−d)2=a2−4d2⇒5d2−2ad=0⇒d(5d−2a)=0⇒5d=2a\begin{aligned} & (a-2 d)(a-d) a(a+d)(a+2 d)=229635 \\ \Rightarrow & a\left(a^2-d^2\right)\left(a^2-4 d^2\right)=229635 ....(i)\\ \because & a-2 d, a-d, a+2 d \text{ are in G.P. } \\ \therefore & (a-d)^2=(a-2 d)(a+2 d) \\ \Rightarrow & (a-d)^2=a^2-4 d^2 \\ \Rightarrow & 5 d^2-2 a d=0 \Rightarrow d(5 d-2 a)=0 \Rightarrow 5 d=2 a \end{aligned}....(ii) Solving equation (i) and (ii), we have, d=6.\mathrm{d}=6 .
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