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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Sequences and Series

NDA 2016 · Mathematics · Sequences and Series

Paragraph: The sixth term of an AP is 2 and its common difference...

Paragraph: The sixth term of an AP is 2 and its common difference is greater than 1.
Question: What is the common difference of the AP so that the product of the first, fourth and fifth terms is greatest?

Options

  1. A8/58/5
  2. B9/59/5
  3. C22
  4. D11/511 / 5

Correct answer

A. 8/58/5

Step-by-step solution

Let first term =a\mathrm{a}& common difference =\mathrm{x}∴a+5x=2⇒a=2−5xLet P=T1×T4×T5⇒P=a(a+3x)(a+4x)⇒P=(2−5x)(2−5x+3x)(2−5x+4x)⇒P=−10x3+34x2−32x+8dpdx=0⇒15x2−34x+16=0⇒(5x−8)(3x−2)=0⇒x=85,[∵x=23<1]\begin{aligned} \therefore & \mathrm{a}+5 \mathrm{x}=2 \Rightarrow \mathrm{a}=2-5 \mathrm{x} \\ & \text{Let } \mathrm{P}=\mathrm{T}_1 \times \mathrm{T}_4 \times \mathrm{T}_5 \\ \Rightarrow & \mathrm{P}=\mathrm{a}(\mathrm{a}+3 \mathrm{x})(\mathrm{a}+4 \mathrm{x}) \\ \Rightarrow & \mathrm{P}=(2-5 \mathrm{x})(2-5 \mathrm{x}+3 \mathrm{x})(2-5 \mathrm{x}+4 \mathrm{x}) \\ \Rightarrow & \mathrm{P}=-10 \mathrm{x}^3+34 \mathrm{x}^2-32 \mathrm{x}+8 \\ & \frac{d p}{d x}=0 \Rightarrow 15 \mathrm{x}^2-34 \mathrm{x}+16=0 \\ & \Rightarrow(5 \mathrm{x}-8)(3 \mathrm{x}-2)=0 \\ \Rightarrow & \mathrm{x}=\frac{8}{5},\left[\because x=\frac{2}{3} < 1\right]\end{aligned}$
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