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

NDA 2016 · Mathematics · Sequences and Series

Paragraph: Let a, x, y, z, b be in AP , where x+y+z=15 . Let a, p,...

Paragraph: Let a,x,y,z,b\mathrm{a}, \mathrm{x}, \mathrm{y}, \mathrm{z}, \mathrm{b} be in AP , where x+y+z=15\mathrm{x}+\mathrm{y}+\mathrm{z}=15. Let a,p,q,r,b\mathrm{a}, \mathrm{p}, \mathrm{q}, \mathrm{r}, \mathrm{b} be in HP , where \mathrmp−1+\mathrmq−1+\mathrmr−1=53\mathrmp^{-1}+\mathrmq^{-1}+\mathrmr^{-1}=\frac{5}{3}
Question: What is the value of aba b ?

Options

  1. A1010
  2. B99
  3. C88
  4. D66

Correct answer

B. 99

Step-by-step solution

Sn=n2(a+l)⇒a+x+y+z+b=52(a+b)a+b+15=52(a+b)⇒a+b=10\mathrm{S}_{\mathrm{n}}=\frac{n}{2}(a+l)\Rightarrow \quad \mathrm{a}+\mathrm{x}+\mathrm{y}+\mathrm{z}+\mathrm{b}=\frac{5}{2}(a+b)\mathrm{a}+\mathrm{b}+15=\frac{5}{2}(a+b)\Rightarrow \quad a+b=10 ...(1) 1a+1p+1q+1r+1b=52(1a+1b)⇒1a+1b+53=52(1a+1b)\begin{aligned} & & \frac{1}{a}+\frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{b}=\frac{5}{2}\left(\frac{1}{a}+\frac{1}{b}\right) \\ \Rightarrow & \frac{1}{a}+\frac{1}{b}+\frac{5}{3}=\frac{5}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\end{aligned}\Rightarrow \quad \frac{3(a+b)}{a b}=\frac{10}{3}....(2). ...(2)⇒3×10ab=103.⇒ab=9\begin{aligned} & \Rightarrow \quad \frac{3 \times 10}{a b}=\frac{10}{3} . \\ & \Rightarrow \quad \mathrm{ab}=9\end{aligned}$
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