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

NDA 2018 · Mathematics · Sequences and Series

If u , vand w (all positive) are the p^ th , q^ th and r^ th terms...

If uu, vand ww (all positive) are the pth ,qth andrth p^{\text{th }}, q^{\text{th }} and r^{\text{th }} terms of a GP, the determinant of the matrix \left|In⁡up1In⁡vq1In⁡wr1\begin{array}{lll}\operatorname{In} u & p & 1 \\ \operatorname{In} v & q & 1 \\ \operatorname{In} w & r & 1\end{array}\right|$ is

Options

  1. A0
  2. B1
  3. C(p−q)(q−r)(r−p)(\mathrm{p}-\mathrm{q})(\mathrm{q}-\mathrm{r})(\mathrm{r}-\mathrm{p})
  4. Dln⁡u×ln⁡v×ln⁡w\ln \mathrm{u} \times \ln \mathrm{v} \times \ln \mathrm{w}

Correct answer

A. 0

Step-by-step solution

Let u,vandwu, v and w are \mathrmpth ,\mathrmqth \mathrmp^{\text{th }}, \mathrmq^{\text{th }} and \mathrmrth \mathrmr^{\text{th }} term of the G.P. with first term ' aa ' and common ratio ' dd '. then, u=a.(d)p−1⇒In⁡u=In⁡(a)+(p−1)In⁡(d)v=a.(d)q−1⇒In⁡v=In⁡(a)+(q−1)In⁡(d)w=a.(d)r−1⇒In⁡w=In⁡(a)+(r−1)In⁡(d)\begin{aligned} & u=a .(d)^{p-1} \Rightarrow \operatorname{In} u=\operatorname{In}(a)+(p-1) \operatorname{In}(d) \\ & v=a .(d)^{q-1} \Rightarrow \operatorname{In} v=\operatorname{In}(a)+(q-1) \operatorname{In}(d) \\ & w=a .(d)^{r-1} \Rightarrow \operatorname{In} w=\operatorname{In}(a)+(r-1) \operatorname{In}(d) \end{aligned} Now, \operatorname{In} u-\operatorname{In} v=(p-q) \operatorname{In}(d)\operatorname{In} u-\operatorname{In} w=(p-r) \operatorname{In}(d)\left|In⁡up1In⁡vq1In⁡wr1\begin{array}{lll}\operatorname{In} u & p & 1 \\ \operatorname{In} v & q & 1 \\ \operatorname{In} w & r & 1\end{array}\right|\left|In⁡up1In⁡u−In⁡v(p−q)0 In  u−In⁡w(p−r)0\begin{array}{ccc}\operatorname{In} u & p & 1 \\ \operatorname{In} u-\operatorname{In} v & (p-q) & 0 \\ \text{\text{ In } } u-\operatorname{In} w & (p-r) & 0\end{array}\right|\left\{\right.\text{ Applying } \mathrm{R}_2 \rightarrow \mathrm{R}_1-\mathrm{R}_2\text{ and } \mathrm{R}_3 \rightarrow \mathrm{R}_1-\mathrm{R}_3 \}  or, \text{ or, }∣In⁡up1(p−q)In⁡(d)(p−q)0(p−r)In⁡(d)(p−r)0∣=(p−q)(p−r)[In⁡(d)−In⁡(d)]=0\begin{aligned} \left|\begin{array}{ccc} \operatorname{In} u & p & 1 \\ (p-q) \operatorname{In}(d) & (p-q) & 0 \\ (p-r) \operatorname{In}(d) & (p-r) & 0 \end{array}\right| \\ \quad=(p-q)(p-r)[\operatorname{In}(d)-\operatorname{In}(d)]=0 \end{aligned}$
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