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If the line \mathrm{\sqrt{5} x=y} meets the lines \mathrm{x=1, x=2 \ldots x=n}, at points \mathrm{A_{1}, A_{2}, \ldots A_{n}} respectively then \mathrm{\left(\mathrm{OA}_{1}\right)^{2}+\left(\mathrm{OA}_{2}\right)^{2} \ldots .+\left(O A_{n}\right)^{2}} is equal to

Option: 1

\mathrm{3 n^{2}+3 n}


Option: 2

\mathrm{2 n^{3}+3 n^{2}+n}


Option: 3

\mathrm{3 n^{3}+3 n^{2}+2}


Option: 4

\mathrm{(3 / 2)\left(n^{4}+2 n^{3}+n^{2}\right)}


Answers (1)

best_answer

Equation of the line in parametric form is

\frac{x}{\frac{1}{\sqrt{6}}}=\frac{y}{\frac{\sqrt{5}}{\sqrt{6}}}=r.


Any point on the given line can be taken as
\mathrm{\left(\frac{r}{\sqrt{6}}, \frac{\sqrt{5}}{\sqrt{6}} r\right)}.
Now, if \mathrm{r=O A} then the point \mathrm{\left(\frac{r}{\sqrt{6}}, \frac{\sqrt{5}}{\sqrt{6}} r\right)}will also satisfy the equation \mathrm{x=1}.
\mathrm{\Rightarrow \mathrm{r}=\sqrt{6}=\mathrm{OA}_{1}}
Similarly \mathrm{\mathrm{OA}_{2}=2 \sqrt{6}}. Hence

\mathrm{\mathrm{OA}_{1}{ }^{2}+\mathrm{OA}_{2}{ }^{2}+\ldots \ldots \ldots+\mathrm{OA}_{3}{ }^{2}+\ldots \ldots \ldots \ldots+\mathrm{OA}_{n}{ }^{2}}
\mathrm{=\sqrt{6}+2^{2} \sqrt{6}+3^{2} \sqrt{6}+\ldots \ldots \ldots+n^{2} \sqrt{6}}
\mathrm{=6\left(\frac{n(n+1)(2 n+1)}{6}\right)}
\mathrm{=2 n^{3}+3 n^{2}+n}

Hence (B) is the correct answer.
 

Posted by

Irshad Anwar

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