JEE 2011 :: General Questions :: Mathematics

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 Let $a_{1},a_{2},a_{3}.......,a_{100}$ be an arithmetic progression with  $a_{1}=3$  and 

$S_{p}=\sum_{i=1}^pa_{i}, 1\leq p\leq100$ for any integer n with $1\leq n\leq 20$ , let m=5n, If $\frac{S_{m}}{S_{n}}$ does not depend  on n, then $a_{2}$ is 

Answer:

Explanation:

Given $a_{1}=3, m=5n$

 and $a_{1},a_{2},....$ are in AP

 $\therefore$   $\frac{S_{m}}{S_{n}}= \frac{ S_{5n}}{S_{n}}$  is independent of n

 Now, $\frac{ \frac{5n}{2}[2 \times 3+(5n-1)d]}{\frac{n}{2}[2 \times 3+(n-1)d]}$

    $\Rightarrow$   $\frac{f{(6-d)+5n}}{(6-d)+n}$

 independent of n , if 

   $6-d=0 \Rightarrow d=6$

 $\therefore$      $a_{2}=a_{1}+d=3+6=9$

 if d=0

 $\frac{S_{m}}{S_{n}}$ is independent of n

 $\therefore$  $a_{2}=3$

The positive integer value of n > 3 satisfying the equation

$\frac{1}{\sin\left(\frac{\pi}{n}\right)}=\frac{1}{\sin\left(\frac{2\pi}{n}\right)}+\frac{1}{\sin\left(\frac{3\pi}{n}\right)}$ is 

Answer:

Explanation:

 Given ,n>3 $\epsilon$ integer

 and $\frac{1}{\sin\left(\frac{\pi}{n}\right)}=\frac{1}{\sin\left(\frac{2\pi}{n}\right)}+\frac{1}{\sin\left(\frac{3\pi}{n}\right)}$

 $\Rightarrow$  $\frac{1}{\sin\frac{\pi}{n}}-\frac{1}{\sin\frac{3\pi}{n}}=\frac{1}{\sin\frac{2\pi}{n}}$

$\Rightarrow$   $\frac{\sin\frac{3\pi}{n}-\sin\frac{\pi}{n}}{\sin\frac{\pi}{n}.\sin\frac{3\pi}{n}}=\frac{1}{\sin\frac{2\pi}{n}}$

$\Rightarrow$  $2 \cos \left(\frac{2 \pi}{n}\right).\sin \frac{\pi}{n}=\frac{\sin\frac{\pi}{n}.\sin\frac{3\pi}{n}}{\sin\frac{2\pi}{n}}$

$\Rightarrow$  $2 \sin \frac{2 \pi}{n}.\cos \frac{2\pi}{n}= \sin \frac{3 \pi}{n}$

$\Rightarrow$  $\sin \frac{4 \pi}{3}=\sin  \frac{ 3 \pi}{n}$

$\Rightarrow$   $\frac{ 4 \pi}{n}= \pi-\frac{3 \pi}{n} \Rightarrow  \frac{7 \pi}{n} = \pi   \Rightarrow  n=7$

Let a, b and c be three real numbers satisfying [a b c] $\begin{bmatrix}1 & 9 &7 \\8 & 2&7 \\7&3&7 \end{bmatrix}=\begin{bmatrix}0 & 0 &0 \end{bmatrix}$

Let b=6, with  a and c satisfying Eq. (E). If $\alpha$ and $\beta$ are the roots of the quadratic  equation  $ax^{2}+bx+c=0$ 

 then  $\sum_{n=0}^{\infty}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)^{n}$ is 

Answer:

Explanation:

 Given  ,$[a  b c]_{1 \times 3}$  $\begin{bmatrix}1 & 9 &7 \\8 & 2&7 \\7&3&7 \end{bmatrix}=\begin{bmatrix}0 & 0 &0 \end{bmatrix}$

 $\Rightarrow$ $\begin{bmatrix}a+8b+7c  \\9a+2b+3c \\7a+7b+7c\end{bmatrix}=\begin{bmatrix}0 & 0 &0 \end{bmatrix}$

$\Rightarrow$  a+8b+7c=0....(i)

$\Rightarrow$  9a+2b+3c=0 .....(ii)

$\Rightarrow$  a+b+c=0 .....(iii)

 On multiplying Eq.(iii) bt=y 2 , then substract from Eq.(ii) we get

 7a+c=0...(iv)

 Again multiplying Eq,(iii) by 3 , then substract from Eq.(ii) , we get

 6a-b=0....(v)

 $\therefore$  b=6a and c=-7a

 if b=6 , a=1 and c=-7

 $\therefore$ $ax^{2}+bx+c=0$

 $\Rightarrow$    $x^{2}+6x-7=0$

$\Rightarrow$  $(x+7)(x-1)=0$

 $\therefore$ x=1,-7

 $\Rightarrow$  $\sum_{n=0}^{\infty}\left(\frac{1}{1}-\frac{1}{7}\right)^{n}\Rightarrow \sum_{n=0}^{\infty}\left(\frac{6}{7}\right)^{n}$

$\Rightarrow$     $1+\frac{6}{7}+(\frac{6}{7})^{n}+....\infty$

 = $\frac{1}{1-\frac{6}{7}}= \frac{1}{1/7}=7$

Let a, b and c be three real numbers satisfying [a b c]$ \begin{bmatrix}1 & 9 &7 \\8 & 2&7 \\7&3&7 \end{bmatrix}=\begin{bmatrix}0 & 0 &0 \end{bmatrix}$

 Let $\omega$  be a solution of   $x^{3}-1=0$  with  Im $(\omega$) >0.If a=2  with b and c satisfying Eq (E). then the value of $\frac{3}{\omega^{\alpha}}+\frac{1}{\omega^{b}}+\frac{3}{\omega^{c}}$

Answer:

Explanation:

 Given  ,$[a  b c]_{1  \times 3}$  $\begin{bmatrix}1 & 9 &7 \\8 & 2&7 \\7&3&7 \end{bmatrix}=\begin{bmatrix}0 & 0 &0 \end{bmatrix}$

 $\Rightarrow$ $\begin{bmatrix}a+8b+7c  \\9a+2b+3c \\7a+7b+7c\end{bmatrix}=\begin{bmatrix}0 & 0 &0 \end{bmatrix}$

$\Rightarrow$  a+8b+7c=0....(i)

$\Rightarrow$  9a+2b+3c=0 .....(ii)

$\Rightarrow$  a+b+c=0 .....(iii)

 On multiplying Eq.(iii) bt=y 2 , then substract from Eq.(ii) we get

 7a+c=0...(iv)

 Again multiplying Eq,(iii) by 3 , then substract from Eq.(ii) , we get

 6a-b=0....(v)

 $\therefore$  b=6a and c=-7a

 if a=2,b=12 and c=-14

 $\therefore$   $\frac{3}{ \omega^{a}}+\frac{1}{\omega^{b}}+\frac{3}{\omega^{c}}$

 $\Rightarrow$   $\frac{3}{\omega^{2}}+\frac{1}{\omega^{12}}+\frac{3}{\omega^{-14}}$

            $= \frac{3}{ \omega^{2}}+1+3 \omega^{2}$

 $=3 \omega+1+ 3 \omega^{2}$

=$1+3(\omega+\omega^{2})$

=1-3=-2

Let a, b and c be three real numbers satisfying [a b c]  $ \begin{bmatrix}1 & 9 &7 \\8 & 2&7 \\7&3&7 \end{bmatrix}=\begin{bmatrix}0 & 0 &0 \end{bmatrix}$

If the point P(a, b, c), with reference to Eq. (E), lies on the plane 2x + y +z = 1, then the value of 7 a + b + c is

Answer:

Explanation:

 Given  ,$[a  b c]_{1 \times 3}$  $\begin{bmatrix}1 & 9 &7 \\8 & 2&7 \\7&3&7 \end{bmatrix}=\begin{bmatrix}0 & 0 &0 \end{bmatrix}$

 $\Rightarrow$ $\begin{bmatrix}a+8b+7c  \\9a+2b+3c \\7a+7b+7c\end{bmatrix}=\begin{bmatrix}0 & 0 &0 \end{bmatrix}$

$\Rightarrow$  a+8b+7c=0....(i)

$\Rightarrow$  9a+2b+3c=0 .....(ii)

$\Rightarrow$  a+b+c=0 .....(iii)

 On multiplying Eq.(iii) bt=y 2 , then substract from Eq.(ii) we get

 7a+c=0...(iv)

 Again multiplying Eq,(iii) by 3 , then substract from Eq.(ii) , we get

 6a-b=0....(v)

 $\therefore$  b=6a and c=-7a

 As  (a,b,c) lies on 2x+y+z=1

 $\Rightarrow$   2a+b+c=1

$\Rightarrow$  2a+6a-7a=1

$\Rightarrow$  a=1,b=6 and c=-7

 $\therefore$   7a+b+c=7+6-7=6

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