Let the six numbers $a_1, a_2, a_3, a_4, a_5, a_6$, be in AP and $a_1+a_3=10$ If the mean of these six numbers is $\frac{19}{2}$ and their variance is $\sigma^2$, then $8 \sigma^2$ is equal to :
- 105
- 210
- 200
- 220
The Correct Option is B
Approach Solution - 1
Numbers :
Variance mean of squares-square of mean
Approach Solution -2
Step 1: Using the given condition \( a_1 + a_3 = 10 \)
The general terms of an arithmetic progression are:
\( a_1, a_1 + d, a_1 + 2d, a_1 + 3d, a_1 + 4d, a_1 + 5d \)
From the condition \( a_1 + a_3 = 10 \), we get:
\[ a_1 + (a_1 + 2d) = 10 \quad \Rightarrow \quad 2a_1 + 2d = 10 \quad \Rightarrow \quad a_1 + d = 5 \quad \cdots (1) \]
Step 2: Using the mean of the numbers
The mean of the six numbers is:
\[ \text{Mean} = \frac{a_1 + (a_1 + d) + (a_1 + 2d) + (a_1 + 3d) + (a_1 + 4d) + (a_1 + 5d)}{6} \] Simplifying this expression: \[ \text{Mean} = \frac{6a_1 + 15d}{6} = a_1 + \frac{5d}{2} \] Given that the mean is \( \frac{19}{2} \), we have: \[ a_1 + \frac{5d}{2} = \frac{19}{2} \quad \Rightarrow \quad 2a_1 + 5d = 19 \quad \cdots (2) \]
Step 3: Solving equations (1) and (2)
From equation (1), we know:
\[ a_1 = 5 - d \] Substituting this into equation (2): \[ 2(5 - d) + 5d = 19 \quad \Rightarrow \quad 10 - 2d + 5d = 19 \quad \Rightarrow \quad 3d = 9 \quad \Rightarrow \quad d = 3 \] From \( a_1 = 5 - d \), we get: \[ a_1 = 5 - 3 = 2 \]
Step 4: Finding the six numbers
The six numbers in the arithmetic progression are:
\[ a_1 = 2, \quad a_2 = 5, \quad a_3 = 8, \quad a_4 = 11, \quad a_5 = 14, \quad a_6 = 17 \]
Step 5: Calculating variance (\( \sigma^2 \))
The variance formula is:
\[ \sigma^2 = \text{mean of squares} - (\text{square of mean}) \] The mean of squares is: \[ \frac{2^2 + 5^2 + 8^2 + 11^2 + 14^2 + 17^2}{6} = \frac{4 + 25 + 64 + 121 + 196 + 289}{6} = \frac{699}{6} = 116.5 \] The square of the mean is: \[ \left( \frac{19}{2} \right)^2 = \frac{361}{4} = 90.25 \] Therefore, the variance is: \[ \sigma^2 = 116.5 - 90.25 = 26.25 \]
Step 6: Calculating \( 8\sigma^2 \) The final result is:
\[ 8\sigma^2 = 8 \times 26.25 = 210 \]
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
Top JEE Main Straight lines Questions
- If the sum of squares of all real values of \( \alpha \), for which the lines \( 2x - y + 3 = 0 \), \( 6x + 3y + 1 = 0 \) and \( \alpha x + 2y - 2 = 0 \) do not form a triangle \( p \), then the greatest integer less than or equal to \( p \) is ....
- The vertices of a triangle are A(–1, 3), B(–2, 2) and C(3, –1). A new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is :
- Let the angles made with the positive $x$-axis by two straight lines drawn from the point $P(2,3)$ and meeting the line $x+y=6$ at a distance $\sqrt{\frac{2}{3}}$ from the point $P$ be $\theta_1$ and $\theta_2$. Then the value of $(\theta_1+\theta_2)$ is
- For $\alpha, \beta \in R$, suppose the system of linear equations $ x-y+z=5 $ $2 x+2 y+\alpha z=8 $ $3 x-y+4 z=\beta$ has infinitely many solutions Then $\alpha$ and $\beta$ are the roots of
- Let $S_1$ and $S_2$ be respectively the sets of all $a \in R -\{0\}$ for which the system of linear equations $ a x+2 a y-3 a z=1$ $ (2 a+1) x+(2 a+3) y+(a+1) z=2$ $(3 a+5) x+(a+5) y+(a+2) z=3$ has unique solution and infinitely many solutions Then
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
Concepts Used:
Straight lines
A straight line is a line having the shortest distance between two points.
A straight line can be represented as an equation in various forms, as show in the image below:
The following are the many forms of the equation of the line that are presented in straight line-
1. Slope – Point Form
Assume P0(x0, y0) is a fixed point on a non-vertical line L with m as its slope. If P (x, y) is an arbitrary point on L, then the point (x, y) lies on the line with slope m through the fixed point (x0, y0) if and only if its coordinates fulfil the equation below.
y – y0 = m (x – x0)
2. Two – Point Form
Let's look at the line. L crosses between two places. P1(x1, y1) and P2(x2, y2) are general points on L, while P (x, y) is a general point on L. As a result, the three points P1, P2, and P are collinear, and it becomes
The slope of P2P = The slope of P1P2 , i.e.
\(\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}\)
Hence, the equation becomes:
y - y1 =\( \frac{y_2-y_1}{x_2-x_1} (x-x1)\)
3. Slope-Intercept Form
Assume that a line L with slope m intersects the y-axis at a distance c from the origin, and that the distance c is referred to as the line L's y-intercept. As a result, the coordinates of the spot on the y-axis where the line intersects are (0, c). As a result, the slope of the line L is m, and it passes through a fixed point (0, c). The equation of the line L thus obtained from the slope – point form is given by
y – c =m( x - 0 )
As a result, the point (x, y) on the line with slope m and y-intercept c lies on the line, if and only if
y = m x +c