Why did the question ask for $8\sigma^2$ instead of just $\sigma^2$?

This is a classic JEE 'trap' designed to test your attention to detail. Many students calculate $\sigma^2$ correctly but forget to perform the final multiplication. Always re-read the final line of the question before bubbling your answer!

JEE Main 2023 (24 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Statistics: Let the six numbers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ be in A.P. and $a_{1} + a_{3} = 10$ . If the mean of these six numbers is $\frac{19}{2}$ and their variance is $σ^{2}$ , then $8 σ^{2}$ is equal to

Q: Wait, why can't we just use the formula for the variance of an A.P. directly?

You absolutely can! The variance of an A.P. with $n$ terms and common difference $d$ is given by $\sigma^2 = \frac{(n^2-1)d^2}{12}$. However, in the heat of a JEE exam, deriving or remembering this formula can be risky. Calculating it step-by-step as we did ensures you don't make a sign error or misremember the formula. It also builds a deeper intuition for how the data is spread.

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the six numbers as points on a number line with equal spacing. Since they are in an Arithmetic Progression (A.P.), they possess a beautiful symmetry around their mean. The mean of an A.P. is simply the average of the first and last terms, or the average of the middle two terms. Can you see how this symmetry simplifies the problem?

Hint 2

Hint 2 (The Arsenal): To find the variance

σ^{2}

, you need the sum of the squares of the terms and the square of the mean. Recall the variance formula:

σ^{2} = \frac{\sum a _{i}^{2}}{n} - (\overset{x}{ˉ})^{2}

. Since you have an A.P., you can express each term

a_{i}

in terms of the first term

a_{1}

and the common difference

d

. Use the given conditions to form a system of linear equations to solve for

a_{1}

and

d

Hint 3

Hint 3 (The Execution): Start by defining the terms as

a_{1}, a_{1} + d, a_{1} + 2 d, a_{1} + 3 d, a_{1} + 4 d, a_{1} + 5 d

. Use the condition

a_{1} + a_{3} = 10

to get

2 a_{1} + 2 d = 10

, which simplifies to

a_{1} + d = 5

. Then, use the mean

\overset{x}{ˉ} = \frac{19}{2}

to find the sum of the terms, which leads to

2 a_{1} + 5 d = 19

. Solve this system to find

a_{1}

and

d

Visualized Solution

Defining the A.P. Sequence

Let the six numbers be $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ in A.P.
First term = $a_{1}$ , Common difference = $d$
Terms: $a_{1}, a_{1} + d, a_{1} + 2 d, a_{1} + 3 d, a_{1} + 4 d, a_{1} + 5 d$

Using the First Condition

Given: $a_{1} + a_{3} = 10$
We know that $a_{3} = a_{1} + 2 d$

Simplifying the First Equation

Substitute $a_{3}$ : $a_{1} + (a_{1} + 2 d) = 10$
$2 a_{1} + 2 d = 10$
Dividing by 2: $a_{1} + d = 5 — (1)$

Applying the Mean Condition

Given: Mean $\overset{x}{ˉ} = \frac{19}{2}$
Mean formula: $\overset{x}{ˉ} = \frac{Sum of all terms}{6}$

Sum of the A.P.

Sum of 6 terms: $S_{6} = \frac{6}{2} [2 a_{1} + (6 - 1) d]$
$S_{6} = 3 (2 a_{1} + 5 d)$
Mean: $\frac{3 ( 2 a _{1} + 5 d )}{6} = \frac{19}{2}$

Simplifying the Mean Equation

$\frac{3 ( 2 a _{1} + 5 d )}{6} = \frac{19}{2}$
$\frac{2 a _{1} + 5 d}{2} = \frac{19}{2}$
$2 a_{1} + 5 d = 19 — (2)$

Solving for $d$

Equation (1): $a_{1} + d = 5 ⟹ 2 a_{1} + 2 d = 10$
Equation (2): $2 a_{1} + 5 d = 19$
Subtracting: $(2 a_{1} + 5 d) - (2 a_{1} + 2 d) = 19 - 10$
$3 d = 9 ⟹ d = 3$

Solving for $a_{1}$

Substitute $d = 3$ in Equation (1)
$a_{1} + 3 = 5$
$a_{1} = 2$

Identifying the Six Numbers

$a_{1} = 2$
$a_{2} = 2 + 3 = 5$
$a_{3} = 5 + 3 = 8$
$a_{4} = 8 + 3 = 11$
$a_{5} = 11 + 3 = 14$
$a_{6} = 14 + 3 = 17$

The Variance Formula

We need to find the variance $σ^{2}$ .
Formula: $σ^{2} = \frac{\sum a _{i}^{2}}{n} - (\overset{x}{ˉ})^{2}$
Here, $n = 6$ and $\overset{x}{ˉ} = \frac{19}{2}$

Calculating Sum of Squares

$\sum a_{i}^{2} = 2^{2} + 5^{2} + 8^{2} + 1 1^{2} + 1 4^{2} + 1 7^{2}$
$\sum a_{i}^{2} = 4 + 25 + 64 + 121 + 196 + 289$
$\sum a_{i}^{2} = 699$

Computing Variance $σ^{2}$

$σ^{2} = \frac{699}{6} - (\frac{19}{2})^{2}$
$σ^{2} = \frac{233}{2} - \frac{361}{4}$
$σ^{2} = \frac{466 - 361}{4} = \frac{105}{4}$

Final Calculation for $8 σ^{2}$

The question asks for $8 σ^{2}$
$8 σ^{2} = 8 \times \frac{105}{4}$
$8 σ^{2} = 2 \times 105 = 210$
Final Answer: 210

00:00 / 00:00

The Sigma Insight: Variance and Standard Deviation

Solution Diagram

Analyzing the Setup

Imagine six markers placed at equal distances, forming an Arithmetic Progression (A.P.). Let the sequence be defined by the starting term

a_{1}

and the common difference

d

The six terms are represented as:

a_{1}, a_{1} + d, a_{1} + 2 d, a_{1} + 3 d, a_{1} + 4 d, a_{1} + 5 d

The Algebraic Hunt

The problem provides two constraints. First, we are given that

a_{1} + a_{3} = 10

. Since

a_{3} = a_{1} + 2 d

, we substitute to obtain:

a_{1} + (a_{1} + 2 d) = 10 \Rightarrow 2 a_{1} + 2 d = 10 \Rightarrow a_{1} + d = 5

Second, the mean

\overset{x}{ˉ}

of the six terms is

\frac{19}{2}

. The sum of an A.P. is given by

S_{n} = \frac{n}{2} [2 a_{1} + (n - 1) d]

. For

n = 6

S_{6} = 3 (2 a_{1} + 5 d)

Setting the mean

\frac{S _{6}}{6} = \frac{19}{2}

leads to:

\frac{3 ( 2 a _{1} + 5 d )}{6} = \frac{19}{2} \Rightarrow \frac{2 a _{1} + 5 d}{2} = \frac{19}{2} \Rightarrow 2 a_{1} + 5 d = 19

Solving the System

We now have a system of two linear equations: 1)

a_{1} + d = 5

2 a_{1} + 5 d = 19

Subtracting twice the first equation from the second:

(2 a_{1} + 5 d) - 2 (a_{1} + d) = 19 - 10 \Rightarrow 3 d = 9 \Rightarrow d = 3

Substituting

d = 3

back into the first equation:

a_{1} + 3 = 5 \Rightarrow a_{1} = 2

The sequence is revealed as:

2, 5, 8, 11, 14, 17

The Statistical Heart

The variance

σ^{2}

is defined by the formula:

σ^{2} = \frac{\sum a _{i}^{2}}{n} - (\overset{x}{ˉ})^{2}

Calculating the squares of the terms:

2^{2} = 4, 5^{2} = 25, 8^{2} = 64, 1 1^{2} = 121, 1 4^{2} = 196, 1 7^{2} = 289

The sum of these squares is

699

. The mean squared is

(\frac{19}{2})^{2} = \frac{361}{4}

. Thus:

σ^{2} = \frac{699}{6} - \frac{361}{4} = \frac{233}{2} - \frac{361}{4}

Converting to a common denominator:

σ^{2} = \frac{466 - 361}{4} = \frac{105}{4}

The Grand Finale

The question asks for the value of

8 σ^{2}

. Substituting our calculated variance:

8 \times (\frac{105}{4}) = 2 \times 105 = 210

The final result is 210.

Common Traps & FAQs

Wait, why can't we just use the formula for the variance of an A.P. directly?

You absolutely can! The variance of an A.P. with

n

terms and common difference

d

is given by

σ^{2} = \frac{( n ^{2} - 1 ) d ^{2}}{12}

. However, in the heat of a JEE exam, deriving or remembering this formula can be risky. Calculating it step-by-step as we did ensures you don't make a sign error or misremember the formula. It also builds a deeper intuition for how the data is spread.

Is there a faster way to find the sum of squares?

Yes, there is! Because the terms are symmetric around the mean, you can shift the sequence by subtracting the mean from each term. This centers the data at zero, making the sum of squares much easier to compute. But for six terms, the manual squaring method is quite fast and less prone to conceptual errors.

Why did the question ask for

8 σ^{2}

instead of just

σ^{2}

This is a classic JEE 'trap' designed to test your attention to detail. Many students calculate

σ^{2}

correctly but forget to perform the final multiplication. Always re-read the final line of the question before bubbling your answer!

Animated Solution for Mathematics - Statistics: Let the six numbers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ be in A.P. and $a_{1} + a_{3} = 10$ . If the mean of these six numbers is $\frac{19}{2}$ and their variance is $σ^{2}$ , then $8 σ^{2}$ is equal to

Select Answer:

Visualized Solution

Defining the A.P. Sequence

Using the First Condition

Simplifying the First Equation

Applying the Mean Condition

Sum of the A.P.

Simplifying the Mean Equation

Solving for $d$

Solving for $a_{1}$

Identifying the Six Numbers

The Variance Formula

Calculating Sum of Squares

Computing Variance $σ^{2}$

Final Calculation for $8 σ^{2}$

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Algebraic Hunt

Solving the System

The Statistical Heart

The Grand Finale

Similar Questions

If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then $a + b + ab$ is equal to :

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Select Answer:

Visualized Solution

Defining the A.P. Sequence

Using the First Condition

Simplifying the First Equation

Applying the Mean Condition

Sum of the A.P.

Simplifying the Mean Equation

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a1​

Identifying the Six Numbers

The Variance Formula

Calculating Sum of Squares

Computing Variance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } σ2

Final Calculation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 8σ2

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Algebraic Hunt

Solving the System

The Statistical Heart

The Grand Finale

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then a+b+ab is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let 9=x1​<x2​<⋯<x7​ be in an A.P. with common difference d. If the standard deviation of x1​,x2​,…,x7​ is 4 and the mean is xˉ, then xˉ+x6​ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the mean and variance of six observations 7, 10, 11, 15, a, b are 10 and 320​, respectively, then the value of ∣a−b∣ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the mean of the data : 7, 8, 9, 7, 8, 7, λ, 8 is 8, then the variance of this data is :-

Solving for $d$

Solving for $a_{1}$

Computing Variance $σ^{2}$

Final Calculation for $8 σ^{2}$

If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then $a + b + ab$ is equal to :

Let $9 = x_{1} < x_{2} < \dots < x_{7}$ be in an A.P. with common difference $d$ . If the standard deviation of $x_{1}, x_{2}, \dots, x_{7}$ is 4 and the mean is $\overset{x}{ˉ}$ , then $\overset{x}{ˉ} + x_{6}$ is equal to :

If the mean and variance of six observations 7, 10, 11, 15, a, b are 10 and $\frac{20}{3}$ , respectively, then the value of $∣ a - b ∣$ is equal to :

If the mean of the data : 7, 8, 9, 7, 8, 7, $λ$ , 8 is 8, then the variance of this data is :-