If the variance of the terms in an increasing A.P., b1, b2, b3, ..., b11 is 90, then the common diff

Q: Why is the variance formula $\frac{d^2(n^2-1)}{12}$? Can't I just use $\frac{1}{n} \sum (x_i - \bar{x})^2$?

You absolutely can! The standard formula is the definition of variance. However, for an A.P., the terms are symmetric around the mean. Using the standard formula involves a long summation that simplifies to this elegant result. Using the derived formula is a mark of a seasoned JEE aspirant who values efficiency.

Q: Why do we reject $d = -3$?

The problem explicitly states the A.P. is 'increasing'. An increasing sequence requires each term to be larger than the previous one, which mathematically forces the common difference $d$ to be positive. If $d$ were negative, the sequence would be decreasing.

Q: Is there any other way to solve this if I forget the formula?

Yes! You can express the terms as $a, a+d, a+2d, \dots, a+10d$. The mean is $a+5d$. The variance is the average of the squared deviations. It will take longer, but the $a$ terms will cancel out, leaving you with the same result. It is a great exercise in algebraic manipulation!

Question

If the variance of the terms in an increasing A.P., $b_1, b_2, b_3, \dots, b_{11}$ is 90, then the common difference of this A.P. is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are going to look at a problem that seems like a dry statistical calculation but is actually a beautiful dance of symmetry and algebra.

We are dealing with an arithmetic progression (A.P.) of eleven terms: $b_1, b_2, \dots, b_{11}$. Imagine these terms as eleven points evenly spaced on a number line.

Because it is an A.P., the distance between any two consecutive points is constant, defined by the common difference $d$. The problem states that this sequence is increasing, which gives us a vital clue about the sign of $d$.

We are also given that the variance of these terms is $90$. Variance is a measure of how spread out these numbers are from their mean, and in an A.P., this spread is perfectly controlled by $d$.

### The Elegant Shortcut

While you could calculate the variance using the definition $\sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (b_i - \bar{b})^2$, that path is fraught with tedious algebra.

Instead, let us use the powerful result derived for the variance of an A.P. with $n$ terms:

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

This formula exists because the variance of the first $n$ natural numbers is $\frac{n^2-1}{12}$. Since an A.P. is just a scaled and shifted version of natural numbers, the variance scales by $d^2$. This formula is a weapon in your JEE arsenal—memorize it, respect it, and use it to save precious minutes during the exam.

### The Calculation

Now, let us apply our knowledge. We have $n = 11$ and $\sigma^2 = 90$.

Substituting these into our formula, we get:

$$ 90 = \frac{d^2(11^2 - 1)}{12} $$

First, let us handle the square: $11^2 = 121$. Our equation becomes:

$$ 90 = \frac{d^2(121 - 1)}{12} $$

Simplifying the numerator, $121 - 1 = 120$. Now the equation looks much friendlier:

$$ 90 = \frac{d^2(120)}{12} $$

Since $120$ divided by $12$ is exactly $10$, we are left with $90 = 10d^2$. Dividing both sides by $10$, we find $d^2 = 9$.

### The Final Insight

We are at the finish line. Taking the square root of $d^2 = 9$ gives us $d = \pm 3$.

Here is where the conceptual understanding matters. The problem stated the A.P. is increasing. If $d$ were $-3$, the sequence would be decreasing.

Therefore, we must reject the negative root. The common difference is **$d = 3$**.

You have successfully navigated the geometry of the sequence, applied the statistical shortcut, and respected the constraints of the problem. This is the essence of JEE mathematics—not just calculation, but understanding the soul of the problem.

Animated Solution for Mathematics - Sequence and Series: If the variance of the terms in an increasing A.P., $b_{1}, b_{2}, b_{3}, \dots, b_{11}$ is 90, then the common difference of this A.P. is

Enter Numerical Value:

Visualized Solution

Visualizing the A.P. Sequence

The Variance Formula for an A.P.

Substituting the Values

Evaluating the Square

Simplifying the Numerator

Simplifying the Fraction

Isolating $d^{2}$

Solving for $d$

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Setup

The Elegant Shortcut

The Calculation

The Final Insight

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