1. Key Concepts and Formulas

This problem involves calculating the common difference of an Arithmetic Progression (A.P.) given its variance. We will utilize the fundamental definitions and formulas for A.P.s and variance, along with standard summation identities.

• Arithmetic Progression (A.P.): A sequence where the difference between consecutive terms is constant, known as the common difference, $d$. For an A.P. with first term $a$ and $N$ terms, the $i$-th term is $b_i = a + (i-1)d$.

  • Sum of an A.P. ($\sum b_i$): For $N$ terms, $S_N = \frac{N}{2}(\text{first term} + \text{last term})$.
  • Mean of an A.P. ($\bar{b}$): $\bar{b} = \frac{\sum b_i}{N}$. For an odd number of terms, the mean is equal to the middle term.

• Variance ($\sigma^2$): A measure of how spread out the numbers in a data set are. For $N$ observations $x_1, x_2, \dots, x_N$, the variance is given by the computational formula: $\sigma^2 = \frac{\sum_{i=1}^{N} x_i^2}{N} - \left(\frac{\sum_{i=1}^{N} x_i}{N}\right)^2 = \frac{\sum x_i^2}{N} - (\bar{x})^2$

• Standard Summation Formulas: These are essential for simplifying sums of powers of integers:

  • Sum of the first $n$ natural numbers: $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$
  • Sum of the squares of the first $n$ natural numbers: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

2. Step-by-Step Solution

We are given an increasing A.P. with 11 terms ($N=11$), $b_1, b_2, \dots, b_{11}$, and its variance is $\sigma^2 = 90$. Our objective is to determine the common difference, $d$. Since the A.P. is increasing, we know that $d > 0$.

Step 1: Define the Terms of the A.P.

• Why this step? To establish the general form of the terms in the A.P., which is necessary for calculating their sum and sum of squares.

Let the first term of the A.P. be $a$ and the common difference be $d$. The $N=11$ terms of the A.P. are: $b_i = a + (i-1)d$ for $i=1, 2, \dots, 11$. So, $b_1 = a$, $b_2 = a+d$, ..., $b_{11} = a+10d$.

Step 2: Calculate the Sum of the Terms ($\sum b_i$) and the Mean ($\bar{b}$)

• Why this step? Both the sum and the mean are direct components required by the variance formula.

Using the formula for the sum of an A.P., $S_N = \frac{N}{2}(\text{first term} + \text{last term})$: $\sum_{i=1}^{11} b_i = S_{11} = \frac{11}{2}(a + (a + 10d))$ $\sum_{i=1}^{11} b_i = \frac{11}{2}(2a + 10d) = 11(a + 5d)$ Now, calculate the mean ($\bar{b}$): $\bar{b} = \frac{\sum b_i}{N} = \frac{11(a + 5d)}{11} = a + 5d$

• Self-check: For an A.P. with an odd number of terms, the mean is the middle term. Here, with 11 terms, the middle term is $b_6 = a + (6-1)d = a+5d$, which matches our calculation.

Step 3: Calculate the Sum of the Squares of the Terms ($\sum b_i^2$)

• Why this step? This is the second critical component for the variance formula and typically involves more algebraic manipulation using summation identities.

We need to calculate $\sum_{i=1}^{11} b_i^2 = \sum_{i=1}^{11} (a + (i-1)d)^2$. To simplify the application of summation formulas, let $k = i-1$. As $i$ ranges from 1 to 11, $k$ ranges from 0 to 10. $\sum_{i=1}^{11} b_i^2 = \sum_{k=0}^{10} (a + kd)^2$ Expand the squared term: $(a + kd)^2 = a^2 + 2akd + k^2d^2$. $\sum_{k=0}^{10} (a^2 + 2akd + k^2d^2)$ By the linearity property of summation, we can split this into three separate sums: $= \sum_{k=0}^{10} a^2 + \sum_{k=0}^{10} 2akd + \sum_{k=0}^{10} k^2d^2$ Factor out constants from each sum: $= a^2 \sum_{k=0}^{10} 1 + 2ad \sum_{k=0}^{10} k + d^2 \sum_{k=0}^{10} k^2$ Now, apply the standard summation formulas. Note that $\sum_{k=0}^{10} k = \sum_{k=1}^{10} k$ and $\sum_{k=0}^{10} k^2 = \sum_{k=1}^{10} k^2$ because the $k=0$ term in both sums is zero.

• $\sum_{k=0}^{10} 1 = 11$ (summing 1 for 11 terms from $k=0$ to $k=10$).
• $\sum_{k=0}^{10} k = \sum_{k=1}^{10} k = \frac{10(10+1)}{2} = \frac{10 \times 11}{2} = 55$.
• $\sum_{k=0}^{10} k^2 = \sum_{k=1}^{10} k^2 = \frac{10(10+1)(2 \times 10+1)}{6} = \frac{10 \times 11 \times 21}{6} = 5 \times 11 \times 7 = 385$.

Substitute these values back into the expression for $\sum b_i^2$: $\sum b_i^2 = a^2(11) + 2ad(55) + d^2(385)$ $\sum b_i^2 = 11a^2 + 110ad + 385d^2$

Step 4: Substitute into the Variance Formula and Solve for $d$

• Why this step? This is where we combine the given variance with our calculated mean and sum of squares to form an equation that can be solved for the common difference, $d$.

Using the variance formula: $\sigma^2 = \frac{\sum b_i^2}{N} - (\bar{b})^2$. We have $\sigma^2 = 90$, $N=11$, $\sum b_i^2 = 11a^2 + 110ad + 385d^2$, and $\bar{b} = a + 5d$. Substitute these into the formula: $90 = \frac{11a^2 + 110ad + 385d^2}{11} - (a + 5d)^2$ First, simplify the fractional term: $\frac{11a^2 + 110ad + 385d^2}{11} = a^2 + 10ad + 35d^2$ Next, expand the squared mean term: $(a + 5d)^2 = a^2 + 2(a)(5d) + (5d)^2 = a^2 + 10ad + 25d^2$ Now, substitute these simplified expressions back into the variance equation: $90 = (a^2 + 10ad + 35d^2) - (a^2 + 10ad + 25d^2)$ Distribute the negative sign: $90 = a^2 + 10ad + 35d^2 - a^2 - 10ad - 25d^2$ Observe the cancellation of terms involving $a$: $90 = (a^2 - a^2) + (10ad - 10ad) + (35d^2 - 25d^2)$ $90 = 0 + 0 + 10d^2$ $90 = 10d^2$ Solve for $d^2$: $d^2 = \frac{90}{10} = 9$ Take the square root of both sides: $d = \pm \sqrt{9}$ $d = \pm 3$

Step 5: Apply the Condition for an Increasing A.P.

• Why this step? The problem statement provides a crucial condition ("increasing A.P.") that allows us to uniquely determine the value of $d$.

We found two possible values for $d$: $3$ and $-3$. An A.P. is increasing if its common difference $d > 0$. An A.P. is decreasing if its common difference $d < 0$. An A.P. is constant if its common difference $d = 0$. Since the problem states that the A.P. is increasing, we must choose the positive value for $d$. Therefore, $d = 3$.

3. Common Mistakes & Tips

• Summation Limits: Be extremely careful with the starting and ending values of the summation index, especially when changing variables (e.g., from $i-1$ to $k$). Ensure the number of terms remains correct.
• Algebraic Simplification: Expanding squares and distributing negative signs correctly is crucial to avoid errors. The cancellation of terms involving 'a' is a key simplification.
• Variance of A.P.: A useful property to remember is that the variance of an A.P. depends only on the common difference ($d$) and the number of terms ($N$), not on the first term ($a$). This is evident from the cancellation of $a$-related terms in our derivation.
• Problem Constraints: Always refer back to the problem statement for conditions like "increasing A.P." to select the correct solution from multiple possibilities.

4. Summary

We systematically solved the problem by first defining the terms of the A.P. and then calculating its mean and the sum of the squares of its terms using standard summation formulas. We substituted these values into the computational formula for variance. The key algebraic step involved simplifying the expression, which led to the cancellation of terms involving the first term 'a', demonstrating that the variance of an A.P. is independent of its starting term. Finally, we solved for $d^2$ and used the condition that the A.P. is increasing to select the positive value for the common difference.

The final answer is $\boxed{3}$.