If the variance of the terms in an increasing `A.P., b_(1),b_(2), b_(3), ...,b_(11)` is 90, then the common difference of this A.P. is ______

Answer

To find the common difference of the increasing arithmetic progression (A.P.) given that the variance of its terms is 90, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Terms of the A.P.**:
The terms of the A.P. can be expressed as:
\[ b_1 = a, \quad b_2 = a + d, \quad b_3 = a + 2d, \ldots, \quad b_{11} = a + 10d \]
where \( a \) is the first term and \( d \) is the common difference.

2. **Calculate the Mean**:
The mean \( \mu \) of the terms is given by:
\[ \mu = \frac{b_1 + b_2 + b_3 + \ldots + b_{11}}{11} = \frac{11a + (0 + 1 + 2 + \ldots + 10)d}{11} \]
The sum of the first 10 natural numbers is \( \frac{10 \times 11}{2} = 55 \), so:
\[ \mu = \frac{11a + 55d}{11} = a + 5d \]

3. **Calculate the Variance**:
The variance \( \sigma^2 \) is given by:
\[ \sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (b_i - \mu)^2 \]
This can be simplified to:
\[ \sigma^2 = \frac{1}{11} \left( (b_1 - \mu)^2 + (b_2 - \mu)^2 + \ldots + (b_{11} - \mu)^2 \right) \]
Substituting \( b_i \) and \( \mu \):
\[ \sigma^2 = \frac{1}{11} \left( (a - (a + 5d))^2 + (a + d - (a + 5d))^2 + \ldots + (a + 10d - (a + 5d))^2 \right) \]
This simplifies to:
\[ \sigma^2 = \frac{1}{11} \left( (-5d)^2 + (-4d)^2 + (-3d)^2 + (-2d)^2 + (-d)^2 + (0)^2 + (d)^2 + (2d)^2 + (3d)^2 + (4d)^2 + (5d)^2 \right) \]

4. **Calculate Each Term**:
The squared terms are:
\[ 25d^2 + 16d^2 + 9d^2 + 4d^2 + d^2 + 0 + d^2 + 4d^2 + 9d^2 + 16d^2 + 25d^2 \]
Summing these gives:
\[ 2(25 + 16 + 9 + 4 + 1) d^2 = 2 \times 55 d^2 = 110 d^2 \]

5. **Variance Formula**:
Therefore, the variance is:
\[ \sigma^2 = \frac{110 d^2}{11} = 10 d^2 \]
We know from the problem that the variance is 90:
\[ 10 d^2 = 90 \]

6. **Solve for d**:
Dividing both sides by 10:
\[ d^2 = 9 \]
Taking the square root:
\[ d = 3 \quad \text{(since the A.P. is increasing, we take the positive root)} \]

### Final Answer:
The common difference \( d \) of the A.P. is **3**.