Let `a_1,a_2,....a11` are in incresing A.P.and if varience of these number is `90` then value of common difference of A.P. is.

Answer

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

### Step 1: Define the terms of the A.P.
Let the first term of the A.P. be \( a \). The terms of the A.P. can be expressed as:
\[ a_1 = a, \quad a_2 = a + d, \quad a_3 = a + 2d, \quad \ldots, \quad a_{11} = a + 10d \]

### Step 2: Calculate the mean \( \bar{x} \)
The mean \( \bar{x} \) of the terms can be calculated as:
\[ \bar{x} = \frac{a_1 + a_2 + \ldots + a_{11}}{11} = \frac{a + (a + d) + (a + 2d) + \ldots + (a + 10d)}{11} \]
This simplifies to:
\[ \bar{x} = \frac{11a + (0 + 1 + 2 + \ldots + 10)d}{11} = \frac{11a + 55d}{11} = a + 5d \]

### Step 3: Calculate the variance
The variance \( \sigma^2 \) is given by:
\[ \sigma^2 = \frac{\sum_{i=1}^{11} (a_i - \bar{x})^2}{11} \]
Substituting \( \bar{x} = a + 5d \):
\[ \sigma^2 = \frac{(a - (a + 5d))^2 + (a + d - (a + 5d))^2 + \ldots + (a + 10d - (a + 5d))^2}{11} \]
This can be expressed as:
\[ \sigma^2 = \frac{(-5d)^2 + (-4d)^2 + (-3d)^2 + (-2d)^2 + (-d)^2 + 0^2 + (d)^2 + (2d)^2 + (3d)^2 + (4d)^2 + (5d)^2}{11} \]
Calculating the squares:
\[ = \frac{25d^2 + 16d^2 + 9d^2 + 4d^2 + d^2 + 0 + d^2 + 4d^2 + 9d^2 + 16d^2 + 25d^2}{11} \]
\[ = \frac{(25 + 16 + 9 + 4 + 1 + 0 + 1 + 4 + 9 + 16 + 25)d^2}{11} \]
Calculating the sum:
\[ = \frac{110d^2}{11} = 10d^2 \]

### Step 4: Set the variance equal to 90
Given that the variance is \( 90 \):
\[ 10d^2 = 90 \]
Solving for \( d^2 \):
\[ d^2 = 9 \]
Taking the square root:
\[ d = 3 \quad \text{or} \quad d = -3 \]

### Step 5: Determine the common difference
Since the A.P. is increasing, we take the positive value:
\[ d = 3 \]

Thus, the common difference of the A.P. is \( \boxed{3} \).