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

We have an increasing arithmetic progression whose eleven terms are written as $$b_1,\,b_2,\,b_3,\ldots ,b_{11}$$.

Let the first term be denoted by $$a$$ and the common difference by $$d$$.

Thus the general term is $$b_k = a + (k-1)d$$ for $$k = 1,2,\ldots ,11$$.

For an A.P. with $$n$$ terms, the mean (average) of all the terms is given by the formula $$\mu = \dfrac{\text{first term} + \text{last term}}{2}.$$ Here the last term is $$a + 10d$$ because the eleventh term has index $$k = 11$$. So we obtain $$\mu = \dfrac{a + \bigl(a + 10d\bigr)}{2} = \dfrac{2a + 10d}{2} = a + 5d.$$

We now recall the general formula for the variance of the $$n$$ terms of an arithmetic progression. If the terms are $$a,\,a+d,\,a+2d,\ldots ,a+(n-1)d$$, the variance (taking the divisor as $$n$$, i.e. population variance) is $$\sigma^2 = \dfrac{d^{2}\,(n^{2}-1)}{12}.$$ This formula is obtained by expanding each squared deviation $$\bigl(a+(k-1)d-\mu\bigr)^2$$, adding them for all $$k$$ from $$1$$ to $$n$$, and finally dividing by $$n$$.

In our problem $$n = 11$$, so we substitute $$n = 11$$ into the formula: $$\sigma^2 = \dfrac{d^{2}\,\bigl(11^{2}-1\bigr)}{12}.$$

We compute the square and the subtraction inside the brackets: $$11^{2} = 121,\quad 121 - 1 = 120.$$ Hence $$\sigma^2 = \dfrac{d^{2}\,\times 120}{12}.$$

Simplifying the fraction $$\dfrac{120}{12}$$ gives $$10$$, so $$\sigma^2 = 10\,d^{2}.$$

The question states that the variance of the eleven terms is $$90$$. We therefore set $$10\,d^{2} = 90.$$

Dividing both sides by $$10$$ gives $$d^{2} = 9.$$

Taking the positive square root (because the progression is increasing and thus $$d > 0$$) yields $$d = 3.$$

So, the answer is $$3$$.