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(A) Let $a$ be the first term and $d$ be the common difference of the given $A.P.$ Since the $A.P.$ is increasing,$d > 0$.The terms are $a, a+d, a+2d,

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$3$

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(A) Let $a$ be the first term and $d$ be the common difference of the given $A.P.$ Since the $A.P.$ is increasing,$d > 0$.The terms are $a, a+d, a+2d, \ldots, a+10d$.The mean $\bar{X} = \frac{1}{11} \sum_{i=0}^{10} (a + id) = a + \frac{d}{11} 	imes \frac{10 	imes 11}{2} = a + 5d$.The variance is given by $\sigma^2 = \frac{1}{n} \sum (x_i - \bar{X})^2$.$\sigma^2 = \frac{1}{11} \sum_{i=0}^{10} (a + id - (a + 5d))^2 = \frac{1}{11} \sum_{i=0}^{10} ((i-5)d)^2 = \frac{d^2}{11} \sum_{i=0}^{10} (i-5)^2$.Calculating the sum: $(-5)^2 + (-4)^2 + (-3)^2 + (-2)^2 + (-1)^2 + 0^2 + 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 25 + 16 + 9 + 4 + 1 + 0 + 1 + 4 + 9 + 16 + 25 = 110$.So,$90 = \frac{d^2}{11} 	imes 110 = 10d^2$.$d^2 = 9 \Rightarrow d = \pm 3$.Since the $A.P.$ is increasing,$d = 3$.

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

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