# If the variance of the terms in on increasing AP. $b_{1},b_{2},b_{3}.....b_{4}$ is 90 then common difference of this A.P is ___ Option: 1 -3 Option: 2 6 Option: 3 3 Option: 4 Both (a) and (c)

$\sum \frac{(x i-\mu)^{2}}{n}=90$

$\mu =\frac{\frac{11}{2}[2 a+10 d]}{11}=a+5 d \\ \sum \frac{(x i-\mu)^{2}}{n} =\frac{\sum x i^{2}}{n}-\mu^{2}$

$\\=\frac{\sum(a+(r-1) d)^{2}}{11}-(a+5 d)^{2} \\ =a^{2}+\frac{10 \times 11 \times 21}{6 \times 11} d^{2}+\frac{2 a d \times 10 \times 11}{2 \times 11} -\left(a^{2}+25 d^{2}+10 ad\right) \\\\ 10 d^{2}=90 \\\\ d= \pm3$

But it is given that AP is increasing

Hence d = 3

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