Verified Solution • 2025–26 Session moderate

QJEEM20Mab9e8: JEE Main 2020 Mathematics - Class 11 JEEM20Mab9e8 Solution: If the variance of the terms in an increasing A.P., b 1 , b 2 , …

If the variance of the terms in an increasing A.P.,
b1 , b2 , b3 ,....,b11 is 90, then the common difference of this A.P. is_______.

Step-by-Step Verified Solution

Let the common difference = d

and b1=a{b_1} = a
b2=a+d{b_2} = a + d
b3=a+2d{b_3} = a + 2d
... b11=a+10d{b_{11}} = a + 10d

Variance = ∑ai211−(∑ai11)2=90{{\sum {a_i^2} } \over {11}} - {\left( {{{\sum {{a_i}} } \over {11}}} \right)^2} = 90

⇒a2+(a+d)2+...+(a+10d)211−(a+(a+d)+...+(a+10d)11)2=90\Rightarrow {{{a^2} + {{\left( {a + d} \right)}^2} + ... + {{\left( {a + 10d} \right)}^2}} \over {11}} - {\left( {{{a + \left( {a + d} \right) + ... + \left( {a + 10d} \right)} \over {11}}} \right)^2} = 90

⇒11[11a2+385d2+110ad]−[11a+55d]2=10890\Rightarrow 11\left[ {11{a^2} + 385{d^2} + 110ad} \right] - {\left[ {11a + 55d} \right]^2} = 10890

⇒1210d2=10890\Rightarrow 1210{d^2} = 10890

⇒d2=9\Rightarrow {d^2} = 9

⇒d=±3\Rightarrow d = \pm 3

As A.P is increasing so d should be positive

∴\therefore d = 3

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