JEE MAIN - Mathematics (2020 - 2nd September Evening Slot - No. 12)
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_______.
b1 , b2 , b3 ,....,b11 is 90, then the common difference of this A.P. is_______.
Answer
3
Explanation
Let the common difference = d
and $${b_1} = a$$
$${b_2} = a + d$$
$${b_3} = a + 2d$$
... $${b_{11}} = a + 10d$$
Variance = $${{\sum {a_i^2} } \over {11}} - {\left( {{{\sum {{a_i}} } \over {11}}} \right)^2} = 90$$
$$ \Rightarrow {{{a^2} + {{\left( {a + d} \right)}^2} + ... + {{\left( {a + 10d} \right)}^2}} \over {11}} - {\left( {{{a + \left( {a + d} \right) + ... + \left( {a + 10d} \right)} \over {11}}} \right)^2} = 90$$
$$ \Rightarrow 11\left[ {11{a^2} + 385{d^2} + 110ad} \right] - {\left[ {11a + 55d} \right]^2} = 10890$$
$$ \Rightarrow 1210{d^2} = 10890$$
$$ \Rightarrow {d^2} = 9$$
$$ \Rightarrow d = \pm 3$$
As A.P is increasing so d should be positive
$$ \therefore $$ d = 3
and $${b_1} = a$$
$${b_2} = a + d$$
$${b_3} = a + 2d$$
... $${b_{11}} = a + 10d$$
Variance = $${{\sum {a_i^2} } \over {11}} - {\left( {{{\sum {{a_i}} } \over {11}}} \right)^2} = 90$$
$$ \Rightarrow {{{a^2} + {{\left( {a + d} \right)}^2} + ... + {{\left( {a + 10d} \right)}^2}} \over {11}} - {\left( {{{a + \left( {a + d} \right) + ... + \left( {a + 10d} \right)} \over {11}}} \right)^2} = 90$$
$$ \Rightarrow 11\left[ {11{a^2} + 385{d^2} + 110ad} \right] - {\left[ {11a + 55d} \right]^2} = 10890$$
$$ \Rightarrow 1210{d^2} = 10890$$
$$ \Rightarrow {d^2} = 9$$
$$ \Rightarrow d = \pm 3$$
As A.P is increasing so d should be positive
$$ \therefore $$ d = 3
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