Verified Solution • 2025–26 Session moderate

QJEEM21Ma4c9b: JEE Main 2021 Mathematics - Class 11 JEEM21Ma4c9b Solution: Let in a series of 2n observations, half of them are equal to a …

Let in a series of 2n observations, half of them are equal to a and remaining half are equal to −- a. Also by adding a constant b in each of these observations, the mean and standard deviation of new set become 5 and 20, respectively. Then the value of a2 + b2 is equal to :

Multiple Choice Options

  • A 425
  • B 250
  • C 925
  • D 650

Step-by-Step Verified Solution

Given series

(a, a, a, ........ n times), ( −- a, −- a, −- a, ...... n times)

Now x‾\overline x = ∑xi2n=0{{\sum {{x_i}} } \over {2n}} = 0

as, xi →\to xi + b

then x‾\overline x →\to x‾\overline x + b

So, x‾\overline x + b = 5 ⇒\Rightarrow b = 5

No change in S.D. due to change in origin

Standard deviation ( σ\sigma ) = ∑i=12n(xi−x‾)22n\sqrt {{{\sum\limits_{i = 1}^{2n} {{{({x_i} - \overline x )}^2}} } \over {2n}}}

=∑i=12nxi22n=2na22n=a2= \sqrt {{{\sum\limits_{i = 1}^{2n} {x_i^2} } \over {2n}}} = \sqrt {{{2n{a^2}} \over {2n}}} = \sqrt {{a^2}}

∴\therefore 20=a2⇒a=2020 = \sqrt {{a^2}} \Rightarrow a = 20

∴\therefore a2 + b2 = 425

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