The S.D. of the series ${\rm{a}},{\rm{\;}}a + d,a + 2\;d, \ldots ,a + 2nd,$ is

$\frac{{n\left( {n + 1} \right)}}{3}{d^2}$

$\frac{{n\left( {n - 1} \right)}}{3}{d^2}$

JEE MATHEMATICS MCQ: STATISTICS

1 Easy

1 of 1

Application +4 | -1

The S.D. of the series

\mathrm{a},\mathrm{\;}a + d,a + 2\;d, \ldots ,a + 2nd,

View answer and explanation

Correct option: B

\sqrt {\frac{{n\left( {n + 1} \right)}}{3}} d

Option B (

\sqrt {\frac{{n\left( {n + 1} \right)}}{3}} d

) is correct because it's the correct formula for the standard deviation of an arithmetic series.

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