The arithmetic mean of the first $n$ odd numbers is

n

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\frac{n}{2}

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\frac{n - 1}{2}

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\frac{n + 1}{2}

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Solution

The correct option is A $n$
First $n$ odd natural numbers are
$1, 3, 5, . . ., (2 n - 1)$
So, the required mean $= \frac{1 + 3 + 5 + . . . + (2 n - 1)}{n} = \frac{\frac{n}{2} [1 + 2 n - 1]}{n} = \frac{n^{2}}{n} = n$

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Similar questions

The arithmetic mean of nC0,nC1,nC2....nCn is
(A)2^n/n
(B)2^n-1/n
(C)2^n/n+1
(D)2^n-1/n+1

\frac{n + 1}{2}

\frac{n - 1}{2}

\frac{n}{2}

\frac{n}{2} + 1

T h e a r i t h m e t i c m e a n o f t h e s e r i e s^{n} c_{0},^{n} c_{1},^{n} c_{2} . . . . .,^{n} c_{n} i s :

$1 + 3 + 5 + . . . . . + (2 n - 1) = n^{2}$ i.e., the sum of first n odd natural numbers is $n^{2}$ .

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