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Terms in an Arithmetic Progression

The arithmeti...

Question

The arithmetic mean (average) of the first $n$ positive integers is

\frac{n}{2}

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\frac{n^{2}}{2}

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n

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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 C $\frac{n + 1}{2}$
The sum of the first $n$ positive integers is $s = \frac{n (n + 1)}{2}$ .

$∴ A . M . = \frac{s}{n}$ and hence the E will be correct answer.

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Similar questions

\frac{n + 1}{2}

\frac{n - 1}{2}

\frac{n}{2}

\frac{n}{2} + 1

n \geq 1

k

l, n

a_{1}, a_{2}, . . ., a_{n}

n

h_{1}, h_{2}, . . . ., h_{n}

i = 1, 2, . . . ., n

j = 1, 2, . . . ., n

p (i, j) = a_{i} h_{j}

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

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 :

For all positive integers n, show that

$^{2 n} C_{n} +^{2 n} C_{n - 1} = \frac{1}{2} (^{2 n + 2} C_{n + 1})$

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