The A.M. of $1, 3, 5, . . ., (2 n - 1)$ is-

n + 1

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n + 2

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n^{2}

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n

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Solution

The correct option is B $n$
We know that

$A . M = \frac{s u m o f a l l t e r m s}{n u m b e r o f t e r m s}$

Now, we will calculate sum of all the terms
$1 + 3 + 5 + . . . . . + (2 n - 1) = \frac{n}{2}$ [ $1^{s t}$ term $+$ last term]

$= \frac{n}{2} [1 + 2 n - 1] = \frac{2 n^{2}}{2} = n^{2}$

So, $A . M = \frac{n^{2}}{n} = n$

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{lim}_{n \to \infty} (\frac{n}{n^{2}} + \frac{n}{n^{2} + 1} + \frac{n}{n^{2} + 2^{2}} + . . . . . . + \frac{n}{2 n^{2} - 2 n + 1})

is equal to

Q. The

A . M .

of the observations

1.3.5, 3.5, 7, 9, ., (2 n - 1) (2 n + 1) (2 n + 3)

(\forall n \in N)

$= {lim}_{n \to \infty} [\frac{1}{n} + \frac{1}{\sqrt{n^{2} + n}} + \frac{1}{\sqrt{n^{2} + 2 n}} + \dots + \frac{1}{\sqrt{n^{2} + (n - 1) n}}]$ is equal to [RPET 2000]

Q. If

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is equal to

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The A.M. of 1,3,5,...,(2n−1) is-

The correct option is B nWe know thatA.M=sum of all termsnumber of termsNow, we will calculate sum of all the terms1+3+5+.....+(2n−1)=n2 [1st term + last term]=n2[1+2n−1]=2n22=n2So, A.M=n2n=n

The A.M. of $1, 3, 5, . . ., (2 n - 1)$ is-