Find the mean of first $n$ natural numbers.

n

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{n}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2 \times n + 1}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{n + 1}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{n + 1}{2}$
Mean, $u = \frac{(1 + 2 + 3... + n)}{n}$
$\Rightarrow u = \frac{1}{n} \times \frac{1}{2} (n) (n + 1)$
$∴ u = \frac{1}{2} (n + 1)$
Ans-Option $D$

Suggest Corrections

Similar questions

Q. 1 + 2 + 3 + ... + n =

\frac{n (n + 1)}{2}

i.e. the sum of the first n natural numbers is

\frac{n (n + 1)}{2}

Q. The sum of first n odd natural numbers is

(a) 2n − 1

(b) 2n + 1

(c) n²

(d) n² − 1

Question 87

The sum of squares of first n natural numbers is given by $\frac{1}{6} n (n + 1) (2 n + 1)$ or $\frac{1}{6} (2 n^{3} + 3 n^{2} + n)$ . Find the sum of squares of the first 10 natural numbers.

Q. Statement 1: The variance of first

n

even natural numbers is

\frac{n^{2} - 1}{4}

Statement 2: The sum of first

n

natural numbers is

\frac{n (n + 1)}{2}

and the sum of squares of first

n

natural numbers is

\frac{n (n + 1) (2 n + 1)}{6}

Prove that: 1 + 2 + 3 + ......... + n = $\frac{n (n + 1)}{2}$ i.e., the sum of the first n natural numbers is $\frac{n (n + 1)}{2}$ .

Join BYJU'S Learning Program

Find the mean of first n natural numbers.

The correct option is D n+12Mean, u=(1+2+3...+n)n⇒u=1n×12(n)(n+1)∴u=12(n+1)Ans-Option D

Find the mean of first $n$ natural numbers.

The correct option is D $\frac{n + 1}{2}$
Mean, $u = \frac{(1 + 2 + 3... + n)}{n}$
$\Rightarrow u = \frac{1}{n} \times \frac{1}{2} (n) (n + 1)$
$∴ u = \frac{1}{2} (n + 1)$
Ans-Option $D$