The sum of first n natural numbers is | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(d) We know that natural numbers are $1,\;2,\;3,\;4,\;........n$ are in $A.P.$

Now sum $ = \frac{n}{2}[2 	imes 1 + (n - 1)1] = \frac{n}{2}(n + 1)$

Vedclass · Accepted Answer

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

Vedclass · Answer

(d) We know that natural numbers are $1,\;2,\;3,\;4,\;........n$ are in $A.P.$

Now sum $ = \frac{n}{2}[2 	imes 1 + (n - 1)1] = \frac{n}{2}(n + 1)$.

The sum of first $n$ natural numbers is

Similar Questions

The sum of all natural numbers between $1$ and $100$ which are multiples of $3$ is

A man deposited $Rs$ $10000$ in a bank at the rate of $5 \%$ simple interest annually. Find the amount in $15^{\text {th }}$ year since he deposited the amount and also calculate the total amount after $20$ years.

Let $S_{n}$ be the sum of the first $n$ terms of an arithmetic progression. If $S_{3 n}=3 S_{2 n}$, then the value of $\frac{S_{4 n}}{S_{2 n}}$ is:

If three numbers be in $G.P.$, then their logarithms will be in

Different $A.P.$'s are constructed with the first term $100$,the last term $199$,And integral common differences. The sum of the common differences of all such, $A.P$'s having at least $3$ terms and at most $33$ terms is.