The sum of the first n natural numbers is: | vedclass

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Question

(C) The sequence of the first $n$ natural numbers is $1, 2, 3, \dots, n$.This is an Arithmetic Progression $(AP)$ where the first term $a = 1$ and the

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The sequence of the first $n$ natural numbers is $1, 2, 3, \dots, n$.This is an Arithmetic Progression $(AP)$ where the first term $a = 1$ and the common difference $d = 1$.The sum of the first $n$ terms of an $AP$ is given by the formula $S_n = \frac{n}{2}[2a + (n-1)d]$.Substituting the values $a = 1$ and $d = 1$ into the formula:$S_n = \frac{n}{2}[2(1) + (n-1)(1)]$$S_n = \frac{n}{2}[2 + n - 1]$$S_n = \frac{n(n+1)}{2}$.Thus, the sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$.

Column $A$	Column $B$
$(A_{1}) \quad 2, -2, -6, -10, \ldots$	$(B_{1}) \quad \frac{2}{3}$
$(A_{2}) \quad a = -18, n = 10, a_{n} = 0$	$(B_{2}) \quad -5$
$(A_{3}) \quad a = 0, a_{10} = 6$	$(B_{3}) \quad 4$
$(A_{4}) \quad a_{2} = 13, a_{4} = 3$	$(B_{4}) \quad -4$
	$(B_{5}) \quad 2$
	$(B_{6}) \quad \frac{1}{2}$
	$(B_{7}) \quad 5$

The sum of the first $n$ natural numbers is:

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