Home › Karnataka SSLC (Class 10) › Mathematics › Arithmetic Progressions › The sum of the first $n$ natural numbers $1 + 2 …
The sum of the first $n$ natural numbers $1 + 2 + 3 + \cdots + n$ equals
A$\dfrac{n(n-1)}{2}$
B$n^2$
C$\dfrac{n(n+1)}{2}$
D$\dfrac{n(n+1)(2n+1)}{6}$
Answer & Solution
Correct answer: C. $\dfrac{n(n+1)}{2}$
1. These form an AP with a = 1, d = 1, last term n.
2. $S_n = \tfrac{n}{2}(a + l) = \tfrac{n}{2}(1 + n)$.
3. So $1 + 2 + \cdots + n = \dfrac{n(n+1)}{2}$.
_Source: Karnataka SSLC (KSEEB) Class 10 Mathematics, Ch1 'Arithmetic Progressions' (sum, a=d=1)_
Related questions
For the AP $10,\ 7,\ 4,\ \ldots$, the term that first becomes negative is theThe sum $2 + 4 + 6 + \cdots + 100$ (the first 50 even numbers) isA man's starting salary is ₹8000 with an annual increment of ₹500. His salary in the 6th yIf the sum of the first $n$ terms of a sequence is $S_n = n^2$, its $n$th term $a_n$ isIn an AP, the 3rd term is 12 and the 7th term is 24. The first term $a$ and common differeThe common difference of an AP whose $n$th term is $a_n = 3n + 2$ isHow many terms are there in the finite AP $7,\ 13,\ 19,\ \ldots,\ 205$?If the first term of an AP is 6 and the common difference is $-3$, the AP is