Series 1, 2, 3...

Solution 1: The sum of the first $n$ positive integers is given by $\frac{n(n+1)}{2}$ . Thus the sum is $\frac{(10)(11)}{2} = 55$ .

Solution 2: The sequence is an arithmetic progression. The sum of an arithmetic progression is given by $\frac{n(2a + (n-1)d)}{2}$ , where $a$ is the first term, $d$ is the common difference and $n$ is the number of terms. In this case $a = 1$ , $d=1$ and $n=10$ . Thus the sum is $\frac{n(2a + (n-1)d}{2} = \frac{10(2(1) + (10-1)(1)}{2} = 5(11) = 55$ .

Solution 3: Listing out the terms and adding them, we have $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55$ .