#### Answer

3660

#### Work Step by Step

Method 1- -
Sum of first 60 positive even integers = 2 + 4 + 6 + 8 + 10 . . . . .+ 120
First term $a_{1}$ = 2
Common difference d = 2
$60^{th}$ term = 120
Sum of n terms = $\frac{n}{2}$($a_{1}$ + $a_{n}$)
Sum of first 60 positive even integers = $\frac{60}{2}$(2 + 120)
= 30$\times$122 = 3660
Method 2- -
2 + 4 + 6 + 8 + 10 . . . . .+ 120 = 2(1 + 2 + 3 + 4 + 5 +......+60)
First find sum of (1 + 2 + 3 + 4 + 5 +......+60) and then multiply the sum by 2
First term of series $a_{1}$ = 1
Common difference in series d = 1
$60^{th}$ term = 60
Sum of n terms = $\frac{n}{2}$($a_{1}$ + $a_{n}$)
Sum of first 60 natural numbers= $\frac{60}{2}$(1 + 60)
= 30$\times$61 = 1830
2 + 4 + 6 + 8 + 10 . . . . .+ 120 = 2(1 + 2 + 3 + 4 + 5 +......+60) = 2 $\times$ 1830 = 3660