Step 1: Understanding the Question:
The question requires us to find the sum of the first 90 even natural numbers. The sequence of even natural numbers starts from 2, 4, 6, 8, and so on. This forms an arithmetic progression (A.P.) where the first term is 2 and the common difference is 2. Step 2: Key Formula or Approach:
The direct formula for the sum of the first $n$ even natural numbers is $S = n(n+1)$.
Alternatively, use the A.P. sum formula: $S_n = \frac{n}{2} \times (2a + (n-1)d)$. Step 3: Detailed Explanation:
The first even natural number is $a = 2$.
The common difference is $d = 2$.
The number of terms is $n = 90$.
The $n$-th term (last term) would be $L = 2n = 2 \times 90 = 180$.
Let's use the standard A.P. formula: $S_{90} = \frac{90}{2} \times (2 + 180)$.
$S_{90} = 45 \times 182$.
Calculating this: $45 \times 180 = 8100$. And $45 \times 2 = 90$. So, $8100 + 90 = 8190$.
Alternatively, let's use the direct formula for the sum of the first $n$ even natural numbers: $Sum = n(n+1)$.
Substitute $n = 90$ into the formula: $Sum = 90 \times (90 + 1)$.
$Sum = 90 \times 91$.
To compute $90 \times 91$, we can do $(90 \times 90) + 90 = 8100 + 90 = 8190$.
Step 4: Final Answer:
The sum of the first 90 even natural numbers is 8190.