Answer
The sum of first 100 even integer is 10100.
Work Step by Step
The provided Expression is:\[2+4+6+\ldots +200\]
Here, \[{{a}_{1}}=2,{{a}_{2}}=4,{{a}_{3}}=6\ \text{and}\ {{a}_{4}}=8\]
As,
\[\begin{align}
& {{a}_{2}}-{{a}_{1}}=4-2 \\
& =2 \\
& {{a}_{3}}-{{a}_{2}}=6-4 \\
& =2 \\
& {{a}_{4}}-{{a}_{3}}=8-6 \\
& =2
\end{align}\]
This implies common difference‘d’ between two consecutive terms is: 2
Therefore,
The provided sequence is arithmetic.
Here \[{{a}_{1}}\] and \[{{a}_{n}}\]is first and the nth term of the provided A.P and n is natural number.
To find the sum of the first ten terms, 100th and 1st term of the sequence are needed.
To find 100th term of the provided sequence formula to be used is:\[{{a}_{n}}={{a}_{1}}+\left( n-1 \right)d\]
Put \[n=100,{{a}_{1}}=1,d=1\] in the above formula, the 100th term of the sequence is:
\[\begin{align}
& {{a}_{100}}=2+\left( 100-1 \right)2 \\
& =2+198 \\
& =200
\end{align}\]
Therefore,
The required sum of the provided sequence is:
\[\begin{align}
& {{s}_{10}}=\frac{100}{2}\left( {{a}_{1}}+{{a}_{100}} \right) \\
& =\frac{100}{2}\left( 2+200 \right) \\
& =50\left( 202 \right) \\
& =10100
\end{align}\]
