Question

Find the sum of the first 50 even positive integers.

Answer 1

Hint: First, we have to find the 50 even positive numbers which can be easily done by dividing the number by 2 and getting remainder 0 is the even number. Then, after getting 50 even numbers we can use the formula of summation of numbers which is calculated by \[{{S}_{n}}=\dfrac{n}{2}\left( a+{{T}_{n}} \right)\] where ‘a’ is 1st term, n is total number of numbers and \[{{T}_{n}}\] is the last term.

Complete step-by-step solution:
Here, we are asked to find 50 even positive integers. So, we can write it as 2, 4, 6, 8, ……..100.
Now, we have a total 50 even numbers. Here, the first term i.e. a is 2. The difference between 2 consecutive numbers is \[d=4-2=2\] . Here, last term i.e. \[{{T}_{n}}={{T}_{50}}=100\] .
So, now we have all the data needed for summation of 50 even integers.
Using the formula of summation, we get
\[{{S}_{n}}=\dfrac{n}{2}\left( a+{{T}_{n}} \right)\]
\[{{S}_{50}}=\dfrac{50}{2}\left( 2+100 \right)\]
\[{{S}_{50}}=\dfrac{50}{2}\left( 102 \right)\]
\[{{S}_{50}}=50\times 51\]
\[{{S}_{50}}=2550\]
Thus, summation of the first 50 even positive integers is 2550.

Note: Be careful while reading what is asked in question. Sometimes, chances of mistakes are not reading even integers and simply calculating 1st 50 positive numbers which results in wrong answers. Also, be careful while using summation formula instead of using ${{n}^{th}}$ term finding formula which is \[{{T}_{n}}=a+\left( n-1 \right)d\] where is the common difference between two consecutive numbers.