Answer:
The statements are incorrect as: The sum of even numbers from 1 to 100(i.e. 2550) is not double\twice of the sum of odd numbers from 1 to 100(i.e. 2500).
Step-by-step explanation:
We know that sum of an Arithmetic Progression(A.P.) is given by:
![S_{n}=\frac{n}{2}\times (a+a_{n})](https://tex.z-dn.net/?f=S_%7Bn%7D%3D%5Cfrac%7Bn%7D%7B2%7D%5Ctimes%20%28a%2Ba_%7Bn%7D%29)
where 'n' denotes the "number" of digits whose sum is to be determined, 'a' denotes the first digit of the series and '
' denote last digit of the series.
Now the sum of even numbers i.e. 2+4+6+8+....+100 is given by the use of sum of the arithmetic progression since the series is an A.P. with a common difference of 2.
![=\frac{50}{2}\times (2+100)](https://tex.z-dn.net/?f=%3D%5Cfrac%7B50%7D%7B2%7D%5Ctimes%20%282%2B100%29)
![=25\times 102\\=2550](https://tex.z-dn.net/?f=%3D25%5Ctimes%20102%5C%5C%3D2550)
Hence, sum of even numbers from 1 to 100 is 2550.
Also the series of odd numbers is an A.P. with a common difference of 2.
sum of odd numbers from 1 to 100 is given by: 1+3+5+....+99
.
Hence, the sum of all the odd numbers from 1 to 100 is 2500.
Clearly the sum of even numbers from 1 to 100(i.e. 2550) is not double of the sum of odd numbers from 1 to 100(i.e. 2500).
Hence the statement is incorrect.