**1. Find the sum of even natural numbers 1 to 100.**Ans- There are 50 even numbers between 1 and 100. Now, according to formulae, n(n+1) = 50x51= 2550.

Similarly, you can find the sum of odd natural numbers between 1 and 100.

**2. Find the sum of even numbers from 41 to 100.**Ans- firstly you should find the sum of even numbers from 1 to 100 = 2550

Then, find the sum of even natural’s number from 1 to 40= 20x21= 420

Now, the sum of even numbers from 41 to 100= 2550-420= 2130

**3. Find the sum of
the square of natural numbers 1 to 20.**

Ans- according to question, find 1^{2}+2^{2}+3^{2}….
20^{2} = n(n+1)(2n+1)/6

Now put n=20,

=
20(21)(41)/6 = 2870

**4. Find the sum of
the square of natural numbers from 11 to 20.**

Ans- according to question, find 11^{2}+12^{2}+13^{2}…..
20^{2} = sum of the square from 1 to 20 – sum of the square from 1 to
10.

Now, the sum of the square from 1 to 20 =2870

And, the sum of the square from 1 to 10 = 385

Hence, the sum of square from 11 to 20 = 2870-385= 2485.

**5. Find the sum of
the square of natural numbers – 4**^{2} + 6^{2} + 8^{2} +
10^{2} + 12^{2} + 14^{2} + 16^{2} + 18^{2}
+ 20^{2}.

Ans- In these types of question, you can see *2*^{2}
is common so take out the 2^{2} and then add the remaining values.

Therefore, 2^{2} (2^{2} + 3^{2} + 4^{2}
+ 5^{2} + 6^{2} + 7^{2} + 8^{2} + 9^{2}
+ 10^{2}) = 4x384 = 1536.

**6. Find the sum of
digits from 1 to 100.**

Ans- In these questions, one has to count each and every
digit occurring from 1 to 100.

Now, from 1 to 100= ‘0’ occurs 11 times; ‘1’ occurs 21
times; and ‘2-9’ occurs 20 times each.

Hence, the sum of digits from 1 to 100 = (11x0) + (21x1) + (44x20)
= 901.

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