FIND SUM OF NATURAL NUMBERS BETWEEN TWO NUMBERS

Question 1 :

Find the sum of all natural numbers between 300 and 500 which are divisible by 11.

Solution :

To find the first number greater than 300 and divisible by 11, we use the following shortcut.

308 + 319 + 330 +........

308 + 319 + 330 +........+495

s_n  =  (n/2)[a+l]

n  =  [(l-a)/d]+1

n  =  [(495-308)/11]+1

n  =  (187/11)+1

n  =  17+1

n  =  18

s₁₈  =  (18/2)[308+495]

s₁₈  =  9(803)

s₁₈  =  7227

Question 2 :

Find the  sum of all natural numbers between 100  and 200 which are not divisible by 5.

Solution :

Sum of natural numbers between 100 and 200 which are not divisible by 5.

  =  Sum of natural number from 100 to 200 - Sum of natural numbers which is divisible by 5 between 100 to 200.

=  (100+101+...........+200)-(100+105+110+....+200)

Number of terms in the series 

100+101+...........+200

n  =  [(200-100)/1] + 1

n  =  101

Number of terms in the series 

100+105+110+....+200

n  =  [(200-100)/5] + 1

n  =  21

Sum of series 100+101+...........+200

=  (101/2) (100+200) 

=  101(150)

=  15150

Sum of series 100+105+110+....+200

=  (21/2) (100+200) 

=  21(150)

=  3150

The required sum is  =  15150 - 3150

=  12000

So, the sum of natural numbers 12000.

Question 3 :

Solve 1 + 6 + 11 + 16 + ..........  + x = 148

Solution :

1 + 6 + 11 + 16 + ..........  + x = 148

n  =  [(l-a)/d] + 1

n  =  [(x-1)/5] + 1

n  =  (x+4)/5

s_n  =  (n/2)[a+l]

148  =  ((x+4)/10)(1+x)

1480  =  x²+ 5x+4

x²+ 5x+4-1480  =  0

x²+ 5x-1476  =  0

(x+41)(x-36)  =  0

x  =  36

So, the last term is 36.

Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.