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

sn  =  (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

sn  =  (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.

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