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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