Question

Find the sum of all natural numbers lying between 100-1000 which are multiple of 5.

Answer 1

a¹ (First term) = 100 + 5
                       = 105
aⁿ ( last term ) = 1000 - 5
                       = 995 
d ( Common
      difference) = 5
n ( N0: of terms ) = [(aⁿ - a¹) / d] + 1
                            = (995 -105) /5 ] + 1
                            = (890/5) + 1
                            = 178 + 1
                            = 179//
               ∴  NO: OF TERMS (n) = 179

Xⁿ -------> Sum of all terms 
 
Xⁿ =X / 2 [ a¹ + aⁿ ]
     = 179/2 [ 105 + 995 ]
     = 179/2 ( 1100) 
     = 179 × 550
     = 98450 //
 
         ∴ SUM OF ALL NATURAL NUMBERS ( Xⁿ ) =  98450  

Answer 2

The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995.

This sequence forms an A.P.

Here, first term, a = 105

Common difference, d = 5

Here,

a+(n−1)d = 995

=> 105+(n−1)5 = 995

=> (n−1)5 = 995−105 = 890

=> n−1 = 178

=> n = 179

Sn = n/2[2a+(n−1)d]

∴ Sn = 179/2[2×(105)+(179−1)×(5)]

= 179/2[2(105)+(178)(5)]

= 179[105+(89)5]

= (179)[105+445]

= 179×550

= 98450