Question

Find the sum of all natural numbers lying between 100 & 1000 which are multiples of 5.
Plz solve this.....

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  

please mark it as brainliest........

Answer 2

105,110,........995

As its in AP

let n be no of terms

So 995 = 105 + ( n-1)5

995 = 105 + 5n -5

995 = 100 +5n

895 = 5n

n= 179

So Sum = 179/2 ( 2× 105 + 178×5)

= 179/2 ( 210 + 890)

= 179/2 ( 1100)

= 179× 550

= 98450