Question

Find the sum of all natural numbers lying between 100 and 1000, which are
multiples of 5. need fully solved answer​

Answer 1

Answer:

98450

Step-by-step explanation:

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