Question

find the sum of natural numbers between 101 and 999 which are divisible by both 2 and 5

Answer 1

Hello friend......

Question :- Find the sum of natural nos. between 101 and 999 which are divisible by both 2 and 5.

Answer :- x = 110 ( first number to be divided by 2 and 5 )

difference, d= 10

y = 990 (last no to be divided by 2 and 5)

xy=(x+y-1)d
990 = 110 + 10x -10
880 + 10 = 10x
890 = 10x
x= 890/10
x= 89

number of natural numbers between 101 and 999 is 89.

Answer 2

Answer:

No of terms(n) = 89

Sum (S) = 48950

Explanation:

By Arithmetic Progression,

First term that is divisible by 2 and 5 both is 110.

So a = 110

110,120,130,140..........990

Last term T = 990

So for no .of terms use :

a + (n -1)d where d is = 10

T = 110 + (n - 1)10

990 - 110 = 10n - 10

880 + 10 = 10n

10n = 890

n = 89

Now for Sum use:

n/2(2a + (n -1) d)

89/2(2(110) + (88)10)

This calculation will give :

S = 48950

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