Question

Find the sum of all three digit numbers which leave remainder 2 when divided by 5
(A) 98910
(B) 9820
(C) 9830
(D) 9840​

Answer 1

Answer:

Step-by-step explanation:

a (first term)= 102

d (Common difference)= 5

I(last term ) = 997

l= an = a + (n – 1) d

997= 102 + (n – 1) × 5

5 (n – 1) = 997 – 102 = 895

n-1= 885/5

(n – 1) = 179

n = 179 +1

n = 180

Sn = n/2 [a+l]

Sn= 180/2 [ 102+ 997]

Sn= 90 × 1099

Sn= 98910

Answer 2

Three digit numbers which left leave reminder 2 when divided by 5 are →

102,107,112,..........,997

This is a sequence of AP( Airthmatic Progression)

Here,

let , first term is a = 102

common difference(d)= 107-102= 5

(L)last term is 997

We know that,

Last term (L) = a + (n -1) d

997 = 102 + (n -1)× 5

997 - 102= (n -1)×5

(n -1)= 895 /5

(n -1)= 179

n = 179+1

n= 180

This sequence have 180 terms .

Hence,

Sum of these 180 terms is →

Sn = n/2 { first term(a) + last term(L)}

Sn = 180/2{ 102+997}

Sn= 90 { 1099}

Sn= 98910

Therefore,

Sun of all three digit numbers which is divisible by 5 and leave reminder 2 is

98910

This is the required solution.