Question

Find the sum of all the 3 digit natural no/s which on division by 7 leaves a remainder 3

Answer 1

smallest 3 digit number which leave the remainder 3 when divided by 5 =103

largest 3 digit number which leave the remainder 3 when divided by 5 =998

3 digit number after 103 which leave the remainder 3 when divided by 5 =108

an ap is formed: 103,108........998

first term=103

common difference=5

let the total no. of three digit numbers which leave the remainder 3 when divided by 5=n

998=103+(n-1)5

180=n

we know that the formula for sum in AP = no of terms*(first term + last term)/2

sum of ap=180*(103+998)/2 = 90*1101

=99090

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Answer 2

smallest 3 digit number which leave the remainder 3 when divided by 5 =103

largest 3 digit number which leave the remainder 3 when divided by 5 =998

3 digit number after 103 which leave the remainder 3 when divided by 5 =108

an ap is formed: 103,108........998

first term=103

common difference=5

let the total no. of three digit numbers which leave the remainder 3 when divided by 5=n

998=103+(n-1)5

180=n

we know that the formula for sum in AP = no of terms*(first term + last term)/2

sum of ap=180*(103+998)/2 = 90*1101

=99090