Question

find the sum of all the three digit number which can leave the remainder 3 when divided by 7​

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
sum of ap=180/2 *(103+998)=90*1101
=99090

Numbers can be represented by 5x+3 103+108+......998 >>>>> 180 terms Sum = 103+108+......998 = 100+3+105+3+.....995+3 = 5*(20+21+...199)+3*180 = 5*19710 +3*180 = 99090

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Answer 2

a = 101

l = 997

997 = 101+(n-1)7

896 = (n-1)7

n-1 = 128

n = 129

sum = 129/2(101+997)

= 129/2(1018)

= 129(509)

= 65661

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