Question

how many 3 digit numbers are there which leaves a remainder 2 on division by 7.find the sum of all numbers?​

Answer 1

Answer:

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so now the two digit number when divided by 7 leaves remainder as 2 begins from 100,107,114,so on up to 996

thus so as all these numbers are divided by same number and moreover leaves same remainder ie 2 we would get common difference between each of those numbers as constant

so as common difference between each of these numbers is constant

these numbers form a sequence of ap ie of arithmetic progression

so thus we need to find=n

ie total such numbers

so now here for above ap

a=t1=100 d=7 tn=996

so now applying formula

tn=a+(n-1)d

we get

996=100+[(n-1)×7]

ie 896=7n-7

ie 7n=903

so n=129

thus there are in total 129 numbers which on dividing by 7 leaves remainder as 2