 Hello friends and welcome again to another problem-solving session on sequence and series We have taken a question which says find the sum of all three digit natural numbers, which are multiple of seven Interesting question, right? So we have to just find out What are three digit? Multiples of seven are you have to find out the first term the last term if you know Then we will be able to find out the Sum of the or the required sum So find the sum of all three digit it says right natural number, which are multiple of seven So let's say our sum is SN so this is what we need to find out a1 a2 a3 and An right what are the criteria on a1 a2 a3 an all our three digit number as well as multiple of seven So we have to find out a1. What is a1? three digit three digit Three digit right multiple of seven First that is it is first three digit multiple of seven Instead of a we should write first three digit multiple of seven and What is that if you see first three digit first three digits start from hundred so hundred hundred one two three Four five if you see hundred not five one zero five it is fifteen times seven So hence one not five is first Three digit multiple of seven. What will be a n guys? a n why do we need a n by the way? Can we not do away with it? We will not be able to because for the sum formula if you see SN requires what? n by 2 twice a Plus n minus 1d so in both the in in in two in finding SN we do require n We do require n Correct, so n is important and for that we need to know the a n right so what is last? Seventh or three digit number which is divisible by seven so nine nine nine clearly not so if you see If you divide this by seven you will get Seven one the seven twenty nine so four and Two and remainder is five so seven into one forty two plus five is nine nine nine Right, that means if I remove five from so this is right nine nine nine is equal to Seven into one forty two plus five Okay, so hence what will be So if you see nine nine nine Minus five is seven times one four two which is nine nine four correct There's seven times one four two, so nine nine four appears to be the last three digit seven multiple of seven Isn't it because after this if the next seven next multiple of seven is one 001 Right, this is the next Multiple of seven, but this is greater than thousand. This is four digit number So hence we have to we can clearly say that highest is nine nine four so a n is nine nine four Right, so the last term is known now known so how to find out n easy it is so a n is nothing but a first term that is one not five Plus n minus one so whatever is n minus one and common difference now if you Check the series. What will the series be like? So let me write few terms of the series multiples of seven so one not five one one two one one nine One two six and so on and so forth till nine nine four clearly common difference D is seven correct D is seven Now and a n is nine nine four so nine nine four is equal to one not five plus And minus one times seven So nine nine four minus one not five plus or rather Yeah, plus seven is equal to Seven in Am I right so hence if you now calculate you will get how much this is One so basically one not five right so eight eight nine Eight eight nine is nine nine four minus one plus seven is seven and so eight ninety six Divide by seven is n Right, so n is equal to one two eight 128 Yeah, so n is 128 so we could find out and so what to do now s n How much nothing but n by two either you adopt a plus two a plus n minus one D or I will write first term Plus last term correct So that means this is one twenty eight by two Into one twenty eight by two first term was one not five and the last term is nine nine four nine nine four so that means this is four sixty four Into one not five plus nine nine four is is nine nine ten one zero nine nine Correct one zero nine nine so if the product will be this product will come out to be seven zero three three six Sixty four into one one nine is one zero nine nine is Seven zero three three six Correct, so this is what is the answer