find the middle term of the sequence formed bya all three digit number which leave a remainder 3 and when divided by 4.also find sum of all numbers on both sides of the middle term seperately..
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The English grammar and sentence of this question is not clear. So it takes a long time to understand the question.
Clue: All Three digit numbers divided by 4 gives 3 as REMAINDER
So the numbers are somewhere between 100 till 999 (all are three digits)
100 is exactly divisible by 4, 101 leaves remainder 1, 102 leaves remainder 2, 103 leaves remainder 3, 104 is exactly divisible by 4, 105 leaves remainder 1, 106 leaves remainder 2, 107 leaves remainder 3, 108 is exactly divisible by 4 and so on.........
From the other end 999 leaves remainder 3 when divided by 4. So that is the last term
So the sequence starts from 103,107,111..........999
999 is the last term
To find out total number of terms in the progression use formula 999= a + (n-1) d
So 999=103+(n-1)X4
Therefore n=225
There are 112 numbers before and 112 numbers and 113th number is = 103+(113-1)X4=551
Sum of number before 551
There are 112 numbers starting with 103
Answer is 36400
Clue: All Three digit numbers divided by 4 gives 3 as REMAINDER
So the numbers are somewhere between 100 till 999 (all are three digits)
100 is exactly divisible by 4, 101 leaves remainder 1, 102 leaves remainder 2, 103 leaves remainder 3, 104 is exactly divisible by 4, 105 leaves remainder 1, 106 leaves remainder 2, 107 leaves remainder 3, 108 is exactly divisible by 4 and so on.........
From the other end 999 leaves remainder 3 when divided by 4. So that is the last term
So the sequence starts from 103,107,111..........999
999 is the last term
To find out total number of terms in the progression use formula 999= a + (n-1) d
So 999=103+(n-1)X4
Therefore n=225
There are 112 numbers before and 112 numbers and 113th number is = 103+(113-1)X4=551
Sum of number before 551
There are 112 numbers starting with 103
Answer is 36400
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