Find sum of all three digit positive numbers which are divisible by 5 and
leaves the remainder 2
Answers
To solve this we will need to use two formulas.
The sum of the A.P and the formula to find the last term in the AP to find the value of n ( total number of integers in the AP)
The smallest 3 digit number that will leave a remainder of 2 when divided by 3 is 101 denoted by - T1
The smallest 3 digit number that will leave a remainder of 2 when divided by 3 is 998 denoted by - Tn
The difference between the values is 3 as 101, 104, 107 satisfy the condition ( 3n+2) denoted by - d
Now let us calculate the total number of integers that satisfy the condition.
Tn=T1+(n−1)d
998 = 101 + (n - 1) * 3
Therefore, n = 300
Now to find the sum of AP, in this GMAT Progression
Sum of an Arithmetic Progression (AP) = [first term + last term2] * n
where 'n' is the number of terms in the sequence, which we calculated to be 300
Substituting the values, we get,
Sum of the AP is [101+998]∗300 = 164,850 - Answer
Hope this helps!
Answer refer to the attachment....
