find sum of all three digit positive numbers which are divisible by 5 and leaves remainder 2
Answers
Step-by-step explanation:
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