find sum of all three digit positive numbers which are divisible by 5 and leaves remainder 2​

find sum of all three digit positive numbers which are divisible by 5 and leaves remainder 2​

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  1. 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

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