Question

Sum of all natural numbers less than 500 and they are multiple of 7 also

Answer 1

Answer:

Okay, let a ( First Term) be 7 ( as it is the smallest number divisible by 7) and d ( common difference) be 7 and l ( last term ) be 497 ( as it is the highest number below 500 to be divisible by 7)

As we need to find the sum of series 7,14,21,28……..497( which is in arithmetic progression) we need to find the number of terms. So let no.of terms be n. Then the equation goes:

a+(n−1)d=497

7+(n−1)7=497

7+7n−7=497

7n=497

n=497/7

n=71

Let Sum of series mentioned above be S. Now,

S=n/2(a+l)

S=71/2(7+497)

S=71/2(504)

S=71(252)

S=17892

So the sum is 17892 (numbers divisible by 7)

But some digits divisible by 7 are also divisible by 11 in an arithmetic progression of 77, 144, 221, 298…… 462 and it it's necessary to find the number of numbers in this series. So the equation for this goes:

77+(n−1)77=462

77+77n−77462

77n=462

n=462/77

n=6

As the number of terms in the series 77,144,221……462 is 7, we can find it's sum. The equation for this goes:

6/2(77+462)

3(539)

1617

So the sum of the latter series has to be deducted from the former series. So,

THE REQUIRED SUM: 17892–1617 = 16275