Question

The sum of
6
66 consecutive integers is
519
519519.
What is the third number in this sequence?

Answer 1

Step-by-step explanation:

Let, the first number is x

so,

1st number = x

2nd number = x+1

3rd number = x+2

4th number = x+3

.................................

.................................

666th number = x+665

now,

sum of the numbers

=x + (x+1) + (x+2) + (x+3)+........+(x+665)

=x + x+1 + x+2 + x+3 +......+ x+665

=666x + (1+2+3+.....+665)

$= 666x + \frac{665(665 - 1)}{2} \\ = 666x + \frac{665 \times 664}{2} \\ = 666x + 220780$

according to the condition,

$666x + 220780 = 519519519 \\ > 666x = 519519519 - 220780 \\ > 666x = 519298739 \\ > x = \frac{519298739}{666} = 779729$

so,

3rd number =x+2=779729+2=779731