The sum of 6 consecutive integers is 393. What is the third number In this sequence?

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Answer 1 · 2016-12-10T15:06:29

65

Let's define the first integer as $x$ $x$ . Then the next five consecutive integers would be:

$x + 1$ $x + 1$ , $x + 2$ $x + 2$ , $x + 3$ $x + 3$ , $x + 4$ $x + 4$ and $x + 5$ $x + 5$ .

The sum of these six integers is 393 so we can write:

$x + x + 1 + x + 2 + x + 3 + x + 4 + x + 5 = 393$ $x + x + 1 + x + 2 + x + 3 + x + 4 + x + 5 = 393$

$6 x + 1 + 2 + 3 + 4 + 5 = 393$ $6x + 1 + 2 + 3 + 4 + 5 = 393$

$6 x + 15 = 393$ $6x + 15 = 393$

$6 x + 15 - 15 = 393 - 15$ $6x + 15 - 15 = 393 - 15$

$6 x + 0 = 378$ $6x + 0 = 378$

$\frac{6 x}{6} = \frac{378}{6}$ $(6x)/6 = 378/6$

$x = 63$ $x = 63$

Because the first integer is 63 then the third would be $x + 2$ $x + 2$ or $63 + 2 = 65$ $63 + 2 = 65$