Question #1f547

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Question #1f547

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Answer 1 · 2018-01-13T02:36:16

Set the first number to some variable $n$ $n$ and write all the other $4$ $4$ numbers in terms of $n$ $n$ , equate the sum to $310$ $310$ , solve for $n$ $n$ , then solve for the third number in terms of $n$ $n$ to get $62$ $62$ .

Explanation:

Let's set the first even number to $n$ $n$ . Since we have $5$ $5$ consecutive numbers, the second should be $n + 2$ $n + 2$ , the third is $n + 4$ $n + 4$ , fourth is $n + 6$ $n + 6$ , and finally fifth is $n + 8$ $n + 8$ . We know that they add up to $310$ $310$ :

$n + (n + 2) + (n + 4) + (n + 6) + (n + 8) = 310$ $n + (n + 2) + (n + 4) + (n + 6) + (n + 8) = 310$

Let's simplify, adding up like terms (add up all the $n$ $n$ s, and add up all the numbers):

$5 n + 20 = 310$ $5n + 20 = 310$

Subtract each side by $20$ $20$ :

$5 n + 20 - 20 = 310 - 20$ $5n + 20 - 20 = 310 - 20$

$5 n = 290$ $5n = 290$

Now we can divide by $5$ $5$ :

$\frac{5 n}{5} = \frac{290}{5}$ $(5n)/5 = 290/5$

$n = 58$ $n = 58$

Since we know the first number, $n = 58$ $n = 58$ , and we know that the third number is $n + 4$ $n + 4$ , simply figure out what $n + 4$ $n + 4$ is:

$n + 4 = 58 + 4 = 62$ $n + 4 = 58 + 4 = 62$

Therefore the third number in the sequence is $62$ $62$ .

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Question #1f547

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