Determine x so that x, 1/2x+7, and 3x-1 are the first three terms of an arithmethic sequence?

Question

Determine x so that x, 1/2x+7, and 3x-1 are the first three terms of an arithmethic sequence?

1 Answer

Impact of this question

4286 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-01T20:59:56

Common difference between the terms must be equal, therefore

$t_{2} - t_{1} = t_{3} - t_{2}$ $t_2 -t_1 = t_3 - t_2$

$\frac{1}{2} x + 7 - x = 3 x - 1 - (\frac{1}{2} x + 7)$ $1/2x +7 -x = 3x-1-(1/2x +7)$

$- \frac{1}{2} x + 7 = 3 x - 1 - \frac{1}{2} x - 7$ $-1/2x + 7 = 3x - 1 - 1/2x - 7$

$15 = 3 x$ $15 =3x$

$x = 5$ $x=5$

We can confirm by plugging in $5$ $5$ for $x$ $x$ to see that the terms of the sequence are $5, \frac{19}{2}, 14$ $5, 19/2, 14$ which do indeed for an arithmetic sequence.

Hopefully this helps!

Science

Math

Humanities

... and beyond

Determine x so that x, 1/2x+7, and 3x-1 are the first three terms of an arithmethic sequence?

1 Answer

Related questions

Impact of this question