# arrow_back Find x using quadratic 7;10;x;22;31

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$x=15$

Explanation:

Given that the sequence is quadratic then it implies that the 2nd order difference is constant.

The sequence is:

$7;10;x;22;31$

First Difference i.e. $a_{n+1}-a_n$

Let:

$a=10-7=3$

$b=x-10$

$c=22-x$

$d=31-22=9$

Second Difference:

Since the sequence is quadratic, let the constant difference be $y$

$b-3=y$

$c-b=y$

$d-c=9-c=y$

Using the above information (simultaneous equations), we can solve for $y$.

Making $c$ subject of formula in $9-c=y$ gives $c=9+y$.

Substituting for $c$ in $c-b=y$ gives $b=9-2y$

Substituting for $b$ in $b-3=y$ gives $y=2$.

Now that we know that $y=2$ we can deduce the value of $x$ by rewriting the first differences that include an $x$

First Differences

$x-10 =5$ therefore $x=15$

$22-x = 7$ therefore $x=15$

by Gold Status
(30,543 points)

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