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

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Quadratic patterns

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Best answer

Answer:

\(x=15\)

 

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 \(x=15\)

 

 

 

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