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6;6;9;15;... are the first four terms of a quadratic pattern. Write down the value of the fifth term (T5) of the pattern.
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The sequence $6;6;9;15;\ldots$ has a constant second difference of 3 thus it is a quadratic sequence.

The first differences are $0,3,6,\dots$ respectively.

To get the general term of a quadratic sequence we need to use the formula

$T_n = an^2+bn+c$

where $a;b;c$ are obtained from the following simultaneous equations:

$2a=3$

$\therefore a=\dfrac{3}{2}$

$5a + b = 3$

$5(\frac{3}{2})+b=3$

$b=\dfrac{-9}{2}$

$4a+2b+c=6$

$4(\frac{3}{2})+2(\frac{-9}{2})+c=6$

$\therefore c=9$

Substituting $a,b,c$ in $T_n=an^2+bn+c$ gives:

$(\frac{3}{2})25+(\frac{-9}{2})5+9$

$=\dfrac{75-45+18}{2} = 24$

by Platinum (130,878 points)

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