#MAASL00262

a) Use Geometric Sequence Formula to find the value of the term: $u_{n}=u_{1}r^{n-1}$

$u_{9}=24(\frac{1}{4})^{9-1}$

$u_{9}=24(\frac{1}{4})^{8}$

$u_{9}=3.662\times10^{-4}$

b) Use Sum of a geometric sequence formula: $S_{n}=\frac{u_{1}}{1-r}$

$S_{n}=\frac{24}{1-\frac{1}{4}}$

$S_{n}=32$

c) Use the arithmetic sequence formula to algebraically solve: $u_{n}=u_{1}+(n-1)d$

We are given the sixth term, so set $u_{6}=-2$

$-2=-12+(6-1)d$

$10=5d$

$d=2$

d) Consider the sum of an arithmetic series formula: $S_{n}=\frac{n}{2}(u_{1}+u_{n})$

However, we have no expression for $u_{n}$

First, derive the full expression for $u_{n}$

. $u_{n}=u_{1}+(n-1)d$

$u_{n}=-12+(n-1)2$

Simplify as much as possible $u_{n}=-12+2n-2$

$u_{n}=2n-14$

Substitute back into sum of an arithmetic series formula: $S_{n}=\frac{n}{2}(-12+2n-14)$

$S_{n}=\frac{n}{2}(-26+2n)$

Factor to simplify and have the ability to cancel 2 $S_{n}=\frac{n}{2}2(-13+n)$

$S_{n}=n(-13+n)$

$S_{n}=-13n+n^{2}$

e) Sum of infinite geometric sequence from part b: $32$

Use the expression for the sum of the arithmetic formula in part d $32=2(-13n+n^{2})$

$16=-13n+n^{2}$

$0=n^{2}-13n-16$

Use the quadratic formula: $n=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

Set $a=1$

; $b=-13$

; $c=-16$

$n=\frac{13\pm\sqrt{(-13)^{2}-4(1)(-16)}}{2(1)}$

$n=\frac{13\pm\sqrt{169+64}}{2}$

$n=\frac{13\pm\sqrt{233}}{2}$

$n= 14.132$

; $n=-1.132$

Ignore the negative $n=14.132$

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