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A convergent geometric series consisting of only positive terms has first term $a$, constant ratio $r$ and $n^{\text {th }}$ term, $T_{n}$, such that $\sum_{n=3}^{\infty} \mathrm{T}_{n}=\frac{1}{4}$.

1. If $\mathrm{T}_{1}+\mathrm{T}_{2}=2$, write down an expression for $a$ in terms of $r$.
2. Calculate the values of $a$ and $r$.
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1.

\begin{array}{l}
a+a r=2\\
\begin{aligned}
a(1+r) &=2 \\
a &=\frac{2}{1+r}
\end{aligned}\\
\text { OR/OF }\\
\frac{a}{1-r}-2=\frac{1}{4}\\
\begin{aligned}
4 a-8(1-r) &=1-r \\
4 a-8+8 r &=1-r \\
4 a &=9-9 r \\
a &=\frac{9-9 r}{4}
\end{aligned}
\end{array}

OR

\begin{aligned}
S_{n} &=\frac{a\left(r^{n}-1\right)}{r-1} \\
2 &=\frac{a\left(r^{2}-1\right)}{r-1} \\
2 &=\frac{a(r-1)(r+1)}{r-1} \\
2 &=a(r+1) \\
a &=\frac{2}{r+1}
\end{aligned}

OR

\begin{aligned}
\frac{a r^{2}}{1-r} &=\frac{1}{4} \\
a &=\frac{1-r}{4 r^{2}}
\end{aligned}

2.

\begin{aligned}
S_{\infty} &=T_{1}+T_{2}+\sum_{n=3}^{\infty} \mathrm{T}_{n} \\
S_{\infty} &=2+\frac{1}{4} \\
\frac{a}{1-r} &=2+\frac{1}{4} \\
\left(\frac{2}{1+r}\right) \times\left(\frac{1}{1-r}\right) &=\frac{9}{4} \\
\frac{2}{1-r^{2}} &=\frac{9}{4} \\
8 &=9-9 r^{2} \\
9 r^{2} &=1 \\
r &=\frac{1}{3} \\
a &=\frac{3}{2}
\end{aligned}
by Diamond (88,180 points)

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