A convergent geometric series consisting of only positive terms has first term $a$, constant ratio $r$ and $n^{th}$ term, $T_n$, such that \( \sum_{m=3}^{\infty} T_m = \frac{1}{4} \) 3.1 If $T_1 + T_2 = 2$, write down an expression for $a$ in terms of $r$ - NSC Mathematics - Question 3 - 2017 - Paper 1

We know from the question that ( \sum_{m=3}^{\infty} T_m = \frac{1}{4} ).
The sum of the series starting from the third term is:

$S = T_3 + T_4 + ... = \frac{T_3}{1 - r},$
where $T_3 = ar^2$. This results in:

$S = \frac{ar^2}{1 - r} = \frac{1}{4}.$
Substituting our expression for $a$:

$\frac{\frac{2r^2}{1 + r}}{1 - r} = \frac{1}{4}.$
Cross-multiplying yields:

$8r^2 = 1 - r(1 + r)$
which simplifies to:

$9r^2 + r - 1 = 0.$
Using the quadratic formula:

$r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4(9)(-1)}}{2(9)} = \frac{-1 \pm \sqrt{37}}{18}.$
The quadratic roots will give potential values for $r$. Now substituting back into our equation for $a$ with $r$:

1. Calculate $r$, then use it to find $a$ as $a = \frac{2}{1 + r}$ where we check the validity of both values derived from $r$.