'n Konvergente meetkundige reeks wat slegs uit positiewe terme bestaan, het eerste term $a$, constante verhouding $r$ en $n$'de term, $T_n$, sodat \[ \sum_{n=3}^{\infty} T_n = \frac{1}{4} \] 3.1 Indien $T_1 + T_2 = 2$, skryf 'n uitdrukking vir $a$ in terme van $r$ neer - NSC Mathematics - Question 3 - 2017 - Paper 1

Using the sum of the infinite series, we know:

$$S = \frac{T_1}{1 - r} = \frac{a}{1 - r}$$

Since we have already established that:

$$S = T_1 + T_2 + \sum_{n=3}^{\infty} T_n = T_1 + T_2 + \frac{T_3}{1 - r}$$

Thus:

$$S = 2 + \frac{T_3}{1 - r}$$

From the problem statement:

$$S = \frac{1}{4}$$

So equating the two:

$$2 + \frac{T_3}{1 - r} = \frac{1}{4}$$

We know that:

$$T_3 = ar^2 = \frac{2r^2}{1+r}$$

Substituting into the equation gives:

$$2 + \frac{\frac{2r^2}{1+r}}{1 - r} = \frac{1}{4}$$

Multiply through by $(1 - r)(1 + r)$ to eliminate the fractions and solve for $r$:

1. Rearranging gives us:
   • The expression simplifies to give:
   • After solving, we find two values: $r = \frac{1}{3}$ or $r = 1$. Typically, we consider $r < 1$ for convergence, so:
   • $r = \frac{1}{3}$.
2. Substitute $r$ back into the expression for $a$:

a = \frac{2}{1 + \frac{1}{3}} = \frac{2}{\frac{4}{3}} = \frac{3}{2}$$ Thus, the solutions are: - $a = \frac{3}{2}$ - $r = \frac{1}{3}$