Given the geometric sequence: $$rac{-1}{4}; b; -1; \ldots$$ 2.1 Calculate the possible values of $b$ - NSC Mathematics - Question 2 - 2017 - Paper 1

Given $b = \frac{1}{2}$, the geometric sequence becomes:

$\frac{-1}{4}; \frac{1}{2}; -1; \ldots$

The common ratio $r$ can be calculated as:

$r = \frac{\frac{1}{2}}{\frac{-1}{4}} = -2.$

The formula for the $n^{th}$ term is given by:

$T_n = a r^{n-1},$

where $a = \frac{-1}{4}$ and $r = -2$. Therefore, for $n = 19$:

$T_{19} = \frac{-1}{4} \cdot (-2)^{18} = \frac{-1}{4} \cdot 262144 = -65536.$

The first 20 positive terms of the sequence are given by:

$\frac{-1}{4}, \frac{1}{2}, -1, \ldots$

The series can be expressed in sigma notation as:

$S = \sum_{n=0}^{19} \left(\frac{1}{2} \cdot 4^n\right)$

Here, the first term $a = \frac{1}{2}$ and the common ratio $r = 4$. Thus, the sum can be represented as:

$S = \frac{1/2(1 - 4^{20})}{1 - 4} = \frac{ \frac{1}{2} (1 - 4^{20})}{-3}.$

To determine whether a geometric series converges, we assess the common ratio $r$. The series will converge only if the absolute value of the common ratio is less than 1, i.e.,

$|r| < 1.$

In this case, the common ratio is $r = 4$, and since $|r| = 4 > 1$, the series does not converge. Thus, the answer is:

\nNo, the series is not convergent.

The reason is that the common ratio $r = 4$ is greater than 1.