Given that: \( \sum_{k=1}^{\infty} 4.3^{2-k} \) 2.1 Is the above geometric series is convergent or divergent. Justify ye 2.2 If further given that \( \sum_{k=p}^{\infty} 4.3^{2-k}=\frac{2}{9} \), determine the value of \( p \).

To determine if the series \( \sum_{k=1}^{\infty} 4.3^{2-k} \) is convergent or divergent, we can first recognize that this is a geometric series. The general form of a geometric series is \( \sum_{k=0}^{\infty} ar^k \), where \( |r| < 1 \) ensures convergence. In this case, we can rewrite the series as: \[ \sum_{k=1}^{\infty} 4.3^{2-k} = 4 \cdot 3^{2} \sum_{k=0}^{\infty} (3^{-1})^k = 12 \cdot \sum_{k=0}^{\infty} (1/3)^k \] Here, the common ratio \( r = \frac{1}{3} \), which is less than 1, so the series converges. Now for the second part, given that \( \sum_{k=p}^{\infty} 4.3^{2-k}=\frac{2}{9} \), we can find the value of \( p \). We first need to express the series starting from \( p \): \[ \sum_{k=p}^{\infty} 4 \cdot 3^{2-k} = 4 \cdot 3^{2} \sum_{k=0}^{\infty} (3^{-1})^{k+p-1} = 4 \cdot 9 \cdot (3^{-p+1}) \cdot \frac{1}{1 - \frac{1}{3}} \] The sum converges to: \[ \sum_{k=p}^{\infty} 4 \cdot 3^{2-k} = 4 \cdot 9 \cdot 3^{-p+1} \cdot \frac{3}{2} = 54 \cdot 3^{-p+1} \] Setting this equal to \( \frac{2}{9} \): \[ 54 \cdot 3^{-p+1} = \frac{2}{9} \] Solving for \( p \): \[ 3^{-p+1} = \frac{2/9}{54} = \frac{1}{243} \] Since \( 243 = 3^5 \), we have: \[ -p+1 = -5 \implies p = 6 \] Thus, the value of \( p \) is 6.