APPLICATION QUESTION 2 Given that: \( \sum_{k=1}^{\infty} 4.3^{2-k} \) 2.1 Is the above geometric series is convergent or divergent. Justify your answer 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, you can identify the common ratio. Rewriting the series gives us \( \sum_{k=1}^{\infty} 4 \cdot (3^{2}) \cdot (3^{-k}) = 4.3^{2} \sum_{k=1}^{\infty} (3^{-1})^{k} \). The common ratio of \( r = \frac{1}{3} \) (less than 1) indicates that the series converges. To find \( p \) from \( \sum_{k=p}^{\infty} 4.3^{2-k}=\frac{2}{9} \), first express the series starting from \( p \): \( \sum_{k=p}^{\infty} 4 \cdot 3^{2} (3^{-1})^{k} = 4.3^{2} \sum_{n=0}^{\infty} (3^{-1})^{n+p-1} \). This expression simplifies to \( \frac{4 \cdot 9}{2} \cdot (3^{-p+1}) \), leading to a nifty equation you can solve for \( p \) by equating with \( \frac{2}{9} \). After some calculations, you’ll find \( p = 4 \).