products view our website. See reverse forelonulusite. Girade 12 March Control I ent 2025 Q(1s110) arsen the series \( 16+12+8+\ldots \) 1.1. Write down the next two terms of the series. 11.2 Determine the sum of the first 61 terms of the series. Given the sigma below: \[ \sum_{n=k}^{\infty} 4\left(\frac{1}{3^{n}}\right)=6 \] 1.2.1 Prove that the series above converges. 1.2.2 Calculate the value of \( k \) [12]

The series \( 16 + 12 + 8 + \ldots \) follows a clear pattern where each term decreases by 4. So, the next two terms would be 4 and 0. To find the sum of the first 61 terms, we recognize this as a finite arithmetic series where the first term is 16 and the common difference is -4. The sum can be found using the formula \( S_n = \frac{n}{2} \times (2a + (n-1)d) \) where \( n = 61 \), \( a = 16 \), and \( d = -4 \). Moving on to the given sigma notation, we see that it represents an infinite geometric series. The sum of such a series converges if the common ratio, \( r \), satisfies \( |r| < 1 \). In this case, it’s \( \frac{1}{3} \), which indeed does satisfy that condition! Thus, we can confidently say that the series converges. To find \( k \), we can manipulate the equation given by the sum formula for geometric series \( S = \frac{a}{1-r} \). Where \( a \) is the first term of the series which begins from \( k \), we can express the sum as \( 4 \left(\frac{1}{3^k}\right) \frac{1}{1 - \frac{1}{3}} = 6 \) leading us to deduce that \( k = 1 \) through simple algebra. Happy calculating!