Round off your answers to two decimal places where necessary. Determine the sum of the first 12 terms of the series \( 36+24+16+\ldots \) Determine the sum of the first 10 terms of the series \( 125+100+80+64+ \) Determine the sum of the series \( 54+18+6+\ldots+\frac{2}{81} \). 4 Determine the sum of the series \( 3125+625+125+25+\ldots+\frac{1}{625} \). Determine the fifth term of the series if \( \mathrm{S}_{n}=\frac{243}{2}\left(1-\left(\frac{1}{3}\right)^{n}\right) \).

To find the sums and terms of the specified series, we’ll evaluate each one step-by-step. 1. The series \( 36 + 24 + 16 + \ldots \) has a first term \( a = 36 \) and a common ratio \( r = \frac{24}{36} = \frac{2}{3} \). The sum of the first \( n \) terms of a geometric series is given by the formula: \[ S_n = a \frac{1 - r^n}{1 - r} \] For the first 12 terms: \[ S_{12} = 36 \cdot \frac{1 - \left(\frac{2}{3}\right)^{12}}{1 - \frac{2}{3}} = 36 \cdot \frac{1 - \left(\frac{2}{3}\right)^{12}}{\frac{1}{3}} = 108 \left(1 - \left(\frac{2}{3}\right)^{12}\right) \approx 108 \cdot (1 - 0.00752) \approx 108 \cdot 0.99248 \approx 107.53 \] 2. The second series \( 125 + 100 + 80 + 64 + \ldots \) has a first term \( a = 125 \) and a common ratio \( r = \frac{100}{125} = 0.8 \). For the first 10 terms: \[ S_{10} = 125 \cdot \frac{1 - 0.8^{10}}{1 - 0.8} = 125 \cdot \frac{1 - 0.1073741824}{0.2} = 125 \cdot \frac{0.8926258176}{0.2} = 125 \cdot 4.463129088 \approx 557.89 \] 3. The series \( 54 + 18 + 6 + \ldots + \frac{2}{81} \) has a first term \( a = 54 \) and a common ratio \( r = \frac{18}{54} = \frac{1}{3} \). To find how many terms are there, we set \( a_n = \frac{2}{81} \): \[ a_n = 54 \left(\frac{1}{3}\right)^{n-1} = \frac{2}{81} \] Solving for \( n \): \[ \left(\frac{1}{3}\right)^{n-1} = \frac{2}{81 \cdot 54} \implies \left(\frac{1}{3}\right)^{n-1} = \frac{2}{4374} = \frac{1}{2187} \implies n-1 = 7 \implies n = 8 \] For the sum of 8 terms: \[ S_8 = 54 \cdot \frac{1 - \left(\frac{1}{3}\right)^8}{1 - \frac{1}{3}} = 54 \cdot \frac{1 - \frac{1}{6561}}{\frac{2}{3}} = 81 \cdot \left(1 - 0.0001524158\right) \approx 81 \cdot 0.9998475842 \approx 80.96 \] 4. For the series \( 3125 + 625 + 125 + 25 + \ldots + \frac{1}{625} \), similar logic applies with \( a = 3125 \) and \( r = \frac{1}{5} \). We can determine the number of terms: \[ \frac{1}{625} = 3125(\frac{1}{5})^{n-1} \implies n-1 = 5 \implies n = 6 \] The sum of the first 6 terms: \[ S_6 = 3125 \cdot \frac{1 - \left(\frac{1}{5}\right)^6}{1 - \frac{1}{5}} = 3125 \cdot \frac{1 - \frac{1}{15625}}{\frac{4}{5}} = 3906.25 \cdot \left(1 - 0.000064\right) \approx 3906.25 \cdot 0.999936 \approx 3905.00 \