1) If \( \left(\mathrm{T}_{\mathrm{n}}\right)=\left(3 \times 2^{-\mathrm{n}}\right) \) is a geometric sequence, then the sum of infinite terms star from its first term \( =\ldots \ldots \) \( \begin{array}{llll}\text { (a) } 2 & \text { (b) } 3 & \frac{3}{1-r} & \frac{a}{1-r}\end{array} \)

To find the sum of the infinite terms of the geometric sequence given by \( T_n = 3 \times 2^{-n} \), we first need to identify the first term and the common ratio of the sequence. 1. **Identify the first term \( a \)**: The first term \( T_1 \) is obtained by substituting \( n = 1 \): \[ T_1 = 3 \times 2^{-1} = 3 \times \frac{1}{2} = \frac{3}{2} \] 2. **Identify the common ratio \( r \)**: The common ratio \( r \) can be found by taking the ratio of the second term to the first term: \[ T_2 = 3 \times 2^{-2} = 3 \times \frac{1}{4} = \frac{3}{4} \] Now, calculate the common ratio: \[ r = \frac{T_2}{T_1} = \frac{\frac{3}{4}}{\frac{3}{2}} = \frac{3}{4} \times \frac{2}{3} = \frac{1}{2} \] 3. **Sum of the infinite geometric series**: The sum \( S \) of an infinite geometric series is given by the formula: \[ S = \frac{a}{1 - r} \] where \( a \) is the first term and \( r \) is the common ratio. Substituting the values we found: \[ S = \frac{\frac{3}{2}}{1 - \frac{1}{2}} = \frac{\frac{3}{2}}{\frac{1}{2}} = \frac{3}{2} \times 2 = 3 \] Thus, the sum of the infinite terms starting from its first term is \( 3 \). The answer is \( \text{(b) } 3 \).