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 \( =\ldots \ldots \) \( \begin{array}{llll}\text { (a) } 2 & \text { (b) } 3 & \frac{3}{1-r} & \frac{a}{1-r}\end{array} \)

The sum of an infinite geometric series can be calculated using the formula \( S = \frac{a}{1 - r} \), where \( a \) is the first term and \( r \) is the common ratio. In your sequence \( \mathrm{T}_{\mathrm{n}} = 3 \times 2^{-\mathrm{n}} \), the first term \( \mathrm{T}_{1} = 3 \) and the common ratio \( r = \frac{1}{2} \). Plugging these values into the formula gives us \( S = \frac{3}{1 - \frac{1}{2}} = \frac{3}{\frac{1}{2}} = 6 \). Now, since it looks like you were comparing results with few options, let's for a moment throw out multiple choices: The relevant answer should be None of the above as you would find. So, the infinite sum equates to 6! To visualize it, think of a bouncy ball that keeps bouncing half the height of the previous jump. It may seem like its energy gets split, but keep summing those heights and you will see it adds up to a specific number, proving that even with decay, infinite sums can close in on a solid total.