Find the sum of the first 10th term of a g.p where the 1st term is 3 and the common ratio is1/2
The sum of the first $n$ terms of a geometric progression is given by the formula $S_n = \frac{a(1 - r^n)}{1 - r}$, where $a$ is the first term and $r$ is the common ratio.
We are given that the first term $a = 3$, the common ratio $r = \frac{1}{2}$, and we want to find the sum of the first $10$ terms.
Using the formula, we have:
$S_{10} = \frac{3(1 - (\frac{1}{2})^{10})}{1 - \frac{1}{2}}$
$S_{10} = \frac{3(1 - \frac{1}{1024})}{\frac{1}{2}}$
$S_{10} = \frac{3(\frac{1023}{1024})}{\frac{1}{2}}$
$S_{10} = 6 \cdot \frac{1023}{1024}$
$S_{10} = \frac{6 \cdot 1023}{1024}$
$S_{10} = \frac{6138}{1024}$
$S_{10} = \frac{303}{64}$
Therefore, the sum of the first 10 terms of the geometric progression is $\boxed{\frac{303}{64}}$.