The first three terms of a GP are (k-3), (2k-4), (4k-3). Find (a) the value of k, (b) the common ratio, (c) the sum of the first 10 terms.

Expand both sides:

Subtract $4k^2$ from both sides:

Rearrange the terms to solve for $k$:

The value of $k$ is $\boxed{7}$.

First term ($T_1$): $k-3 = 7-3 = 4$
Second term ($T_2$): $2k-4 = 2(7)-4 = 14-4 = 10$
Third term ($T_3$): $4k-3 = 4(7)-3 = 28-3 = 25$
The terms are $4, 10, 25, \dots$

Verify with the next pair of terms:

The common ratio is $\boxed{\frac{5}{2}}$.

$a = 4$
$r = \frac{5}{2}$
$n = 10$

First, calculate the denominator:

Next, calculate $\left(\frac{5}{2}\right)^{10}$:

Now substitute these values back into the sum formula:

Divide the numerator by 3: $9764601 \div 3 = 3254867$.

The sum of the first 10 terms is $\boxed{\frac{3254867}{128}}$.