The sum of the first k terms of a series S is 3k^2 + 5k . Which one...

The sum of the first $k$ terms of the series is given by $S_k = 3k^2 + 5k$.

The $k$-th term of the series, $T_k$, is obtained by subtracting the sum of the first $k-1$ terms from the sum of the first $k$ terms:

$T_k = S_k - S_{k-1}$

Substituting the given expression for $S_k$ and $S_{k-1}$:

$T_k = (3k^2 + 5k) - [3(k-1)^2 + 5(k-1)]$

$T_k = 3k^2 + 5k - [3(k^2 - 2k + 1) + 5k - 5]$

$T_k = 3k^2 + 5k - (3k^2 - 6k + 3 + 5k - 5)$

$T_k = 3k^2 + 5k - (3k^2 - k - 2)$

$T_k = 6k + 2$

To find the common difference, we calculate the difference between consecutive terms:

$T_k - T_{k-1} = (6k + 2) - [6(k-1) + 2]$

$T_k - T_{k-1} = 6k + 2 - (6k - 6 + 2)$

$T_k - T_{k-1} = 6$

Since the difference between consecutive terms is a constant, the terms of the series form an arithmetic progression with a common difference of 6.

Answer: The terms of $S$ form an arithmetic progression with common difference 6.