Ratio of two sums

With a common difference $d=3$ , $S_k$ can be expressed as $S_k = \frac k2 (2a + (k-1)d) = \frac k2 (2a + 3(k-1)) = \frac k2 (2a + 3k - 3)$ .

Since the value of $\dfrac{S_{3n}}{S_n}$ does not depend on $n$ , then $\dfrac{S_{3n}}{S_n}$ must be a constant for all $n$ . Thus, $\dfrac{S_3}{S_1} = \dfrac{S_6}{S_2} \quad \Leftrightarrow \quad S_2 S_3 = S_1 S_6 = (2a + 3)(3a + 9) = a \cdot 3 \cdot (2a + 15) \quad \Leftrightarrow \quad a = 1.5 .$

The $100^\text{th}$ term can be expressed as $T_{100} = a + (100 - 1) \times d = 1.5 + 99\times3 = \boxed{298.5}$ .

Since $\dfrac {S_{3n}}{S_n}$ is independent of $n$ , let it be equal to a constant $A$ . Then

$\begin{aligned} \frac {S_{3n}}{S_n} & = A \\ \frac {\frac {3n}2 \left(2a_1+3(3n-1)\right)}{\frac n2 \left(2a_1+3(n-1)\right)} & = A & \small \color{#3D99F6} \text{where }a_1 \text{ is the first term of the sequence.} \\ \frac {6a_1+27n-9}{2a_1+3n-3} & = A \\ 6a_1+27n-9 & = 2Aa_1+3An-3A \\ (6-2A)a_1 + {\color{#3D99F6}(27-3A)}n & = 9 - 3A & \small \color{#3D99F6} \text{For }\frac {S_{3n}}{S_n} \text{ to be independent of }n \implies A = 9 \\ \implies (6-18)a_1 + {\color{#3D99F6}0} & = 9 - 27 \\ -12 a_1 & = 18 \\ \implies a_1 & = \frac 32 \end{aligned}$

Then the 100th term of the sequence $a_{100} = a_1 + 3(100-1) = \frac 32 + 297 = \boxed{298.5}$ .

Let the first term of the sequence be a. Then S(3n)/S(n)= (6a+27n-9)/(2a+3n-3)= 9-((12a-18)/(2a+3n-3)). For this to be independent of n, a must equal (3/2) or 1.5, which yields the 100th. term as 1.5+3(100-1)= 298.5