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Let the common difference of the arithmetic progression be $d$ , then $a_2 = a_1 +d$ , $S_3 = \dfrac 32(2a_1+2d) = 3(a_1+d)$ , and $a_{14} = a_1 + 13d$ . For geometric progression, we have $S_3^2 = a_2a_{14}$ :

$\begin{aligned} 3^2(a_1+d)^2 & = (a_1+d)(a_1+13d) \\ 9a_1^2 + 18a_1d + 9d^2 & = a_1^2 + 14a_1d + 13d^2 \\ 8a_1^2 + 4a_1d - 4d^2 & = 0 \\ 2a_1^2 + a_1d - d^2 & = 0 \\ (2a_1-d)(a_1+d) & = 0 \end{aligned}$

$\implies \begin{cases} d = 2a_1 & \implies a_2 = 3a_1, & S_3 = 9a_1 & a_{14} = 27a_1 & \small \blue{\text{in GP}} \\ d = -a_1 & \implies \red{a_2 = 0}, & \red{S_3 = 0} & a_{14} = 2=-12a_1 & \small \red{\text{not in GP}} \end{cases}$

Therefore the sum of $\dfrac {S_{30}}{a_8} = \dfrac {\frac {30}2(2a_1+29d)}{a_1+7d} = \dfrac {15\times 60a_1}{15a_1} = \boxed{60}$ .

If k is the common difference of the arithmetic progression, we have (by definitions)

a2 = a1+k

S3 = 3*(a1+k)

Hence, the common quotient of the geometric progression is 3. Thus, (a14)/(a2) = 3*3 = 9, so that:

(a1 + 13*k) / (a1 + k) = 9, or

a1 = k/2

Thus, (S30) / (a8) = (30 a1 + k 29 30/2) / (a1 + 7 k) = (15 k + 15 29k) / (k/2 + 7*k) = 60