In an [increasing sequence](https://artofproblemsolving.com/wiki/index.php/Increasing_sequence) of four positive integers, the first three terms form an [arithmetic progression](https://artofproblemsolving.com/wiki/index.php/Arithmetic_progression), the last three terms form a [geometric progression](https://artofproblemsolving.com/wiki/index.php/Geometric_progression), and the first and fourth terms differ by $30$. Find the sum of the four terms.
Denote the first term as $a$, and the common difference between the first three terms as $d$. The four numbers thus are in the form $a,\ a+d,\ a+2d,\ \frac{(a + 2d)^2}{a + d}$. 
Since the first and fourth terms differ by $30$, we have that $\frac{(a + 2d)^2}{a + d} - a = 30$. Multiplying out by the denominator, 
\[(a^2 + 4ad + 4d^2) - a(a + d) = 30(a + d).\] 
This simplifies to $3ad + 4d^2 = 30a + 30d$, which upon rearranging yields $2d(2d - 15) = 3a(10 - d)$.
Both $a$ and $d$ are positive integers, so $2d - 15$ and $10 - d$ must have the same sign. Try if they are both [positive](https://artofproblemsolving.com/wiki/index.php/Positive) (notice if they are both [negative](https://artofproblemsolving.com/wiki/index.php/Negative), then $d > 10$ and $d < \frac{15}{2}$, which is a contradiction). Then, $d = 8, 9$. Directly substituting and testing shows that $d \neq 8$, but that if $d = 9$ then $a = 18$. Alternatively, note that $3|2d$ or $3|2d-15$ implies that $3|d$, so only $9$ may work. Hence, the four terms are $18,\ 27,\ 36,\ 48$, which indeed fits the given conditions. Their sum is $\boxed{129}$.

Postscript
As another option, $3ad + 4d^2 = 30a + 30d$ could be rewritten as follows:

$d(3a + 4d) = 30(a + d)$

$d(3a + 3d)+ d^2 = 30(a + d)$

$3d(a + d)+ d^2 = 30(a + d)$

$(3d - 30)(a + d)+ d^2 = 0$

$3(d - 10)(a + d)+ d^2 = 0$

This gives another way to prove $d<10$, and when rewritten one last time:

$3(10 -d)(a + d) = d^2$

shows that $d$ must contain a factor of 3.

-jackshi2006
EDIT by NealShrestha:
Note that once we reach $3ad + 4d^2 = 30a + 30d$ this implies $3|d$ since all other terms are congruent to $0\mod 3$.