Glamorous Geometric Series

Because all of the terms in the geometric sequence are integers, the common ratio must be rational. Thus, we can write the sequence as $a, a\left(\frac{m}{n} \right), a \left(\frac{m}{n} \right)^2, a \left(\frac{m}{n} \right)^3, a \left(\frac{m}{n} \right)^4$ , where $m,n$ are relatively prime, and $m>n$ (because the sequence must be strictly increasing). Now, the most important property of $a$ is that $n^4$ must divide it. Noting that $n=1$ does not lend itself to any possible values for $a$ , we begin our casework with $n=2$ .

If $n=2$ , then $16 \mid a$ , and $a>50$ . Thus, the smallest possible value for $a$ is $64$ . We need only to confirm that there exists an integer $m$ such that $m>n$ and that $\frac{64}{2^4} \cdot m^4=4m^4<500$ . Indeed, $m=3$ works. and so $a=64$ is a solution. We continue this process for $n=2$ , finding $a=80,96$ are also possible starting values. However, if $a>=112$ , we find that $m$ must be an integer less than $3$ , and so there are no more solutions for this case.

Continuing with $n=3$ in a similar fashion, we find $a=81$ to be the smallest multiple of $81$ greater than $50$ . Easily, $m=4$ is the only value of $m$ that satisfies all conditions. Again, note that for $a>=162$ , $m$ must be an integer less than $4$ , which does not satisfy $m>n$ , and so there are no more solutions.

Considering $n=4$ , it follows that the smallest possible value for $m$ is $5$ ; however, this yields that $e=625 \cdot \frac{a}{n^4}$ (with $\frac{a}{n^4}$ being an integer), which clearly is greater than $500$ . Thus, there are no more solutions.

From this casework, the solutions are $a=64, 80, 96, 81$ , which sum to $\boxed{321}$ .

Since, it is a geometric sequence, there is a common ratio between the consecutive terms of the sequence. Lets assume the common ratio is $r$ .

Firstly, we will prove that $1 < r < 2$ .

Since, $a < b$ , it is apparent that $r = \frac{b}{a} > 1$ .

$r<2$ because, if $r \geq 2$ , then, the minimum possible value of the fifth term of the sequence $ar^4 = 50 \times 2^5 = 800 > 500$ which contradicts with the given conditions.

So, $1 < r < 2$ .

It is also apparent that $r$ is rational number, because, $a,b,c,d,e \epsilon N$ and the product of a irrational number and a natural number is always irrational.

So, we can write that $r = \frac{x}{y}$ where $x,y \epsilon N$ and they are co-prime and $x > y$ .

Now, for $a,b,c,d$ and $e$ to be integers, $a$ has to be divisible by $y^4$ . Otherwise, $ar^4 = a\frac{x^4}{y^4}$ would not be an integer.

Since, $a$ is divisible by $y^4$ , we can see that the minimum value of $\frac {a}{y^4} = 1$ and therefore the minimum value of $ar^4 = a \times \frac{x^4}{y^4} = x^4$ .
And since, the value of $e = ar^4$ cannot exceed $500$ , therefore $x^4 < 500$ and so $x < \sqrt[4]{500} \approx 4.7$ .So, the maximum possible value of $x = 4$ .

We can also see that the minimum value of $x =3$ . Cause, we said earlier that $x>y$ and $x,y$ are positive integers.If $x\leq2$ , then the only possible values of $x$ are $2$ and $1$ . [Since, $x$ is a positive integer.]
But, we said that $x>y$ . But, there is no positive integer less than 1. And since $y$ has to be a positive integer, so there is no possible value of $y$ when $x =1$ .Similarly, when $x=2$ , $y=1$ .But, then, $r = 2$ . But, we have proved that $1< r <2$ So, $x > 2$ . So, the minimum value of $x$ is $3$ .

So, we can say that $3 \leq x \leq 4$ .

So, the only possible values of $x$ are $4$ and $3$ .

Now, when $x = 4$ , $3$ is the only possible value for $y$ .
[ Since, $y > 1$ and $y<x$ and $x$ and $y$ are co-prime.]
So, $r = \frac {4}{3}$ .
So, the minimum value of $a = 3^4$ . This is the only value of $a$ . Because, since $a$ is divisible by $3^4$ , we can write $a = k \times 3^4$ where $k \epsilon N$ .
Then $ar^4 = k \times 4^4 < 500$
or, $k < \frac{500}{4^4}$
or, $k < 1\frac{61}{64}$
Since, $k$ is a natural number, the only possible value of $k$ is 1.

In the same way, When $x = 3$ , the only value of $y$ is $2$ .
So, the only value of $r$ is $\frac{3}{2}$ .
Since, $a$ is divisible by $2^4$ , we can write that $a = k \times 2^4$ where $k \epsilon N$ .
So, $e = ar^4 = 2^4 \times k \times \frac {3^4}{2^4} =k \times 3^4$
So, $k \times 3^4 < 500$
or, $k < \frac{500}{3^4}$
or, $k < 6\frac{14}{81}$
So, the maximum value of $k$ is $6$ .
On the other hand, $k \times 2^4 >50$
or, $k > 3\frac{1}{8}$
So, the minimum value of $k$ is $4$ .
So, the sum of all the possible values of $a$ in this case is $\displaystyle \sum_{k=4}^6 {(k \times 2^4})$ .

So, the sum of all the possible values of $a = 3^4 +\displaystyle \sum_{k=4}^6 {(k \times 2^4}) = 15 \times 2^4 +3^4 = 321$ .
So, the answer is $321$ .

Suppose the common ratio is a positive integer, say $x$ ,where $x \neq 1$ , or that $x \geq 2$ . Then $e=a*x^{4} > a*2^{4}=16a > 800 >e$ , a contradiction.

So the common ratio $x$ is not an integer. However it is surely a rational number, because $x = \frac {b}{a}$ . Let $x= \frac {p}{q}$ in its reduced form, where $q \neq 1$ and $p>q$ , since the G.P. is increasing.

So $e= \frac {a*p^{4}}{q^{4}}$ , so that $q^{4}$ must be a factor of $a$ . Suppose $q \geq 4$ , then clearly,

$\frac {p}{q} \geq \frac {q+1}{q} \geq (1+\frac {1}{q}) \geq \frac {5}{4}$ . Also $a \geq q^{4} \geq 256$

Thus $e \geq 256* (\frac {5}{4})^{4} \geq 625$ , a contradiction.

Thus $q=2,3$ . If $q= 2$ , then $p=3$ is the only possibility, otherwise $\frac {p}{q}$ will be $2$ or more. So $x=\frac {3}{2}$ , and $16$ is a factor of $a$ . Clearly $e = \frac {81a}{16} < 500$ , or $a<99$ , and thus $a=64,80,96$ .

If $q=3$ , then $a \geq 81$ and $p=4,5$ . But for $p=5$ , $e = a*x^{4} \geq \frac {81*5^4}{3^4} = 625 \geq e$ , a contradiction. So $p=4$ . Now $e= \frac {256a}{81} < 500$ , or $a< 159$ . But since $81$ is a factor of $a$ , $a=81$ .

Thus the sum of all possible values of $a$ are $64+80+96+81 = 321$ .

Given: geometric series of integers increasing between 50 and 500. Terms (a,b,c,d,e) could be written as (a, a r, a r^2, a r^3, a r^4) with r being the ratio between terms in the geometric series.

r must be greater than 1 for it to be an increasing geometric series. r^4 must be less than 10 for it to be between 50 and 500 (which have a ratio of 10)

Roughly, this means that r^2 < 3 and r < 1.75. In summary, 1 < r < 1.75

In order for a,b,c,d,e to all be integers, the ratio must be a rational number. Let us call this ratio x/y a must be evenly divisible by y^4 or not all a,b,c,d,e will be integers

Now, we must remember that a * (x/y)^4 must be less than 500. If x is 5 or greater, we will be multiplying by 625 in the numerator. a/(y^4) will be one or greater or else not all of a,b,c,d,e will be integers. This means that x must be smaller than 5.

If x=2, then the only possible ratio greater than 1 is 2/1, which does not satisfy r^4 < 10.

Thus, x=3 or x=4.

If x=3, y=2 (if y=1, r^4 is not less than 10). Thus, a must be a power of 16 (so that a is evenly divisible by y^4), and it must be greater than 50. a = 64, 80, 96 all satisfy this with a*r^4 < 500

If x=4, y=3 (if y=2, r^4 is not less than 10). Thus, a must be a power of 81 (so that a is evenly divisible by y^4). a = 81 satisfies a*r^4 < 500

64+80+96+81 = 321

We have that $e=ar^4$ , where $r$ is the common ratio. Since $2^4\cdot 50=800>500$ , $r<2$ . Also, $a$ must be divisible by a fourth power, since $r$ is not an integer. $a$ can only be divisible by $16,81,256$ , since all other fourth powers are too big. Similarly, the numerator of $r$ can only be $2,3,4$ . This gives two values of $r$ : $\dfrac{3}{2}$ and $\dfrac{4}{3}$ .

If $r=\dfrac{3}{2}$ , $16|a$ . If $a\ge 112$ , $e\ge 567$ , so $a=64,80,96$ .

If $r=\dfrac{4}{3}$ , $81|a$ . If $a\ge 162$ , $e\ge 512$ , so $a=81$ .

The answer is $64+80+96+81=\boxed{321}$ .

Let the terms be $a$ , $ar$ , $ar^2$ , $ar^3$ & $ar^4$ . [where $r$ is the common ratio with $r>1$ ]

So, $a > 50$ and $ar^4 < 500 \Rightarrow r < 2$

So, $r$ is a rational of form $\frac{p}{q}$ with $1 < q < p < 2q$ . [ $p,q$ integers]

So, last term $e = ar^4 = \frac{ap^4}{q^4}$

$\Rightarrow a=kq^4>50$ [all terms are integers], so $2 \leq q \leq 3$ [ $q \leq 3$ as $q < p < 5$ as stated below]

Also, $e = kp^4 < 500$ with $k \geq 1$ . So $3 \leq p \leq 4$

With $q = 2, p = 3$ , we have $3 < k < 7$ giving 3 solutions with corresponding a's : $4*2^4, 5*2^4, 6*2^4$ .

With $q = 3, p = 4$ , we have $0 < k < 2$ giving 1 solution with corresponding a: $3^4$ .

So required sum is $321$ .

Rewrite the form of the sequence terms in terms of $a$ and $r$ , the common ratio of the geometric sequence. Thus $50 < a < ar < ar^2 < ar^3 < ar^4 < 500$ . Clearly, for all of the terms to remain integers, $a$ must have as a factor the fourth power of some integer, say $f$ .

$f$ can either be 2, 3, or 4 since $2^4 = 16$ , $3^4 = 81$ and $4^4 = 256$ ; note that $f = 5$ will cause $a = 625 > 500$ . The common ratio $r$ is then a fraction $\frac{f+1}{f}$ - it is sufficient to consider this as we are only concerned with which values of $a$ satisfy the condition and we don't want $ar^4$ to exceed 500.

For $f = 2$ , it becomes apparent that the valid solutions of $a$ are $4 \times 2^4$ , $5 \times 2^4$ and $6 \times 2^4$ , as larger constant factors will cause $ar^4 > 500$ . For $f = 3$ , $a = 3^4$ is the only possible solution as larger constant factors will also cause $ar^4 > 500$ . Finally for $f = 4$ , we cannot get any solutions for $a$ as we will always have $ar^4 > 500$ .

Thus our solutions for $a$ are $64$ , $80$ , $96$ and $81$ . Adding these up gives the desired answer $321$ .

First, $1 < r < 2$ since the sequence is increasing and $2^4 = 16 > 10$ . Thus, $r$ must be a fractional number. This requires that $a$ have at least four of some prime factor.

Say $a$ has four (or more) factors of 2. The smallest such that $a > 50$ is $16 * 4$ . The lowest $r$ with $2$ in the denominator is $\frac{3}{2}$ , and $64 * (\frac {3}{2})^4$ is $4 * 3^4 = 324$ . Thus, $64$ works, as well as $80$ and $96$ , but not $112$ .

Repeating with $3$ , we find that only $81$ works.

We can confirm that $4$ and all further factors do not work for $a$ since, at minimum, $e = ar^4 \rightarrow e = x^4 (\frac {x+1}{x})^4 \rightarrow e = (x + 1)^4$ , and $(x+1)^4 > 500$ for all $x \geq 4$ .

$64 + 80 + 96 + 81 = 321$ .

As a,b,c,d,e are integers and 1<r<2

Only for r= 3/2 and 4/3 we get all integer values for a b c d e.

a= 64+80+96+81= 321

Let $a = k, b = kr, c = kr^2, d=kr^3$ and $e = kr^4$ where $r$ is the common ratio.

$r$ can't be $1$ , since $a < b < c < d < e$ .

Also $r$ can't be $2$ , since $e < 500$ .

Hence $1<r<2$ . Since $a,b,c,d$ and $e$ are integers, the possible common ratio's are : $r =\frac{4}{3}$ and $\frac{3}{2}$ . For $r =\frac{4}{3}\, a=81$ and for $r =\frac{3}{2}, a = 64$ or $80$ or $96$ .

Answer $= 81 + 64 + 80 + 96 =\boxed{321}$ .

e = a * r^4

also e/a upper limit is 10 so r^4 < 10 implies r < 2

if r= 3/2, e/a = 81/16 min value of a = 64 ... for which e = 324 ... under the allowable limit also, a = 80 ... for which e = 405 ... is under the allowable limit if a = 96 ... for which e = 486 But if a = 112 ... e>500

if r= 4/3 ... e/a = 256/81 min value of a = 81 ... for which e = 256 there is no other multiple of 81 which keeps e<500

if r = 5/3 ... e/a = 625/81 ... that itself takes e>500 ... so we cannot have 5 or any other larger number in the numerator of 'r'

thus the only possible values of 'a' are: 64, 80, 96, 81 ... sum 321