Integers a,b,c,d and e satisfy 50<a<b<c<e and a,b,c,d,e from a geometric sequence what is the sum of all possible distinct values of a?
No way to tell. All you have said is that 50<a.
I think there is something missing here.
Crazy question.
Agree with you, there has to be more
As it stands:
b/a = c/b and c/b = d/c and d/c = e/d
b^2 = ac , c^2 = bd , d^2 = ed
then b^2 c^2 d^2 = abcde
I then ran this "primitive" BASIC program
10 for a = 50 to 100
20 for b = 51 to 100
30 for c = 52 to 100
40 for d = 53 to 100
50 for e = 54 to 100
60 if (b*c*d)^2 = (a*b*c*d*e) then print a;b;c;d;e
70 next e
80 next d
90 next c
100 next b
110 next a
and got no output after about 6 million loops
I think I missed the obvious
Since a,b,c,d,e are integers in GS, the smallest value of r is 2
a - b - c - d - e
51 102 204 408 816
52 104 ...
53 ....
infinite number of choices for a
so...
still a silly question
Forget my previous post,
r could be 1.5
my program line 60 should say:
60 if (b*c*d)^2 = (a*b*c^2*d*e) then print a;b;c;d;e
and I got
64 96 144 216 324
80 120 180 270 405
96 144 216 324 486
The answer is 321. Possible values of a are:
64(r=1.5)
80(r=1.5)
81(r=4/3)
96(r=1.5)
To find the sum of all possible distinct values of a, we need to determine the values of a, b, c, d, and e that satisfy the given conditions.
Let's denote the geometric sequence as a, ar, ar^2, ar^3, and ar^4, where r is the common ratio.
From the given conditions, we have:
1. 50 < a < b < c < e - This implies that a must be less than 50 and e is the largest term.
2. Since a and e are the first and last terms of the sequence, they satisfy the condition of being a geometric sequence. So, a * r^4 = e. We can rewrite e as a * r^4.
3. Since a, b, c, and d are consecutive terms, we can set up the following equations:
b = a * r
c = a * r^2
d = a * r^3
Putting all these conditions together, we can simplify:
50 < a < a * r < a * r^2 < a * r^3 * r^4
Dividing all terms by a, we get:
1 < r < r^2 < r^3 * r^4
Now, we need to find all possible values of r that satisfy these conditions. To do so, we can consider different cases:
Case 1: If r > 1, then r^2 > r > 1, which means r^2 > r^3 > r^4. In this case, there are no possible solutions.
Case 2: If r = 1, then we have a * r^4 = a, which means a = a. In this case, a can take any value greater than 50, so there are infinite solutions.
Case 3: If 0 < r < 1, then r^2 < r < 1, which means r^2 < r^3 < r^4. In this case, there are no possible solutions.
Therefore, the only case that yields possible solutions is when r = 1. In this case, a can take any value greater than 50.
So, the sum of all possible distinct values of a is infinite since there are infinite solutions in this case.