If a, b, c, and d are positive real numbers such that a,b,c,d form an increasing arithmetic sequence and a, b, d form a geometric sequence, then a/d is
A) 1/12 B) 1/6 C) 1/4 D) 1/3 E) 1/2
Let the common difference of the arithmetic sequence be $d$. Then $b=a+d$ and $d=b+d$, so $3d=b$ and $d=b/3$. Also, since $a$, $b$, $d$ form a geometric sequence, we know that $(a+d)/a=a/d=b/a$. Substituting $d=b/3$, we get $(a+b/3)/a=a/(b/3)=3a/b$. Therefore, $a/d=3a/b=\boxed{\textbf{(D)}\ 1/3}$.
Let's assume the common difference in the arithmetic sequence is "d1" and the common ratio in the geometric sequence is "r".
Since a, b, c form an increasing arithmetic sequence, we have:
b = a + d1
c = a + 2d1
Since a, b, d form a geometric sequence, we have:
b/a = d/a
d = ba/a = b
Substituting the value of d in terms of b:
d = b
b = a + d1
a + d1 = b
We can now rewrite the geometric sequence:
b/a = b/(a + d1)
Cross-multiplying, we get:
b(a + d1) = ba
Expanding the equation:
ab + bd1 = ab
Subtracting ab from both sides:
bd1 = 0
Since d1 is positive (as stated in the problem), we can conclude that b must be 0. However, this contradicts the given information that a, b, c, and d are positive real numbers.
Therefore, there is no solution for the given conditions.