Question

A strictly increasing sequence of positive integers a, b, c satisfies the property that $\frac{a+b}{2}$, $\frac{a+c}{3}$ form a geometric sequence. What is the smallest possible value of a + b + c?

• A. 8
• B. 9
• C. 10
• D. 11

Answer 1

Answer: (C)

10

Answer 2

We are given that $\frac{a+b}{2}$, $b$, and $\frac{a+c}{3}$ form a geometric progression. This means that

$b^2 = \frac{a+b}{2} \cdot \frac{a+c}{3}$, which simplifies to $6b^2 = (a+b)(a+c)$.

We want to find the smallest possible value of $a+b+c$ with $a, b, c$ being strictly increasing positive integers.

Let's try small values for $a$ and $b$.

If $a=1$ and $b=2$, then $6(2^2) = (1+2)(1+c)$, so $24 = 3(1+c)$, which gives $1+c = 8$, so $c=7$. The sequence is $1, 2, 7$, and $a+b+c = 1+2+7 = 10$.

We verify that $\frac{a+b}{2} = \frac{1+2}{2} = \frac{3}{2}$, $b=2$, and $\frac{a+c}{3} = \frac{1+7}{3} = \frac{8}{3}$. The common ratio is $\frac{2}{3/2} = \frac{4}{3}$ and $\frac{8/3}{2} = \frac{4}{3}$. Since the common ratios are equal, the condition is satisfied.

If $a=1$ and $b=3$, then $6(3^2) = (1+3)(1+c)$, so $54 = 4(1+c)$, which gives $1+c = \frac{54}{4} = \frac{27}{2}$, so $c = \frac{25}{2}$, which is not an integer.

If $a=2$ and $b=3$, then $6(3^2) = (2+3)(2+c)$, so $54 = 5(2+c)$, which gives $2+c = \frac{54}{5}$, so $c = \frac{44}{5}$, which is not an integer.

If $a=1$ and $b=4$, then $6(4^2) = (1+4)(1+c)$, so $96 = 5(1+c)$, which gives $1+c = \frac{96}{5}$, so $c = \frac{91}{5}$, which is not an integer.

If $a=2$ and $b=4$, then $6(4^2) = (2+4)(2+c)$, so $96 = 6(2+c)$, which gives $2+c = 16$, so $c = 14$. The sequence is $2, 4, 14$, and $a+b+c = 2+4+14 = 20$.

If $a=3$ and $b=4$, then $6(4^2) = (3+4)(3+c)$, so $96 = 7(3+c)$, which gives $3+c = \frac{96}{7}$, so $c = \frac{75}{7}$, which is not an integer.

Thus, the smallest possible value of $a+b+c$ is $1+2+7 = 10$.