Let three real numbers $a, b, c$ be in arithmetic progression and $a+1, b, c+3$ be in geometric progression. If $a10$ and the arithmetic mean of $a, b$ and $c$ is 8, then the cube of the geometric mean of $a, b$ and $c$ is

Question

Let three real numbers $a, b, c$ be in arithmetic progression and $a+1, b, c+3$ be in geometric progression. If $a>10$ and the arithmetic mean of $a, b$ and $c$ is 8, then the cube of the geometric mean of $a, b$ and $c$ is

Aicharya · Accepted Answer

$\begin{aligned} & 2 b=a+c, b^2=(a+1)(c+3) \ & \frac{a+b+c}{3}=8 ightarrow b=8, a+c=16 \ & 64=(a+1)(19-a)=19+18 a-a^2 \ & a^2-18 a-45=0 ightarrow(a-15)(a+3)=0,(a>10) \ & a=15, c=1, b=8 \ & \left((a b c)^{1 / 3}ight)^3=a b c=120\end{aligned}$

Let three real numbers $a, b, c$ be in arithmetic progression and $a+1, b, c+3$ be in geometric progression.…

Solution