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

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

Accepted Answer

$$\begin{aligned}
& 2 b=a+c \quad \text{.... (1)}\
& b^2=(a+1)(c+3) \quad \text{.... (2)}\
& \frac{a+b+c}{3}=8 \quad \text{.... (3)}
\end{aligned}$$

$$\begin{aligned}
\Rightarrow & \frac{3 b}{3}=8 \
& b=8 \
\Rightarrow \quad & a c+3 a+c+3=64
\end{aligned}$$

$$\begin{aligned}
& 3 a+c+a c=61 \quad \text{... (4)}\
& a+c=16 \
& c=16-a
\end{aligned}$$

from equation (4)

$$\begin{aligned}
& 3 a+16-a+a(16-a)=61 \
& \Rightarrow \quad(a-15)(a-3)=0 \
& \quad a=15(a>10) \
& \Rightarrow \quad a=15, b=8, c=1 \
& \left((a \cdot b \cdot c)^{\frac{1}{3}}\right)^3=15 \times 8 \times 1=120
\end{aligned}$$

Solutions · 1