Sequences & Series
AP and GP — Cube of Geometric Mean
nta_pyq_2024_apr
Grade 11
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
Step-by-Step Solution
Key Concept: $2b=a+c$, $b^2=(a+1)(c+3)$. AM $=8\Rightarrow b=8$, $a+c=16$. $(a+1)(19-a)=64\Rightarrow a^2-18a-45=0\Rightarrow a=15$ (since $a>10$), $c=1$.
$a=15$, $b=8$, $c=1$. $(abc)=(abc)^3/3..$ wait: cube of GM $=(abc)=120$.
Correct Answer: 3