Sequences & Series
Sequence and Series
Allen Star Batch
Grade 11
Question:
Let 'p' be the first of 'n' arithmetic means between two positive numbers and 'q' be first of 'n' harmonic means between same two numbers. The $\frac{q}{p}$ can lie in interval(s):
$(-\infty, 1]$
$\left[1, \left(\frac{n+1}{n-1}\right)^2\right]$
$\left[\left(\frac{n-1}{n+1}\right)^2, \left(\frac{n+1}{n-1}\right)^2\right]$
$\left[\left(\frac{n+1}{n-1}\right)^2, \infty\right)$
Step-by-Step Solution
Key Concept: The discriminant of a quadratic constraint determines the feasible range for $q$ in terms of $p$ and $n$.
Given $p = a + n\left(\frac{b-a}{n+1}\right) = \frac{nb+a}{n+1}$, we find $q = \frac{(n+1)ab}{na+b}$ by using the harmonic mean relationship. After algebraic manipulation, the discriminant condition $D \geq 0$ yields $(q-p)\left(q(n-1)^2 - p(n+1)^2\right) \geq 0$, which simplifies to $(q-p)\left(q(n-1)^2 - p(n+1)^2\right) \geq 0$. This gives the range $q \in (-\infty, p] \cup \left[\left(\frac{n+1}{n-1}\right)^2 p, \infty\right)$.
Correct Answer: 1,4