Sequences & Series
Sequence and Series
star_batch_jee_advanced_2025
Grade 11
Question:
if $p^q$, $q^r$ and $r^p$ terms of an H.P. be respectively $x$, $y$, $z$, then $(p-q)xy + (q-r)yz + (r-p)xz=$
$xyz + pqr$
$pqr$
$xyz$
$0$
Step-by-Step Solution
Key Concept: Terms in AP with reciprocals satisfy a linear dependence relation when weighted by consecutive differences.
Given $\frac{1}{x} = a + (p-1)d$, $\frac{1}{y} = a + (q-1)d$, and $\frac{1}{z} = a + (r-1)d$, we find $\frac{1}{x} - \frac{1}{y} = (p-q)d$ and similar expressions. Multiplying by respective variables and substituting yields $xy(p-q) = \frac{y-x}{d}$, $yz(q-r) = \frac{z-y}{d}$, and $zx(r-p) = \frac{x-z}{d}$. Adding these three equations gives $(p-q)xy + (q-r)yz + (r-p)zx = \frac{1}{d}(y-x+z-y+x-z) = 0$.
Correct Answer: 4