Sequences and Series 3 Question 16
16. If the $m$ th, $n$th and $p$ th terms of an AP and GP are equal and are $x, y, z$, then prove that $x^{y-z} \cdot y^{z-x} \cdot z^{x-y}=1$.
(1979, 3M)
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Solution:
- Let $a, d$ be the first term and common difference of an $AP$ and $b, r$ be the first term and common ratio of a GP. Then,
$$ \begin{aligned} & x=a+(m-1) d \text { and } x=b r^{m-1} \\ & y=a+(n-1) d \text { and } y=b r^{n-1} \\ & z=a+(p-1) d \text { and } z=b r^{p-1} \end{aligned} $$
Now, $x-y=(m-n) d, y-z=(n-p) d$
and $\quad z-x=(p-m) d$
Again now, $x^{y-z} \cdot y^{z-x} \cdot z^{x-y}$
$$ \begin{aligned} & =\left[b r^{m-1}\right]^{(n-p) d} \cdot\left[b r^{n-1}\right]^{(p-m) d} \cdot\left[b r^{p-1}\right]^{(m-n) d} \\ & =b^{[n-p+p-m+m-n] d} \cdot r^{[(m-1)(n-p)+(n-1)(p-m)+(p-1)(m-n)] d} \\ & =b^{0} \cdot r^{0}=1 \end{aligned} $$
