$p, q, r$ are three numbers in G.P. Prove that the first term of an A.P. whose $p t h, q t h, a n d r t h$ terms are in H.P. is to be the common difference as $q + 1 : 1$

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Solution

$q^{2} = p r$
as p,q,r are in G.P.
$T_{p}, T_{q}, T_{r}$ of an A.P. are in H.P.
$∴ \frac{1}{T_{p}}, \frac{1}{T_{q}}, \frac{1}{T_{r}}$ are in A.P.
$\frac{1}{T_{q}} - \frac{1}{T_{p}} = \frac{1}{T_{r}} - \frac{1}{T_{q}}$
$\frac{T_{p} - T_{q}}{T_{p}} = T_{q} - T_{r} T_{r}, \frac{(p - q) d}{a + (p - 1) d} + \frac{(q - r) d}{a + (r - 1) d}$
or $a (p + r - 2 q) = d [(p - 1) (q - r) - (r - 1) (p - q)]$
or $\frac{a}{d} = \frac{p q - p r - + r - (r p - r q - p + q)}{p + r - 2 q}$
$= \frac{p (q + 1) + r (q + 1) - 2 p r - 2 q}{p + r - 2 q}$
Put $p r =$ $q^{2}$ in numerator
$∴ \frac{a}{d} = (q + 1) \frac{(p + r - 2 q)}{p + r - 2 q} = q + 1$

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