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# $p, q, r$ are three unequal numbers in AP. If $q-p$, $r-q$ and $p$ are in GP, the $p\;:\;q\;:\;r$ equals

$(a)\;1:3:4\qquad(b)\;1:2:3\qquad(c)\;2:3:5\qquad(d)\;1:2:4$

Answer : (b) $\;1:2:3$
Explanation : $p,q,r \;are\;in\;AP$
$q-p=r-q$
$q-p\;,r-q\;,p\;are\;in\;GP$
$(r-q)^2=p\;(q-p)$
$(q-p)^2=p\;(q-p)$
$q-p=p$
$q=2p$
$r=3p$
$p:q:r=1:2:3.$