# If $\large\frac{a_{2}a_{3}}{a_{1}a_{4}}$ = $\large\frac{a_{2}+a_{3}}{a_{1}+a_{4}}$ = $3\;(\large\frac{a_{2}-a_{3}}{a_{1}-a_{4}})$, then, $a_{1}$, $a_{2}$, $a_{3}$, $a_{4}$ are in

$(a)\;AP\qquad(b)\;GP\qquad(c)\;HP\qquad(d)\;None\;of\;these$

Answer : (c) HP
Explanation : $\large\frac{a_{2}a_{3}}{a_{1}a_{4}}=\large\frac{a_{2}+a_{3}}{a_{1}+a_{4}}$
$\large\frac{a_{1}+a_{4}}{a_{1}a_{4}}=\large\frac{a_{2}+a_{3}}{a_{2}a_{3}}$
$\large\frac{1}{a_{4}}+\large\frac{1}{a_{1}}=\large\frac{1}{a_{2}}+\large\frac{1}{a_{3}}$
$\large\frac{1}{a_{4}}-\large\frac{1}{a_{3}}=\large\frac{1}{a_{2}}-\large\frac{1}{a_{1}}----(1)$
Also , $\;\large\frac{3(a_{2}-a_{3})}{a_{1}-a_{4}}=\large\frac{a_{2}a_{3}}{a_{1}a_{4}}$
Also , $\;\large\frac{3(a_{2}-a_{3})}{a_{2}a_{3}}=\large\frac{a_{1}-a_{4}}{a_{1}a_{4}}$
$=3\;(\large\frac{1}{a_{3}}-\large\frac{1}{a_{2}})=\large\frac{1}{a_{4}}-\large\frac{1}{a_{1}}$
$From \;(1)\;and\;(2)$
$=\large\frac{1}{a_{2}}-\large\frac{1}{a_{1}}=\large\frac{1}{a_{3}}-\large\frac{1}{a_{2}}=\large\frac{1}{a_{4}}-\large\frac{1}{a_{3}}$
$\large\frac{1}{a_{1}}\;,\large\frac{1}{a_{2}}\;,\large\frac{1}{a_{3}}\;\large\frac{1}{a_{4}}\;are\;in\;AP\;.$
$a_{1}\;,a_{2}\;a_{3}\;a_{4}\;are\;in\;HP\;.$
answered Jan 24, 2014 by