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# If $a_{1},a_{2},...a_{n}$ are in HP. , Then $\;a_{1}a_{2}+a_{2}a_{3}+.....+a_{n}\;a_{n-1}$ is equal to

$(a)\;n\;(a_{1}-a_{n})\qquad(b)\;(n-1)(a_{1}-a_{n})\qquad(c)\;(n-1)a_{1}a_{n}\qquad(d)\;na_{1}a_{n}$

Answer : $(c)\;(n-1)a_{1}a_{n}$
Explanation : $a_{1}\;,a_{2}\;....\;a_{n}\;$ are in HP
$\frac{1}{a_{1}}\;,\frac{1}{a_{2}}\;,....\;\frac{1}{a_{n}}\;$ are in AP
$\frac{1}{a_{2}}-\frac{1}{a_{1}}=\frac{1}{a_{3}}-\frac{1}{a_{2}}=\frac{1}{a_{4}}-\frac{1}{a_{3}}=d$
$a_{1}\;a_{2}=\frac{1}{d}\;(a_{1}-a_{2})$
$a_{2}\;a_{3}=\frac{1}{d}\;(a_{2}-a_{3})$
$a_{1}a_{2}+a_{2}a_{3}+\;...\;+a_{n-1}a_{n}=\frac{1}{d}\;[a_{1}-a_{2}+a_{2}-a_{3}+....+a_{n-1}-a_{n}]$
$=\frac{1}{d}\;(a_{1}-a_{n})$
$\frac{1}{a_{n}}=\frac{1}{a_{1}}+(n-1)d$
$d=\frac{a_{1}-a_{n}}{a_{1}a_{n}(n-1)}$
$S=a_{1}a_{n}(n-1).$