# If $a_{1}, a_{2}, a_{3},.....$ are in HP and $f (k)$ = $\sum_{r=1}^{n}\;a_{r}-a_{k}$, then $\large\frac{a_{1}}{f (1)}$, $\large\frac{a_{2}}{f (2)}$, $\large\frac{a_{3}}{f (3)}$,...,$\large\frac{a_{n}}{f (n)}$ are in

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

Answer : (c) HP
Explanation : $\;f (k)=\sum_{r=1}^{n}\;a_{r}-a_{k}$
$\sum_{r=1}^{n}\;a_{r}=a_{k}+ f (k)$
$a_{1}+f (1)+=a_{2}+f (2)=a_{3}+f (3)+......$
$Now\;a_{1} , a_{2} , a_{3} ....\;are\;in\;HP$
$\large\frac{1}{a_{1}} , \large\frac{1}{a_{2}} , \large\frac{1}{a_{3}},...are\;in\;AP$
$\large\frac{a_{1}+f (1)}{a_{1}}\;,\large\frac{a_{2}+f (2)}{a_{2}}\;\large\frac{a_{3}+f (3)}{a_{3}}\;,..\;are\;in\;AP$
$\large\frac{f (1)}{a_{1}} , \large\frac{f (2)}{a_{2}} , \large\frac{f (3)}{a_{3}},...are\;in\;AP$
$\large\frac{a_{1}}{f (1)}\;,\large\frac{a_{2}}{f (2)}\;,\large\frac{a_{3}}{f (3)}\;...are\;in\;HP$
answered Jan 23, 2014 by