question_answer

If sixth term of a H.P. is \[\frac{1}{61}\] and its tenth term is \[\frac{1}{105},\] then first term of that H.P. is [Karnataka CET 2001]

A) \[\frac{1}{28}\]

B) \[\frac{1}{39}\]

C) \[\frac{1}{6}\]

D) \[\frac{1}{17}\]

Correct Answer: C

Solution :

 of H.P. \[=\frac{1}{61}\,\Rightarrow {{T}_{6}}\] of A.P. = 61 and  of H.P. \[=\frac{1}{105}\Rightarrow \,{{T}_{10}}\] of A.P. = 105, so \[a+5d=61\]    .......(i)  and  \[a+9d=105\]    ?..(ii) From (i) and (ii),\[a=6\] Therefore first term of H.P.\[=\frac{1}{a}=\frac{1}{6}\].