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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
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The correct answer is:
$\frac{1}{6}$
$T_6 \quad=\frac{1}{61} \Rightarrow T_6$ of H.P. $\quad$ A.P. $=61$
and $T_{10}$ of H.P. $=\frac{1}{105} \Rightarrow T_{10}$ of A.P. = 105,
so, $a+5 d=61 \ldots(i)$ and $a+9 d=105 \ldots(ii)$
From (i) and (ii), $a=6$
Therefore first term of H.P. minimise
$=\frac{1}{a}=\frac{1}{6}$
and $T_{10}$ of H.P. $=\frac{1}{105} \Rightarrow T_{10}$ of A.P. = 105,
so, $a+5 d=61 \ldots(i)$ and $a+9 d=105 \ldots(ii)$
From (i) and (ii), $a=6$
Therefore first term of H.P. minimise
$=\frac{1}{a}=\frac{1}{6}$
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