If the sixth term of a H.P. is frac{1}{61} an | vedclass

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(C) Let the first term of the $A.P.$ be $a$ and the common difference be $d.$Since the $n^{th}$ term of a $H.P.$ is the reciprocal of the $n^{th}$ ter

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$\frac{1}{6}$

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(C) Let the first term of the $A.P.$ be $a$ and the common difference be $d.$Since the $n^{th}$ term of a $H.P.$ is the reciprocal of the $n^{th}$ term of an $A.P.,$ we have:$T_6$ of $H.P. = \frac{1}{61} \implies T_6$ of $A.P. = a + 5d = 61$ $(i)$$T_{10}$ of $H.P. = \frac{1}{105} \implies T_{10}$ of $A.P. = a + 9d = 105$ $(ii)$Subtracting equation $(i)$ from $(ii):$$(a + 9d) - (a + 5d) = 105 - 61$$4d = 44 \implies d = 11$Substituting $d = 11$ into equation $(i):$$a + 5(11) = 61$$a + 55 = 61 \implies a = 6$Thus,the first term of the $H.P.$ is $\frac{1}{a} = \frac{1}{6}.$

If the sixth term of a $H.P.$ is $\frac{1}{61}$ and its tenth term is $\frac{1}{105},$ then the first term of that $H.P.$ is

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