If the 7^{th} term of a H.P. is frac{1}{10} a | vedclass

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

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$\frac{1}{49}$

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(D) Let the corresponding $A.P.$ have the first term $a$ and common difference $d$.Since the $n^{th}$ term of $H.P.$ is the reciprocal of the $n^{th}$ term of $A.P.$,we have:$7^{th}$ term of $A.P. = a + 6d = 10$$12^{th}$ term of $A.P. = a + 11d = 25$Subtracting the first equation from the second:$(a + 11d) - (a + 6d) = 25 - 10$$5d = 15 \Rightarrow d = 3$Substituting $d = 3$ into $a + 6d = 10$:$a + 6(3) = 10 \Rightarrow a + 18 = 10 \Rightarrow a = -8$The $20^{th}$ term of the $A.P.$ is $T_{20} = a + 19d = -8 + 19(3) = -8 + 57 = 49$.Therefore,the $20^{th}$ term of the $H.P.$ is the reciprocal of $49$,which is $\frac{1}{49}$.

If the $7^{th}$ term of a $H.P.$ is $\frac{1}{10}$ and the $12^{th}$ term is $\frac{1}{25}$,then the $20^{th}$ term is

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