The 4^{th} term of a H.P. is frac{3}{5} and 8 | vedclass

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(B) If a sequence is in $H.P.$,then its reciprocal is in $A.P.$Let the $A.P.$ be $a, a+d, a+2d, \dots$The $4^{th}$ term of $H.P.$ is $\frac{3}{5}$,so

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$\frac{3}{7}$

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(B) If a sequence is in $H.P.$,then its reciprocal is in $A.P.$Let the $A.P.$ be $a, a+d, a+2d, \dots$The $4^{th}$ term of $H.P.$ is $\frac{3}{5}$,so the $4^{th}$ term of $A.P.$ is $\frac{5}{3}$. Thus,$a + 3d = \frac{5}{3}$.The $8^{th}$ term of $H.P.$ is $\frac{1}{3}$,so the $8^{th}$ term of $A.P.$ is $3$. Thus,$a + 7d = 3$.Subtracting the first equation from the second: $(a + 7d) - (a + 3d) = 3 - \frac{5}{3} \implies 4d = \frac{4}{3} \implies d = \frac{1}{3}$.Substituting $d$ in the first equation: $a + 3(\frac{1}{3}) = \frac{5}{3} \implies a + 1 = \frac{5}{3} \implies a = \frac{2}{3}$.The $6^{th}$ term of the $A.P.$ is $a + 5d = \frac{2}{3} + 5(\frac{1}{3}) = \frac{2}{3} + \frac{5}{3} = \frac{7}{3}$.Therefore,the $6^{th}$ term of the $H.P.$ is the reciprocal of $\frac{7}{3}$,which is $\frac{3}{7}$.

The $4^{th}$ term of a $H.P.$ is $\frac{3}{5}$ and $8^{th}$ term is $\frac{1}{3},$ then its $6^{th}$ term is

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