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

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(B) Let the $H.P.$ be $\frac{1}{a}, \frac{1}{a+d}, \frac{1}{a+2d}, \dots$ where the corresponding $A.P.$ is $a, a+d, a+2d, \dots$ Given that the $4^{t

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$\frac{3}{7}$

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(B) Let the $H.P.$ be $\frac{1}{a}, \frac{1}{a+d}, \frac{1}{a+2d}, \dots$ where the corresponding $A.P.$ is $a, a+d, a+2d, \dots$ Given that the $4^{th}$ term of $H.P.$ is $\frac{3}{5},$ the $4^{th}$ term of $A.P.$ is $\frac{5}{3}.$ Thus,$a + 3d = \frac{5}{3} \dots (1)$ Given that the $8^{th}$ term of $H.P.$ is $\frac{1}{3},$ the $8^{th}$ term of $A.P.$ is $3.$ Thus,$a + 7d = 3 \dots (2)$ Subtracting $(1)$ from $(2),$ we get $(a + 7d) - (a + 3d) = 3 - \frac{5}{3} \implies 4d = \frac{4}{3} \implies d = \frac{1}{3}.$ Substituting $d = \frac{1}{3}$ into $(1),$ we get $a + 3(\frac{1}{3}) = \frac{5}{3} \implies a + 1 = \frac{5}{3} \implies a = \frac{2}{3}.$ The $6^{th}$ term of the $A.P.$ is $a + 5d = \frac{2}{3} + 5(\frac{1}{3}) = \frac{2+5}{3} = \frac{7}{3}.$ Therefore,the $6^{th}$ term of the $H.P.$ is the reciprocal of $\frac{7}{3},$ which is $\frac{3}{7}.$ Hence,the correct option is $B$.

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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