If the 5^{th} term of a H.P. is frac{1}{45} a | vedclass

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(A) Let the corresponding $A.P.$ be $a, a+d, a+2d, \dots$Since the $5^{th}$ term of the $H.P.$ is $\frac{1}{45}$,the $5^{th}$ term of the $A.P.$ is $4

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$1/89$

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(A) Let the corresponding $A.P.$ be $a, a+d, a+2d, \dots$Since the $5^{th}$ term of the $H.P.$ is $\frac{1}{45}$,the $5^{th}$ term of the $A.P.$ is $45$.So,$a + 4d = 45$ $(i)$Since the $11^{th}$ term of the $H.P.$ is $\frac{1}{69}$,the $11^{th}$ term of the $A.P.$ is $69$.So,$a + 10d = 69$ $(ii)$Subtracting $(i)$ from $(ii)$,we get $(a + 10d) - (a + 4d) = 69 - 45$,which gives $6d = 24$,so $d = 4$.Substituting $d = 4$ into $(i)$,we get $a + 4(4) = 45$,so $a + 16 = 45$,which gives $a = 29$.The $16^{th}$ term of the $A.P.$ is $a + 15d = 29 + 15(4) = 29 + 60 = 89$.Therefore,the $16^{th}$ term of the $H.P.$ is $\frac{1}{89}$.

If the $5^{th}$ term of a $H.P.$ is $\frac{1}{45}$ and the $11^{th}$ term is $\frac{1}{69}$,then its $16^{th}$ term will be:

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