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

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(A) Let the corresponding $A.P.$ be $a, a+d, a+2d, \dots$ such that the $n^{th}$ term of $H.P.$ is $1/(a+(n-1)d)$.Given that the ${5^{th}}$ term of $H

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$1/89$

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(A) Let the corresponding $A.P.$ be $a, a+d, a+2d, \dots$ such that the $n^{th}$ term of $H.P.$ is $1/(a+(n-1)d)$.Given that the ${5^{th}}$ term of $H.P.$ is $1/45$,therefore the ${5^{th}}$ term of $A.P.$ is $a + 4d = 45$ $(i)$.Given that the ${11^{th}}$ term of $H.P.$ is $1/69$,therefore the ${11^{th}}$ term of $A.P.$ is $a + 10d = 69$ $(ii)$.Subtracting $(i)$ from $(ii)$:$(a + 10d) - (a + 4d) = 69 - 45$$6d = 24 \implies d = 4$.Substituting $d = 4$ in $(i)$:$a + 4(4) = 45 \implies a + 16 = 45 \implies a = 29$.Now,the ${16^{th}}$ term of the $A.P.$ is $a + 15d = 29 + 15(4) = 29 + 60 = 89$.Thus,the ${16^{th}}$ term of the $H.P.$ is $1/89$.

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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