If the 7^{th} term of a harmonic progression | vedclass

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(D) Let the harmonic progression $(H.P.)$ be $H_1, H_2, H_3, \dots$. The corresponding arithmetic progression $(A.P.)$ is $\frac{1}{H_1}, \frac{1}{H_2

Vedclass · Accepted Answer

$\frac{56}{15}$

Vedclass · Answer

(D) Let the harmonic progression $(H.P.)$ be $H_1, H_2, H_3, \dots$. The corresponding arithmetic progression $(A.P.)$ is $\frac{1}{H_1}, \frac{1}{H_2}, \frac{1}{H_3}, \dots$.Given that the $7^{th}$ term of the $H.P.$ is $8$,the $7^{th}$ term of the corresponding $A.P.$ is $\frac{1}{8}$.Given that the $8^{th}$ term of the $H.P.$ is $7$,the $8^{th}$ term of the corresponding $A.P.$ is $\frac{1}{7}$.Let $a$ be the first term and $d$ be the common difference of the $A.P.$$a + 6d = \frac{1}{8}$  --- $(1)$$a + 7d = \frac{1}{7}$  --- $(2)$Subtracting $(1)$ from $(2)$: $d = \frac{1}{7} - \frac{1}{8} = \frac{8-7}{56} = \frac{1}{56}$.Substituting $d$ in $(1)$: $a + 6(\frac{1}{56}) = \frac{1}{8} \implies a + \frac{3}{28} = \frac{1}{8} \implies a = \frac{1}{8} - \frac{3}{28} = \frac{7-6}{56} = \frac{1}{56}$.The $15^{th}$ term of the $A.P.$ is $a + 14d = \frac{1}{56} + 14(\frac{1}{56}) = \frac{15}{56}$.Therefore,the $15^{th}$ term of the $H.P.$ is the reciprocal of the $15^{th}$ term of the $A.P.$,which is $\frac{56}{15}$.

If the $7^{th}$ term of a harmonic progression is $8$ and the $8^{th}$ term is $7$,then its $15^{th}$ term is

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