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

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(D) Let the harmonic progression $(H.P.)$ be $H_1, H_2, \dots, H_n$. The corresponding arithmetic progression $(A.P.)$ is $A_1, A_2, \dots, A_n$ where

Vedclass · Accepted Answer

$\frac{56}{15}$

Vedclass · Answer

(D) Let the harmonic progression $(H.P.)$ be $H_1, H_2, \dots, H_n$. The corresponding arithmetic progression $(A.P.)$ is $A_1, A_2, \dots, A_n$ where $A_n = \frac{1}{H_n}$.Given that the $7^{th}$ term of $H.P.$ is $8$,the $7^{th}$ term of $A.P.$ is $A_7 = \frac{1}{8}$.Given that the $8^{th}$ term of $H.P.$ is $7$,the $8^{th}$ term of $A.P.$ is $A_8 = \frac{1}{7}$.For an $A.P.$,$A_n = a + (n-1)d$. Thus:$a + 6d = \frac{1}{8}$ (Equation $1$)$a + 7d = \frac{1}{7}$ (Equation $2$)Subtracting Equation $1$ from Equation $2$ gives $d = \frac{1}{7} - \frac{1}{8} = \frac{8-7}{56} = \frac{1}{56}$.Substituting $d$ into Equation $1$: $a + 6(\frac{1}{56}) = \frac{1}{8} \implies a = \frac{1}{8} - \frac{6}{56} = \frac{7-6}{56} = \frac{1}{56}$.The $15^{th}$ term of the $A.P.$ is $A_{15} = a + 14d = \frac{1}{56} + 14(\frac{1}{56}) = \frac{15}{56}$.Therefore,the $15^{th}$ term of the $H.P.$ is $H_{15} = \frac{1}{A_{15}} = \frac{56}{15}$.

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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