 If the 7th term of a harmonic progression is 8 and the 8th term is 7, than its 15th term is:-A). 16.B). 14.C). ${27}/{14}\;$.D). ${56}/{15}\;$. Verified
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Hint: The general term of the HP is given by $\dfrac{1}{a+\left( n-1 \right)d}$ where a is reciprocal of first term of HP and d is common difference in reciprocal of each term are this to get the result.

Now the 7th term of HP is equal to 8.
As described in the hint the general term is given by $\dfrac{1}{a+\left( n-1 \right)d}\ =\ {{T}_{n}}$
Now for the beneath term $n=7$
$\therefore \ {{T}_{7}}=\dfrac{1}{a+6d}$
$\because \ {{T}_{7}}=\ 8$
$\Rightarrow \ \dfrac{1}{a+6d}\ =\ 8$
$\Rightarrow \ a+6d=\ \dfrac{1}{8}$ (1)
Now the 8th term of HP is equal to 7.
$\therefore \,{{T}_{8}}\ =\ 7\ =\dfrac{1}{a+\left( 8-7 \right)d}$
$7\ =\ \dfrac{1}{a+7d}$
$\therefore \ a+7d\ =\ \dfrac{1}{7}$ (2)
Now on subtracting $eq\ \left( 1 \right)\ \text{from}\ \text{eq}\ \left( 2 \right)$
$\left( a+7d \right)-\left( a+6d \right)\ =\ \dfrac{1}{7}-\dfrac{1}{8}$
$a+7d-a-6d=\dfrac{8-7}{56}$
$\therefore \ d\ =\ \dfrac{1}{56}$
Now since we know the value of d, we can compute the value of a by substituting in equation (1), we get
$a\ +\ 6\ \times \ \dfrac{1}{56}=\ \dfrac{1}{8}$
$a\ =\ \dfrac{1}{8}-\dfrac{6}{56}$
$a\ =\ \dfrac{7-6}{56}$
$a\ =\ \dfrac{1}{56}$
Now if we want to compute the 15th term we can substitute the value of $a,d\ and\ n=15$ in general terms.
$\therefore \ {{T}_{15\ }}\ =\ \dfrac{1}{\dfrac{1}{56}+\left( 15-1 \right)\times \dfrac{1}{56}}$
$\Rightarrow \ \ \ {{T}_{15}}=\ \ \dfrac{1}{\dfrac{1}{56}+\dfrac{14}{56}}$
\Rightarrow \ \ \ {{T}_{15}}\ \ \ \ =\dfrac{1}{\begin{align} & 15 \\ & \overline{56} \\ \end{align}}
$\therefore \ {{T}_{15}}\ \ \ =\ \ \dfrac{56}{15}$
$\therefore$ The 15th term of HP is $\dfrac{56}{15}$

The correct option is D.

Note: If you can notice the general term in HP is reciprocal of general term of AP (arithmetic progression).
Therefore series of terms are a HP series when their reciprocal are in AP.