Question

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). ${56}/{15}\;$.

Answer 1

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.

Complete step-by-step answer:
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.