
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
483.3k+ views
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.
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.
Recently Updated Pages
Master Class 11 Accountancy: Engaging Questions & Answers for Success

Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE

The highest possible oxidation states of Uranium and class 11 chemistry CBSE

Find the value of x if the mode of the following data class 11 maths CBSE

Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE

A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE

Trending doubts
Define least count of vernier callipers How do you class 11 physics CBSE

The combining capacity of an element is known as i class 11 chemistry CBSE

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Find the image of the point 38 about the line x+3y class 11 maths CBSE

Can anyone list 10 advantages and disadvantages of friction

Distinguish between Mitosis and Meiosis class 11 biology CBSE
