Compute the 16th term of HP if the 6th and Ilth term of HP are 10 and 18, respectively

To find the 16th term of a Harmonic Progression (HP), we need to first transform the HP into an Arithmetic Progression (AP) by considering the reciprocals of the terms. Let the general terms of the HP be denoted as \( t_n \), where \( t_n \) is the \( n \)-th term of the HP.

In HP:
- The 6th term is \( t_6 = 10 \),
- The 11th term is \( t_{11} = 18 \).

The corresponding terms in the AP will be:
- \( \frac{1}{t_6} = \frac{1}{10} \),
- \( \frac{1}{t_{11}} = \frac{1}{18} \).

Let the terms of the corresponding AP be \( a_n = \frac{1}{t_n} \), and the common difference of the AP be \( d \). The formula for the \( n \)-th term of an AP is given by:

\[
a_n = a_1 + (n-1)d
\]

For the 6th and 11th terms in the AP:
\[
a_6 = a_1 + 5d = \frac{1}{10}
\]
\[
a_{11} = a_1 + 10d = \frac{1}{18}
\]

Now, we have two equations:
1. \( a_1 + 5d = \frac{1}{10} \)
2. \( a_1 + 10d = \frac{1}{18} \)

### Solving the system of equations:
Subtract the first equation from the second:

\[
(a_1 + 10d) - (a_1 + 5d) = \frac{1}{18} - \frac{1}{10}
\]
\[
5d = \frac{1}{18} - \frac{1}{10}
\]

Now, we find the common denominator for the fractions on the right-hand side:

\[
\frac{1}{18} - \frac{1}{10} = \frac{5 - 9}{90} = \frac{-4}{90} = -\frac{2}{45}
\]

So,
\[
5d = -\frac{2}{45}
\]
\[
d = -\frac{2}{45} \times \frac{1}{5} = -\frac{2}{225} = -\frac{2}{225}
\]

Now, substitute the value of \( d \) into the first equation:

\[
a_1 + 5 \times \left( -\frac{2}{225} \right) = \frac{1}{10}
\]
\[
a_1 - \frac{10}{225} = \frac{1}{10}
\]
\[
a_1 = \frac{1}{10} + \frac{2}{45} = \frac{9 + 2}{90} = \frac{11}{90}
\]

### Finding the 16th term:
Now that we have \( a_1 = \frac{11}{90} \) and \( d = -\frac{2}{225} \), we can find the 16th term of the AP:

\[
a_{16} = a_1 + 15d = \frac{11}{90} + 15 \times \left( -\frac{2}{225} \right)
\]
\[
a_{16} = \frac{11}{90} - \frac{30}{225}
\]
Simplify \( \frac{30}{225} \) to \( \frac{2}{15} \), so:
\[
a_{16} = \frac{11}{90} - \frac{2}{15}
\]
Convert \( \frac{2}{15} \) to have a denominator of 90:
\[
a_{16} = \frac{11}{90} - \frac{12}{90} = \frac{-1}{90}
\]

### 16th term of the HP:
Since \( a_{16} = \frac{1}{t_{16}} \), we take the reciprocal to find \( t_{16} \):

\[
t_{16} = -90
\]

Thus, the 16th term of the HP is \( \boxed{-90} \).

Would you like further details or have any questions?

Here are 5 related questions to expand on the concept:
1. What is the relationship between HP and AP?
2. How can you find the nth term of an HP if only the first term and common difference are known?
3. How do you calculate the sum of a harmonic progression?
4. Can a harmonic progression have negative terms, and what does it imply?
5. What is the difference between harmonic mean and arithmetic mean?

**Tip:** Always remember that the reciprocal of a harmonic progression forms an arithmetic progression.

Compute the 16th term of HP if the 6th and 11th term of HP are 10 and 18, respectively.

Learn how to compute the 16th term of a Harmonic Progression when the 6th and 11th terms are given as 10 and 18. This problem involves transforming the HP into an Arithmetic Progression, solving a system of linear equations, and applying key formulas for AP and HP. Suitable for students in Grades 10-12.

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