Harmonic Progression Formulas

Question 1:
The sum of the first n terms of a harmonic progression is given by (S_{n} = \frac{n^{2} + 4}{3n}). What is the common difference between the terms of this progression?

A) (\frac{1}{3})
B) (\frac{2}{3})
C) (\frac{3}{4})
D) (\frac{4}{5})

Solution: A) (\frac{1}{3})
The sum of the first n terms of a harmonic progression is given by (S_n = \frac{n}{a_1} = \frac{n}{\frac{1}{a}} = na), where a is the common difference. Comparing this with the given formula, we have (\frac{n^2 + 4}{3n} = na). Solving for a, we get (a = \frac{1}{3}).

Question 2:
In a harmonic progression, the n_{th} term is (\frac{1}{n+2}). What is the sum of the first 10 terms of this progression?

A) (\frac{185}{99})
B) (\frac{99}{185})
C) (\frac{135}{70})
D) (\frac{70}{135})

Solution: A) (\frac{185}{99})[/latex]
The sum of the first n terms of a harmonic progression is given by (S_{n}= a_{1}
(\frac{1}{1} + \frac{1}{2} + \ldots + \frac{1}{n}), where a_{1} is the first term. Substituting the given n_{th} term, we get (a_{1} = \frac{1}{n+2}). Then, (S_{10} = \frac{1}{n+2} (\frac{1}{1} + \frac{1}{2} + \ldots + \frac{1}{10}) = \frac{185}{99}).

Question 3:
The sum of an infinite harmonic progression is (\frac{5}{3}). What is the sum of the squares of the terms in this progression?

A) (\frac{10}{3})
B) (\frac{25}{9})
C) (\frac{15}{4})
D) (\frac{9}{5})

Solution: A) (\frac{10}{3})
The sum of the squares of the terms in a harmonic progression is given by ( S_{squares} = a_{1}^{2}(\frac{1}{1^{2}} + \frac{1}{2^{2}} + …..) , where a_{1} is the first term. Given the sum of the infinite harmonic progression as (\frac{5}{3}), we can use the result (S_{\infty} = a_{1} (\frac{1}{1} + \frac{1}{2} + \ldots) = \frac{5}{3}) to find a_{1}. Then, S_{squares} = \frac{10}{3}).

Question 4:
In a harmonic progression, the sum of the first 6 terms is 3 times the sum of their reciprocals. What is the sum of the first 12 terms of this progression?

A) 2
B) 3
C) 4
D) 5

Solution:C) 4
Let (S_{n}) be the sum of the first n terms and (S'_{n}) be the sum of the reciprocals of the first n terms. The given condition can be written as ((S_{6}) = (3S'_{6}) ). Using the formula for the sum of the first n terms of a harmonic progression, we have ((S_{6}) = \frac{6}{a})and ((S'_{6}) =\frac{6}{a}). Thus, (\frac{6}{a} = 3 * \frac{6}{a}), which implies (a = 3). Then, ((S_{12}) = \frac{12}{a} = 4).

Question 5:
In a harmonic progression, the sum of the first 5 terms is 6, and the sum of their cubes is 405. What is the first term of this progression?

A) (\frac{1}{3})
B) (\frac{1}{4})
C) (\frac{1}{5})
D) (\frac{1}{6})

Solution: A) (\frac{1}{3})
Let a be the first term and d be the common difference of the harmonic progression. The sum of the first 5 terms is given by ((S_{5}) = 5a + 10d = 6). The sum of their cubes is given by (S_{cubes}) = a^{3} + (a+d)^{3} + (a+2d)^{3} + (a+3d)^{3} + (a+4d)^{3} = 405). Using the value of ((S_{5})), we can solve for d, and then using ((S_{cubes})), we can solve for a, which turns out to be (\frac{1}{3}).