How To Solve Harmonic Progression Questions Quickly

Type 1: n^th term of an HP : $a_{n} = \frac{1}{a+(n-1)d}$

Question 1. If the sum of reciprocals of first 11 terms of an HP series is 110, find the 6th term.

Options:

A. 10

B. $\frac{1}{10}$

C. $\frac{1}{6}$

D. $\frac{1}{5}$

Solution    Reciprocals of first 11 terms of an HP will be AP
                   Therefore, $s_{n} =\frac{n}{2}[2a+(n-1)\times d]$

                   $S_{n}$ = 110
                   n = 11
                   $110 = \frac{11}{2}[2a + (11 − 1) d]$
                   $110 = \frac{11}{2} \times [2a + 10d]$
                   220 = 22a + 110d
                   22a + 110d = 220
                   a + 5d = 10

                  which is the 6^th term of the AP series

                  Therefore, the 6^th term in HP = $\frac{1}{10}$

                  Correct option: B

Question 2 Find the infinite sum of series $\frac{1}{(3^{2}-4)} + \frac{1}{(4^{2}-4)} + \frac{1}{(5^{2}-4)}$ …….

Options:

A. 0.45

B. 0.69

C. 0.87

D. 2

Solution     In the given series,

$\frac{1}{(3^{2}-4)} + \frac{1}{(4^{2}-4)} + \frac{1}{(5^{2}-4)}$

$\frac{1}{(3^{2}-2^{2})} + \frac{1}{(4^{2}-2^{2})} + \frac{1}{(5^{2}-2^{2})}$

We know, that a²-b² = (a-b) (a+b)

S = $\frac{1}{(3-2)(3+2)} + \frac{1}{(4-2)(4+2)} +\frac{1}{(5-2)(5+2)}$ …..

S = $\frac{1}{4}(1-\frac{1}{5}) + \frac{1}{4}(\frac{1}{2} – \frac{1}{6}) + \frac{1}{4}(\frac{1}{3} – \frac{1}{7})$

S = $\frac{1}{4}\times 1.833$

S = 0.45

Correct option: A

Question 3. Identify the 4th and 8th term of the series 6, 4, 3….

Options:

A. $\frac{12}{5}$ and $\frac{13}{9}$

B. $\frac{12}{9}$ and $\frac{13}{5}$

C. $\frac{12}{5}$ and $\frac{12}{9}$

D. $\frac{12}{5}$ and $\frac{12}{7}$

Solution     Consider the series in the form of $\frac{1}{6}, \frac{1}{4}, \frac{1}{3}$

We know that $T_{2} – T_{1} = T_{3} – T_{2} = \frac{1}{12}$

As $\frac{1}{6}, \frac{1}{4}, \frac{1}{3}$ is an AP, Thus 4^th term of the AP = a +3d = $\frac{1}{ 6} + 3 \times \frac{1}{12} = \frac{5}{12}$

Similarly the 8^th term of the series = a +7d = $\frac{1} {6} + 7 \times \frac{1}{12} = \frac{9}{12}$

Thus the 4^th term of an AP = $\frac{12}{5}$ and 8^th term = $\frac{12}{9}$ respectively.

Correct option: C