Question

Which of the following are harmonic progressions?
(i) $1,\frac{1}{4},\frac{1}{7},\frac{1}{10},......$ (ii) $1,\frac{2}{3},\frac{1}{2},\frac{2}{5},........$ (iii) $\frac{1}{2},\frac{1}{6},\frac{1}{18},......$
(iv) $\frac{1}{3},\frac{1}{7},\frac{1}{11},........$ (v) 6, 4, 3, ......... (vi) $1,\frac{1}{2},\frac{1}{4},......$

Answer 1

In harmonic progression, the numerator of each term is set to be $1$ and the set of terms at the denominator forms an A.P.

So, we need to check whether the terms at denominators form an A.P.

(i) $1,\frac{1}{4},\frac{1}{7},\frac{1}{10},......$

The terms at denominator are $1,4,7,10,...$

Now, $4-1=3,7-4=3,10-7=3$

So, they have a common difference and that is $3$.

$\therefore$ The denominators form an A.P.

Thus, the given sequence is a H.P.

(ii) $1,\frac{2}{3},\frac{1}{2},\frac{2}{5},........$

We can write the given sequence as $\frac{1}{1},\frac{1}{\frac{3}{2}},\frac{1}{2},\frac{1}{\frac{5}{2}},........$

The terms at denominator are $1,\frac{3}{2},2,\frac{5}{2},...$

Now, $\frac{3}{2}-1=\frac{1}{2},2-\frac{3}{2}=\frac{1}{2},\frac{5}{2}-2=\frac{1}{2}$

So, they have a common difference and that is $\frac{1}{2}$.

$\therefore$ The denominators form an A.P.

Thus, the given sequence is H.P.

(iii) $\frac{1}{2},\frac{1}{6},\frac{1}{18},......$

The terms at denominator are $2,6,18,...$

Now, $6-2=4,18-6=12$

So, there difference is not common.

$\therefore$ The denominators does not form an A.P.

Thus, the given sequence is not an H.P.

(iv) $\frac{1}{3},\frac{1}{7},\frac{1}{11},........$

The terms at denominator are $3,7,11,...$

Now, $7-3=4,11-7=4$

So, they have a common difference and that is $4$.

$\therefore$ The denominators form an A.P.

Thus, the given sequence is a H.P

(v) 6, 4, 3, .........

We can write the given sequence as $\frac{1}{\frac{1}{6}},\frac{1}{\frac{1}{4}},\frac{1}{\frac{1}{3}},........$

The terms at denominator are $\frac{1}{6},\frac{1}{4},\frac{1}{3}...$

Now, $\frac{1}{4}-\frac{1}{6}=\frac{1}{12},\frac{1}{3}-\frac{1}{4}=\frac{1}{12}$

So, they have a common difference and that is $\frac{1}{12}$.

$\therefore$ The denominators form an A.P.

Thus, the given sequence is H.P.

(vi) $1,\frac{1}{2},\frac{1}{4},......$

The terms at denominator are $1,2,4,...$

Now, $2-1=1,4-2=2$

So, there difference is not common.

$\therefore$ The denominators does not form an A.P.

Thus, the given sequence is not an H.P.

Hence, (i), (ii), (iv) and (v) are harmonic progressions.