Harmonic Progression
Category : 11th Class
The sequence is said to be in H.P. If the reciprocal of its terms gives the A.P. It has got wide application in the field of geometry and theory of sound. The questions are generally solved by inverting the terms and using the property of arithmetic progression.
Three numbers a, b, c are said to be in HP if,
\[\frac{a}{c}=\frac{a-b}{a-c}\]
Harmonic Mean (HM)
If 'a' and 'b' be any two terms, then their harmonic mean is given by \[HM=\frac{2ab}{a+b}\].
Relation between AM, GM, and HM
Since we know that,
\[AM=\frac{a+b}{2},\,\,GM=\sqrt{ab}\,and\,HM=\frac{2ab}{a+b}\]
Then,
\[AM\times HM=\frac{a+b}{2}\times \frac{2ab}{a\times b}=ab={{G}^{2}}\]
\[AM\times HM=G{{M}^{2}}\]
Form the above relation we can say that AM > GM and GM is intermediate value between AM and HM, therefore GM > HM. Hence we can say that AM > GM > HM.
Also the relation between A and G is given by,
\[AM-GM=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{2}}\]
The harmonic mean between two numbers is \[\frac{48}{5}\] and geometric mean is 12. The two numbers are:
(a) (3 & 20)
(b) (2 & 12)
(c) (6 & 24)
(d) (5 & 32)
(e) None of these
Answer: (c)
Explanation
Let the two number be ‘a’ and ‘b’.
Then, \[HM=\frac{2ab}{a+b}\,and\,GM=\sqrt{ab}\]
Putting the value of HM and GM in the above relation we get,
\[\frac{48}{5}=\frac{2ab}{a+b}\,and\,12=\sqrt{ab}\]
On solving these two equations we get, A = 6 & b = 24
If \[{{p}^{th}}\] term of HP is equal to \[{{q}^{th}}\] and the p, then (p + q) term of the series is.
(a) \[\frac{pq}{p+q}\]
(b) \[\frac{p-q}{p+q}\]
(c) \[\frac{p-q}{pq}\]
(d) \[\frac{p+q}{pq}\]
(e) None of these
Answer: (a)
Explanation
Let 'a' and 'd' be the first term and common difference of an AP,
Then the \[{{p}^{th}}\,and\,{{q}^{th}}\] term of the AP is \[{{a}_{p}}(p-1)d\,and\,{{a}_{p}}=a+(q-1)d\]
For HP series the corresponding terms are,
\[\Rightarrow \,\,\frac{1\,}{pq}=a+(p-1)d\,and\,\frac{1}{p}=a+(q-1)d\]
On solving the above equation we get,
\[a=\frac{1}{pq}and\,\,d=\frac{1}{pq}\]
Therefore, \[{{(p+q)}^{th}}\,term\,=\frac{p+q}{pq}\]
Hence \[{{(p+q)}^{th}}\,\] of the HP is given by \[\frac{pq}{p+q}\]
For any two numbers the ratio of HM : GM is 12 :13, and then the ratio of the two numbers is given by:
(a) 3 : 8
(b) 2 : 5
(c) 4 : 9
(d) 5 : 7
(e) None of these
Answer: (c)
Explanation
Let the two number be 'a' and 'b'. Then,
\[HM=\frac{2ab}{a+b\,}and\,GM=\sqrt{ab}\]
\[\Rightarrow \,\,\,\,\frac{HM}{GM}=\frac{\frac{2ab}{a+b}}{\sqrt{ab}}\]
\[\Rightarrow \,\,\,\frac{12}{13}=\frac{2\sqrt{ab}}{a+b}\]
\[\Rightarrow \,\,\,\frac{13}{12}=\frac{a+b}{2\sqrt{ab}}\]
By componendo and dividendo, we get
\[\Rightarrow \,\,\,\frac{25}{1}=\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}}\]
\[\Rightarrow \,\,\,\frac{25}{1}=\frac{{{\left( \sqrt{a}+\sqrt{b} \right)}^{2}}}{{{\left( \sqrt{a}-\sqrt{b} \right)}^{2}}}\]
\[\Rightarrow \,\,\frac{5}{1}=\frac{\left( \sqrt{a}+\sqrt{b} \right)}{\left( \sqrt{a}-\sqrt{b} \right)}\]
\[\Rightarrow \,\,\frac{a}{b}=\frac{9}{4}\]
The number of bricks arranged in a complete pyramid on a square base of side 10 units is given by.
(a) 290
(b) 385
(c) 425
(d) 525
(e) None of these
Answer: (b)
Explanation
The relation used here is given by \[S={{n}^{2}}+{{(n-1)}^{2}}+{{(n-2)}^{2}}+----+1\]
The number of shots arranged in a graveyard in the shape of pyramid whose base is in the form of equilateral triangle of side 8 units is given by.
(a) 100
(b) 140
(c) 120
(d) 64
(e) None of these
Answer: (c)
Explanation
The relation used here is given by,
The number of shot in each layer is \[S=n+(n-1)+(n-2)+----+1\]
Total number of shot is given by, \[S=\frac{n(n+1)(n+2)}{6}\]
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