## Harmonic Progression:

If the reciprocals of the terms of the sequence form an arithmetic progression, then the terms of the given sequence are said to be Harmonic Progression (H.P.).

*n ^{th} term of H.P.*

If the sequence in H.P. is ** 1/a, 1/(a + d), 1/(a + 2d), ………..**, then its n

^{th}term is

**.**

*1/[a + (n – 1)d]*## Harmonic Means:

If several numbers are in H.P., then the numbers lying between the first and the last will be called the Harmonic Means between the first and the last numbers.

If three numbers are in H.P., then the middle number will be called the Harmonic Mean (H.M.) between the first and the last.

Let H be the H.M. between a and b.

∴ a, H, b are in H.P.

1/a, 1/H, 1/b, are in A.P. ⇒ 2/H = 1/a + 1/b ⇒ 2/H = (a + b)/ab ⇒ H/2 = ⇒ H = 2ab/(a + b) |

Example-Insert 4 H.M. between 20 and 5. Let HSolution-_{1}, H_{2}, H_{3}, H_{4} be the 4 H.M. between 20 and 5.∴ 20, H _{1}, H_{2}, H_{3}, H_{4}, 5 are in H.P.⇒ 1/20, 1/H _{1}, 1/H_{2}, 1/H_{3}, 1/H_{4}, 1/5 are in A.P.Let d be the common difference of A.P. d = (1/5 – 1/20)/(4 + 1) ⇒ d = 3/100 ∴ 1/H _{1} = 1/20 + d = 1/20 + 3/100 = 8/1001/H _{2} = 1/20 + 2d = 1/20 + 6/100 = 11/1001/H _{3} =1/20 + 3d = 1/20 + 9/100 =14/100and 1/H _{4} = 1/20 + 12/100 = 17/100∴ 4 H.M.’s are 100/8, 100/11, 100/14, and 100/17. |

Example-The 10^{th }and 14^{th} terms of an H.P. are 2/29 and 2/41 respectively. Find the 20^{th} term. 10Solution-^{th} term of H.P. T_{10} = 2/29∴ corresponding 10 ^{th} term of A.P. a_{10} = 29/214 ^{th} term of H.P. T_{14} = 2/41∴ corresponding 14 ^{th} term of A.P. a_{14} = 41/2Let ‘a’ and ‘d’ be the first term and common difference of A.P. Now, a10 = 29/2 ⇒ a + 9d = 29/2 ……….(i) and a14 = 41/2 ⇒ a + 13d = 41/2 ……….(ii) Subtract (ii) from (i):- -4d = 29/2 – 41/2 ⇒ -4d = -12/2 ⇒ -4d = -6 ⇒ d = 3/2 Substitute d = 3/2 in equation (i):- a + 9 (3/2) = 29/2 ⇒ a + 27/2 = 29/2 ⇒ a = 29/2 – 27/2 ⇒ a = 1 Now, a20 = a + 19d = 1 + 19 (3/2) = 1 + 57/2 = 59/2 ∴ The 20 ^{th} term of H.P. is 2/59. |

Example-If the p^{th}, q^{th,} and r^{th} terms of an H.P. be x, y and z respectively, prove that (q – r) yz + (r – p) zx + ( p – q) xy = 0. pSolution-^{th} term of H.P. = x⇒ p ^{th} term of A.P. = 1/xq ^{th} term of H.P. = y⇒ q ^{th} term of A.P. = 1/yand r ^{th} term of H.P. = z⇒ r ^{th} term of A.P. = 1/zLet a and d be the first term and common difference of A.P. ∴ a + (p – 1) d = 1/x …………(i) Also, a + (q – 1) d = 1/y …………(ii) a + (r – 1) d = 1/z …………(iii) Subtract (ii) from (i):- (p – q) d = 1/x – 1/y ⇒ (p – q) d = (y – x)/xy ⇒ p – q = (y – x)/xyd …………(A) Subtract (iii) from (ii):- (q – r) d = 1/y – 1/z ⇒ (q – r) d = (z – y)/yz ⇒ q – r = (z – y)/dyz …………(B) Subtract (i) from (iii):- (r – p) d = 1/z – 1/x ⇒ (r – p) d = (x – z)/zx ⇒ r – p = (x – z)/dxz …………(C) Now, (q – r) yz + (r – p) zx + (p – q) xy = [(z – y)/dyz] yz + [(x – z)/dxz ] zx + [(y – x)/dxy] xy = (z – y + x – z + y – x)/d = 0/d =0 |

Example-If the ratio of G.M. and H.M. between two numbers is 5 : 4, prove that the numbers are in the ratio 4 : 1. Let ‘a’ and ‘b’ be the two numbers.Solution-Now, G.M./H.M. = 5/4 ⇒ √ab/(2ab/a+b) = 5/4 ⇒ (a + b)/2√ab = 5/4 Apply Componendo and Dividendo:- (a + b + 2√ab)/(a + b – 2√ab) = (5 + 4)/(5 – 4) ⇒ (√a + √b) ^{2}/(√a – √b)^{2} = 9⇒ (√a + √b/√a – √b) ^{2} = (3)^{2}⇒ (√a + √b/√a – √b) = 3 Apply Componendo and Dividendo:- ⇒ (√a + √b + √a – √b)/(√a + √b – √a + √b) = 4/2 ⇒ 2√a/2√b = 2 ⇒ √(a/b) = 2 ⇒ a/b = 4 ∴ a : b = 4 : 1 |

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