## Geometric Mean:

If a and b be the two numbers and G_{1}, G_{2}, G_{3}, ………. G_{n} is the ‘n’ numbers between a and b such that a, G_{1}, G_{2}, G_{3}, ………. G_{n}, b are in Geometric Progression then G_{1}, G_{2}, G_{3}, ………. G_{n} is called the Geometric means between a and b.

*Single Geometric Mean Between a and b:*

The Geometric Mean between a and b is ** √(ab)**.

*Insertion of n Geometric Mean between a and b:*

Let G_{1}, G_{2}, G_{3}, ………. G_{n} be the n geometric means between a and b.

∴ a, G_{1}, G_{2}, G_{3}, ………. G_{n}, b are in geometric progression.

Let ‘r’ be the common ratio of geometric progression.

Now, b = a_{n+2}⇒ b = ar ^{n+2-1}⇒ b = ar ^{n+1}⇒ b/a = r ^{n+1}⇒ r = (b/a)^{1/n+1}∴ G _{1} = arG _{2} = ar^{2}G _{3} = ar^{3}………….. G _{n} = ar^{n} |

*Relation Between Arithmetic Mean and Geometric Mean:*

Let a and b be the two numbers, then A = (a + b)/2 and G = √(ab) Now, A – G = [(a + b)/2] – [√(ab)] ⇒ A – G = [(a + b – 2 √a √b)]/2 ⇒ A – G = [(√a) ^{2} + (√b)^{2} – 2 √a √b]/2⇒ A – G = (√a – √b) ^{2}/2A – G > 0 ⇒ A > G i.e. Arithmetic Mean > Geometric Mean |

If A and G be the Arithmetic Mean and Geometric Mean between two numbers then they satisfy the equation having a and b as its roots given-

x^{2} – 2Ax + G^{2} = 0 |

If A and G be the Arithmetic Mean and Geometric Mean between two numbers then the numbers are-

A ± √(A^{2} – G^{2}) |

Example-Find the value of n so that (a^{n+1} + b^{n+1})/(a^{n} + b^{n}) is the geometric mean between a and b. Geometric Mean between a and b = √(ab)Solution-Also for some values of n, the geometric mean between a and b is [(a ^{n+1} + b^{n+1})/(a^{n} + b^{n})]⇒ (a ^{n+1} + b^{n+1})/(a^{n} + b^{n}) = √(ab)⇒ (a ^{n+1} + b^{n+1})/(a^{n} + b^{n}) = a^{1/2}b^{1/2}⇒ a ^{n+1} + b^{n+1} = a^{n+1/2} b^{1/2} + b^{n+1/2} a^{1/2}⇒ a ^{n+1} – a^{n+1/2} b^{1/2} = b^{n+1/2} a^{1/2} – b^{n+1} ⇒ a ^{n+1/2} (a^{1/2} – b^{1/2}) = b^{n+1/2} (a^{1/2} – b^{1/2}) ⇒ a ^{n+1/2} = b^{n+1/2}⇒ a ^{n+1/2}/b^{n+1/2} = 1⇒ (a/b) ^{n+1/2} = (a/b)^{0}⇒ n + 1/2 = 0 ⇒ n = -1/2 |

Example- Insert 3 geometric means between 1/9 and 9. Let GSolution-_{1}, G_{2}, G_{3} be the 3 geometric means between 1/9 and 9.∴ 1/9, G _{1}, G_{2}, G_{3}, 9 are in geometric progression.Let r be the common ratio of geometric progression. Now, r = (b/a) ^{1/n+1}⇒ r = [9/(1/9)] ^{1/(3+1)}⇒ r = (81) ^{1/4}⇒ r = (3 ^{4})^{1/4}⇒ r = 3 ∴ 3 geometric means are (1/9) 3, (1/9) 3 ^{2}, (1/9) 3^{3}= 1/3, 1, 3 |

Example-The A.M. between two numbers is 34 and their G.M. is 16. Find the numbers. Here, A = 34, G = 16Solution-∴ Numbers are A ± √(A ^{2} – G^{2})= 34 ± √(34 ^{2} – 16^{2})= 34 ± √[(34 -16) (34 + 16)] = 34 ± √(18 x 50) = 34 ± √(9 x 2 x 10) = 34 ± 3 x 10 = 34 ± 30 ∴ Numbers are 64 and 4 |

Example- If the A.M. of two numbers a and b (a > b) is twice as their G.M. then prove that a : b = (2 + √3) : (2 – √3). A = (a + b)/2 ……….(i)Solution-and G = √(ab) ……….(ii) Also A = 2a ⇒ (a + b)/2 = 2 √(ab) ⇒ a + b = 4√(ab) ……….(iii) Now, (a – b) ^{2} = (a + b)^{2} – 4ab⇒ (a – b) ^{2} = [4√(ab)]^{2} – 4ab [using (iii)]⇒ (a – b) ^{2} = 16ab – 4ab⇒ (a – b) ^{2} = 12ab⇒ (a – b) = √(12ab) ⇒ (a – b) = 2√(3ab) ……….(iv) Dividing (iii) by (iv):- (a + b)/(a – b) = [4√(ab)]/[2√3√ab] ⇒ (a + b)/(a – b) = 2/√3 Apply Componendo and Dividendo:- [(a + b) + (a – b)]/[(a + b) – (a – b)] = (2 + √3)/(2 – √3) ⇒ 2a/2b = (2 + √3)/(2 – √3) ⇒ a/b = (2 + √3)/(2 – √3) |

Example-If a, b, c are in A.P., x is the G.M. between a and b, y is the G.M. between b and c. Then prove that b^{2} is the A.M. between x^{2} and y^{2}. a, b, c are in A.P.Solution-⇒ 2b = a + c ……….(i) x is the G.M. between a and b. ⇒ x ^{2} = ab ……….(ii)Also, y is the G.M. between b and c. ⇒ y ^{2} = bc ……….(iii)Add (ii) and (iii):- x ^{2} + y^{2} = ab + bc⇒ x ^{2} + y^{2} = b (a + c)⇒ x ^{2} + y^{2} = b (2b)⇒ x ^{2} + y^{2} = 2b^{2}∴ b ^{2} is the A.M. between x^{2} and y^{2}. |

Example-Find the sum of an infinite G.P. whose first term is 28 and the fourth is 4/49. Let a and r be the first term and common ratio of a G.P. respectively.Solution-Here, a = 28 and a = 4/49 ⇒ ar ^{3} = 4/49⇒ 28r ^{3} = 4/49⇒ r ^{3} = 1/(49 x 7)⇒ r ^{3} = (1/7)^{3}⇒ r = 1/7 ∴ S _{∞} = a/(1 – r)⇒ S _{∞} = 28/(1 – 1/7)⇒ S _{∞} = 28/(6/7)⇒ S _{∞} = (28 x 7)/6⇒ S _{∞} = 98/3 |

Example-The first term of a G.P. exceeds the second term by two and the sum to infinity is 9/2. Find the G.P. Let a be the first term and r be the common ratio of G.P.Solution-Now, a _{1} – a_{2} = 2⇒ a – ar = 2 ⇒ a (1 – r) = 2 ⇒ 1 – r = 2/a ……….(i) Also, S _{∞} = 9/2⇒ a/(1 – r) = 9/2 ⇒ a/(2/a) = 9/2 [using (i)] ⇒ a ^{2} = 9⇒ a = ± 3 When a = 3, equation (i) becomes- 1 – r = 2/3 ⇒ r = 1 – 2/3 ⇒ r = 1/3 G.P. is 3, 3 x 1/3, 3 x 1/3 ^{2}, ……….= 3, 1, 1/3, ………. When a = -3, equation (i) becomes- 1 – r = 2/(-3) ⇒ r = 1 + 2/3 ⇒ r = 5/3 > 1 which is not possible ∴ a ≠ -3 |