## Arithmetic Mean:

If a and b are the two numbers such that a, A_{1}, A_{2}, A_{3}, ……., A_{n}, b are in arithmetic progression then A_{1}, A_{2}, A_{3}, ……., A_{n} be the ‘n’ arithmetic means between a and b.

*Single Arithmetic Mean Between a and b:*

If a and b be the two numbers then the arithmetic mean between a and b is-

A = (a + b)/2 |

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

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

∴ a, A_{1}, A_{2}, A_{3}, ……., A_{n}, b are in arithmetic progression.

Now, b = a_{n} + 2⇒ b = a + (n + 2 – 1) d ⇒ b = a + (n + 1) d ⇒ d = (b – a)/(n + 1)∴ A _{1} = a + d = II^{nd} term of arithmetic progressionA _{2} = a + 2d = III^{rd} term of arithmetic progressionA _{3} = a + 3d = IV^{th} term of arithmetic progression………….. A _{n} = a + nd = (n + 1)^{th} term of arithmetic progression |

Example-Prove that the sum of n arithmetic means between the two numbers is equal to n times the single arithmetic mean between the two numbers. Let a and b be the two numbers and let ASolution-_{1}, A_{2}, A_{3}, ……., A_{n} be the n arithmetic means between a and b.∴ a, A _{1}, A_{2}, ……., A_{n}, b are in arithmetic progression.Now, A _{1} + A_{2} + ……… + A_{n} = n/2 [A _{1} + A_{n}] = n/2 [a + d + a + nd] = n/2 [a + (a + (n + 1) d)] = n/2 [a + b] = n [(a + b)/2] = n (single arithmetic mean between the numbers) |

Example-Insert 4 arithmetic means between 5 and 20. Let ASolution-_{1}, A_{2}, A_{3}, ……., A_{n} be the 4 arithmetic means between 5 and 20.∴ 5, A _{1}, A_{2}, A_{3}, A_{4}, 20 are in arithmetic progression.Here d = (20 -5)/(4 + 1) = 3 ∴ A _{1} = a + d = 5 + 3 = 8A _{2} = a + 2d = 5 + 2 (3) = 11A _{3} = a + 3d = 5 + 3 (3) = 14A _{4} = a + 4d = 5 + 4 (3) = 17∴ 4 arithmetic means between 5 and 20 are 8, 11, 14, 17. |

Example-There are m arithmetic means between 7 and 85 such that (m – 3)^{th} mean: m^{th} mean is as 11 : 24. Find the value of m. Here, a = 7, b = 85Solution-∴ d = (b – a)/(m + 1) = (85 – 7)/(m + 1) = 78/(m + 1) Now, A _{m-3}/A_{m} = 11/24⇒ [a + (m – 3) d]/(a + md) = 11/24 ⇒ 24a + 24 (m – 3) d = 11a + 11md ⇒ 13a = 11md – 24 (m – 3) d ⇒ 13a = [11m – 24 (m – 3)] d ⇒ 13a = (11m -24m +72) [78/(m + 1)] ⇒ 13 (7) = (-13m + 72) [78/(m + 1)] ⇒ 7 (m + 1) = (-13m + 72) 6 ⇒ 7m + 7 = -78m + 432 ⇒ 85m = 425 ⇒ m = 425/85 = 5 |

Example- Find the value of n such that (a^{n} + b^{n})/(a^{n-1} + b^{n-1}) may be the A.M. between the two distinct numbers a and b. Arithmetic Mean between a and b is (a +b)/2.Solution-Also for some value of n, (a ^{n} + b^{n})/(a^{n-1} + b^{n-1}) is the A.M. of a and b.∴ (a ^{n} + b^{n})/(a^{n-1} + b^{n-1}) = (a +b)/2⇒ 2 (a ^{n} + b^{n}) = (a +b) (a^{n-1} + b^{n-1})⇒ 2a ^{n} + 2 b^{n} = a^{n} + ab^{n-1} + ba^{n-1} + b^{n}⇒ a ^{n} + b^{n} = ab^{n-1} + ba^{n-1}⇒ a ^{n} – ba^{n-1} = ab^{n-1} – b^{n}⇒ a ^{n-1} (a – b) = b^{n-1} (a – b)⇒ a ^{n-1} = b^{n-1}⇒ a ^{n-1}/b^{n-1} = 1⇒ (a/b) ^{n-1} = (a/b)^{0}⇒ n – 1 = 0 ⇒ n = 1 |

Example-There are m arithmetic means between 14 and 38. If the ratio of the second and the lat means be as 4 : 7, determine the value of m. Here a = 14, b = 38Solution-∴ d = (b – a)/(m + 1) = (38 – 14)/(m + 1) = 24/(m + 1) Now, A _{2}/A_{m} = 4/7⇒ (a + 2d)/(a + md) = 4/7 ⇒ 7a + 14d = 4a + 4md ⇒ 3a = (4m – 14) d ⇒ 3a = (4m – 14) [24/(m + 1)] ⇒ 3 (14) = 2 (2m – 7) (24)/(m + 1) ⇒ 7 = 8 (2m – 7)/(m + 1) ⇒ 7m + 7 = 16m – 56 ⇒ 9m = 63 ⇒ m = 7 |

Example- If A is the arithmetic mean between a and b, prove that (i) (A – a) ^{2} + (A – b)^{2} = (1/2) (a – b)^{2}(ii) 4 (a – A) (A – b) = (a – b) ^{2} A is the arithmetic mean between a and b.Solution-∴ A = (a +b)/2 (A – a)(i)^{2} + (A – b)^{2}= [(a + b)/2 – a] ^{2} + [(a + b)/2 – b]^{2}= [(b – a)/2] ^{2} + [(a – b)/2]^{2} = 2 [(a – b)/2] ^{2} = 2 (a – b) ^{2}/4= (1/2) (a – b) ^{2}4 (a – A) (A – b) (ii) = 4 [a – (a + b)/2] [(a + b)/2 – b] = 4 [(2a – a – b)/2] [(a + b – 2b/2] = (a – b) (a – b) = (a – b) ^{2} |