Arithmetic Mean in Sequence and Series

Arithmetic Mean:

If a and b are the two numbers such that a, A1, A2, A3, ……., An, b are in arithmetic progression then A1, A2, A3, ……., An 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 A1, A2, A3, ……., An be the n arithmetic means between a and b.

∴ a, A1, A2, A3, ……., An, b are in arithmetic progression.

Now, b = an + 2
⇒ b = a + (n + 2 – 1) d
⇒ b = a + (n + 1) d
d = (b – a)/(n + 1)

∴ A1 = a + d = IInd term of arithmetic progression
A2 = a + 2d = IIIrd term of arithmetic progression
A3 = a + 3d = IVth term of arithmetic progression
…………..
An = 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.

Solution- Let a and b be the two numbers and let A1, A2, A3, ……., An be the n arithmetic means between a and b.

∴ a, A1, A2, ……., An, b are in arithmetic progression.

Now, A1 + A2 + ……… + An
= n/2 [A1 + An]
= 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.

Solution- Let A1, A2, A3, ……., An be the 4 arithmetic means between 5 and 20.

∴ 5, A1, A2, A3, A4, 20 are in arithmetic progression.

Here d = (20 -5)/(4 + 1) = 3

∴ A1 = a + d = 5 + 3 = 8
A2 = a + 2d = 5 + 2 (3) = 11
A3 = a + 3d = 5 + 3 (3) = 14
A4 = 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: mth mean is as 11 : 24. Find the value of m.

Solution- Here, a = 7, b = 85
∴ d = (b – a)/(m + 1) = (85 – 7)/(m + 1) = 78/(m + 1)

Now, Am-3/Am = 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 (an + bn)/(an-1 + bn-1) may be the A.M. between the two distinct numbers a and b.

Solution- Arithmetic Mean between a and b is (a +b)/2.
Also for some value of n, (an + bn)/(an-1 + bn-1) is the A.M. of a and b.

∴ (an + bn)/(an-1 + bn-1) = (a +b)/2
⇒ 2 (an + bn) = (a +b) (an-1 + bn-1)
⇒ 2an + 2 bn = an + abn-1 + ban-1 + bn
⇒ an + bn = abn-1 + ban-1
⇒ an – ban-1 = abn-1 – bn
⇒ an-1 (a – b) = bn-1 (a – b)
⇒ an-1 = bn-1
⇒ an-1/bn-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.

Solution- Here a = 14, b = 38
∴ d = (b – a)/(m + 1) = (38 – 14)/(m + 1) = 24/(m + 1)

Now, A2/Am = 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


Solution- A is the arithmetic mean between a and b.
∴ A = (a +b)/2

(i) (A – a)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

(ii) 4 (a – A) (A – b)
= 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

Trigonometric Ratios of Associated Angles
Trigonometric Ratios of Some Standard Angles
Measurement of Angles in Trigonometry
General Solution of an Equation of the Type (a cos θ + b sin θ = c)
Trigonometric Identities Connecting the Angles of a Triangle
Arithmetic Progression or Arithmetic Sequence– Wikipedia

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