IIT JEE Examples Based on Relationship between AM, GM and HM | JEE AM GM HM Examples

Progression and series is an important topic of IIT JEE Mathematics syllabus. All the three progressions i.e. arithmetic, geometric and harmonic are quite simple but require a lot of practice as they can be a bit confusing. Another concept related with the progressions is that of means. All the three means i.e. arithmetic, geometric and harmonic means are quite interrelated

What exactly is the relationship between AM, GM and HM?

Let us consider two numbers x and y. Then, if x, a, y forms an arithmetic progression, then this ‘a’ is termed as the arithmetic mean. Likewise, if this sequence of x, a, y forms a geometric progression then it is termed a geometric mean and same goes for harmonic mean. Mathematically, the three means can be described as under:

Arithmetic Mean: For two numbers ‘a’ and ‘b’, the arithmetic mean is defined as:

(a+b)/2

√ab

Geometric Mean: Unlike arithmetic mean, the geometric mean takes into account the product of the numbers. For two numbers ‘a’ and ‘b’, the geometric mean is defined as

Harmonic Mean: The harmonic mean of two numbers ‘a’ and ‘b’ is defined as 

If A.M denotes the arithmetic mean, G.M denotes the geometric mean and H.M, the harmonic mean, then the relationship between the three is given by:

A.M × G.M = H.M² 

Their relationship can also be illustrated using the inequality:

A ≥ G ≥ H

Watch this Video for more reference
 

We now consider some of the illustrations based on A.M, G.M and H.M:

Illustration: What is the harmonic mean of the roots of the equation

(5 + √2) x² – (4 + √5) x + 8 + 2√5 = 0?

Solution: Let ‘a’ and ‘b’ be the roots of the quadratic equation. Then by the concepts of sum and product of roots, we have,

a+b = (4 + √5)/(5 + √2)

ab = (8 + 2√5)/(5 + √2)

Hence, the harmonic mean between ‘a’ and ‘b’ is given by

H = 2ab/(a+b)

    = (16 + 4√5)/ (4 + √5)

    = 4.

Illustration: Evaluate: 1² + 2² + 3² + 4² + 5².

Solution: We can write the given expression as Σ⁵_r=1 r² 

Now, this is actually the sum of squares of first five natural numbers

= n(n+1)(2n+1)/6

Here, n =5. So,

Σ⁵_r=1 r² = 5(5+1)(2.5+1)/6 = 55

Illustration: Evaluate: 6² + 7² + 8² + 9² + 10². 

Solution: If we attempt to calculate as it is, it could be very lengthy and hence, we try to express it in some known way so that it becomes easy to apply some formula. Hence,

6² + 7² + 8² + 9² + 10² = 1² + 2² + 3² + …… +10² – (1² + 2² +…… 5²)

                                  = Σ¹⁰_r=1 r² – Σ⁵_r=1 r²

Hence, this is the sum of squares of first 10 and first five natural numbers. So, again using the formula n(n+1)(2n+1)/6, we get

= 10(10+1)(2.10+1)/6 – 5(5+1)(2.5+1)6

= 385 – 55 = 330

Illustration: Evaluate: 1² + 3² + 5² +………+ (2n–1)². 

Solution: We just try to express the given expression in such a form so as to use some formula. 

We know the formula for sum of squares of first n natural numbers, so we try to use that here. 

We can express the given expression as 

1² + 3² + 5² +………+ (2n–1)² = [1² + 2² + 3² + 4² +……+ (2n–1)² + (2n)²] – [(2² +4² +………+ (2n)²]

                                           = (2n)(2n+1)(4n+1)/6 – 4 (1² + 2² + ……+ n²)

                                           = (2n)(2n+1)(4n+1)/6 – 4 . n(n+1)(2n+1)/6

                                           = n(4n²–1)/3

Illustration: Find the sum: 1. 2 + 2. 3 + 3. 4 +…… + n. (n+1)

Solution: The r^th term of the equation can be expressed as

t_r = r(r+1)

   = r² + r

Hence, t₁ = 1² + 1

            t₂ = 2² + 2

            t₃ = 3² + 3

…….        ……..     ……..

            t_n = n² + n

Adding we get

t₁ + t₂ + t₃ +…+ t_n = (1² + 2² + 3² +…+ n²) + (1 + 2 + 3 + …+ n).

Now, using the formulae of sum of first n natural numbers and sum of squares of first n natural numbers, we get

Σⁿ_r=1 t_r = Σⁿ_r=1 r² + Σⁿ_r=1 r

Sn = n(n+1)(2n+1)/6 + n(n+1)/2

     = n(n+1)/6 [2n + 1 + 3]

     = n(n+1)/6 (2n + 4)

     = n(n+1)(n+2)/3

Illustration: Find the sum to n-terms of the series

1/1.2 + 1/2.3 + 1/3.4 +......+ 1/n(n+1)

Solution: The r^th term of the equation is

t_r = 1/r(r+1)

   = 1/r – 1/r+1

t₁ = 1 –1/2

t₂ = 1/2 – 1/3

t₃ = 1/3 – 1/4

t₄ = 1/4 – 1/5

t_n =1/n – 1/n+1

Adding

Σⁿ_r=1 t_r = 1 – 1/n+1 = n/n+1

Factorial:

Factorial of a natural number n is defined as the product of first n natural numbers and it is noted |n or n!

Some Basic results to be remembered:

        |0 = 1                  (by definition)

        |1 = 1

        |2 = 2. |1 = 2

        |3 = 3. |2 = 3.2.1 = 6

        |4 = 4. |3 = 4.3.2.1 =24

        |n = n | (n-1)

Illustration: Evaluate the sum: 1 |1 + 2 |2 + 3 |3 +… to n terms.

Solution: The r^th term is t_r = r |r

                                     = |r + 1 – |r

t₁ = |2 – |1

t₂ = |3 – |2

t₃ = |4 – |3

t_n = |n + 1 – |n

On adding, many terms get cancelled and we obtain

Σⁿ_r=1 t_r = |n + 1 – 1

Illustration: If x, y, z are positive real numbers, such that x + y + z = a, then prove that 1/x + 1/y + 1/z > 9/a.

Solution: Since A.M. > H.M.

Hence, using this we obtain,

x + y + z/3 > 3/x^–1 + y^–1 + z^–1 

⇒ 1/x + 1/y + 1/z > 9/a.

Illustration: Prove that (a + b + c) (ab + bc + ca) > 9abc.

Solution: Using AM > GM, we have

a+b+c/3 > (abc)^1/3 and ab+bc+ca/3 > (ab.bc.ca)^1/3.

Multiplying these two results, we have

(a+b+c/3)(ab+bc+ca/3) > (abc)^1/3 (ab.bc.ca)^1/3

or,(ab+bc+ca)(a+b+c)/9 > (a³b³c³)^1/2

or, (a + b + c)(ab + bc + ca) > 9abc.

Hence, we get the desired result.

askIITians is an online portal where students are free to ask any kind of queries on topics of arithmetic, geometric and harmonic progressions and the interrelationship between the three.
 

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