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Examples Based on Relationship between AM, GM and HM

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 x G.M = H.M2

Their relationship can also be illustrated using the inequality:

     A ≥ G ≥ H

Watch the following video for more on the relationship between the arithmetic and geometric mean

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 Σ5r=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,

Σ5r=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)

                                  = Σ10r=1 r2 – Σ5r=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 rth term of the equation can be expressed as

tr = r(r+1)

   = r2 + r

Hence, t1 = 12 + 1

            t2 = 22 + 2

            t3 = 32 + 3

…….        ……..     ……..

            tn = n2 + n

Adding we get

t1 + t2 + t3 +…+ tn = (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

Σnr=1 tr = Σnr=1 r2 + Σnr=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 rth term of the equation is

tr = 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

tn =1/n – 1/n+1

Adding

Σnr=1 tr = 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 rth term is tr = r |r

                                     = |r + 1 – |r

t1 = |2 – |1

t2 = |3 – |2

t3 = |4 – |3

tn = |n + 1 – |n

On adding, many terms get cancelled and we obtain

Σnr=1 tr = |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.

Related resources:

To read more, Buy study materials of Sequences and Series comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.

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