WHAT ARE YOU WAITING FOR?

# Algebra

Sequence and Series - 3

**Harmonic Progression:**

If numbers are in Harmonic Progression or H.P., then their reciprocals are in A.P. So, if there are three numbers a, b and c in Harmonic Progression, then their reciprocals i.e. 1/a, 1/b and 1/c are in Arithmetic Progression.

For example, consider the numbers 30,40 and 60

2/40 = 1/30 + 1/60

So, these three numbers are in H.P.

Harmonic Mean of two numbers:

For two numbers a and b, their Harmonic Mean or H.M. is given by 2ab/(a+b) or

**Harmonic Mean of n numbers:**

**Example 1**: If the harmonic mean between two quantities is to their geometric means is 12 to 13. Prove that the quantities are in the ratio of 4 to 9.

**Solution:** Let the two numbers be a and b respectively.

Their geometric mean = √ab and their harmonic mean = 2ab/(a+b)

12/13 = 2√ab/(a+b)

or (a+b)/2√ab = 13/12

Using componendo and dividend, we get:

**or a/b = 4/9**

**Example 2:** If b is the harmonic mean between a and c, prove that

1/(b-a) + 1/(b-c) = 1/a + 1/c

**Solution:** Given that b is the harmonic mean between a and c, so

b = 2ac/(a+c)

LHS = 1/(b-a) + 1/(b-c)

Now b-a = {2ac/a+c} - a = a(c-a)/(c+a)

b-c = {2ac/(a+c)} – c = c(a-c)/(a+c)

1/(b-a) + 1/(b-c) = {(c+a)/a(c-a) - (a+c)/c(c-a) } = { (c2-a2)/ac(c-a) } = {(c+a)/ac}

= 1/a + 1/c

Thus, LHS = RHS.

**Arithmetic-geometric Progression:**

An *arithmetic-geometric progression* (AGP) is a progression in which each term can be represented as the product of the terms of an (AP) and a (GP).

For example, consider the following series:

1/5 + 2/5^{2} + 3/5^{3} + 4/5^{4} +……

Here the numerators i.e. 1,2,3,4… are in AP while the denominators 5,5^{2},5^{3},… are in GP and so this series is an AGP.

General form of an AGP :

*The sequences of an AGP can be represented as: ar, (a+d)r ^{2}, (a+2d)r^{3}, ……..*

*How to find the sum of an AGP?*

Let us take an example: Find the sum of the series : 1/7 + 3/7^{2} + 5/7^{3} + 7/7^{4} +…….

S = 1/7 + 3/7^{2} + 5/7^{3} + 7/7^{4} +……. (1)

Multiply each term by the common ratio of the GP, i.e. 1/7 we get

S/7= 1/7^{2} + 3/7^{3} + 5/7^{4} +……….(2)

Subtract (2) from (1), we get:

6S/7 = 1/7 + 2/7^{2} + 2/7^{3} + 2/7^{4} +…….

Now, terms from 2/72 form an infinite GP.

6S/7 = 4/21

**or S = 2/9 **

**Example 3:** Find the sum of the series: ½ + 5/12 + 7/24 + 9/48 +……….

**Solution:** The series can be written as 3/6 + 5/12 + 7/24 + 9/48 +……

Now, this is an AGP with the common ratio of ½

S = 3/6 + 5/12 + 7/24 + 9/48 +…… (1)

Multiply the series by ½ we get

S/2 = 3/12 + 5/24 + 7/48 +….

Subtract (2) from (1), we get:

S/2 = 3/6 + 2/12 + 2/24 + 2/48

**Or S = 5/3**

**Solution:** S= 1 + 4/7 + 9/7^{2} + 16/7^{3} +25/7^{4} (1)

Now, this is not an AGP but can be reduced to an AGP. (Note that the difference in numerators is in AP)

Multiply the series by 1/7 (common ratio)

S/7 = 1/7 + 4/7^{2} + 9/7^{3} + 16/7^{5} (2)

Subtract (2) from (1), we get:

6S/7 = 1 + 3/7 + 5/7^{2} + 7/7^{3} + 9/7^{4} (3)

Now, (3) is an AGP

Multiply the series by 1/7, we get

6S/49 = 1/7 + 3/7^{2} + 5/7^{3} +7/7^{4} (4)

Subtract (4) from (1) we get:

36S/49 = 1 + 2/7 + 2/7^{2} + 2/7^{3} +……

36S/49 = 1 + 1/3

Or S = 49/27

**Thus, option (3) is correct.**

**Example 5 :** Find the sum of n terms of a series whose nth term is given by n(n+1)(n+4)

**Solution:** Here t_{n} = n(n+1)(n+4) =n^{3} + 5n^{2} + 4n

We have to find S = Σt_{n} = Σn^{3} + 5Σn^{2} + 4Σn

= {n(n+1)/2}^{2 } +5{n(n+1)(2n+1)/6} + 4{n(n+1)/2}

This further simplifies to :

**Example 6:** Find the sum to n terms of the series whose nth term is given by n^{2} +2^{n}.

**Solution:** S = Σt_{n} = Σn^{2} +Σ2^{n }

**Example 7:** Find the sum to 100 terms of the series:

1 + (1+2) + (1+2+3) +…………

**Solution:** t_{n} = 1+2+3+…..n = n(n+1)/2

Sum = Σt_{n} = ½ {Σn^{2 }+ Σn}

= ½ { n(n+1)(2n+1)/6 + n(n+1)/2} = ¼ n(n+1){ (2n+1)/3 + 1}

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

**Put n = 100, we get S = (100*101*102)/6 = 171700**

__Series in which Difference of terms is in A.P.__

At times one can witness series in which the difference in terms is in Arithmetic Progression. Now, how to solve such problems? Let’s take an example

**Example 8:** Find the sum of 20 terms of the series:

3 + 8 + 16 + 27 + 41 +……. Upto 20 terms

**Solution: **here we can see that the difference in terms is in A.P. Now always in such cases, the nth term is of the form an^{2 }+bn +c where a, b and c are unknowns whose values we need to calculate first.

Let t_{n} = an^{2}+bn +c

When n =1 , t_{n} = 3, when n =2, t_{n} becomes 8 and when n =3, t_{n} becomes 16

So, 3 = a+b+c (1)

8 = 4a +2b +c (2)

16 = 9a +3b +c. (3)

Solving (1), (2) and (3) we get:

a = 3/2 , b = ½ and c =1

So, t_{n} = (3/2)n^{2} + ½ n + 1

Sum = Σt_{n} = (3/2) Σn^{2} + ½ Σn + Σ1

Now put n=20

**Sum = 4430 **

__Special Series:__

A series in which any term except the first and the last is equal to the sum of its neighbours:

Let’s say we have a series in which t_{n} = t_{n-1} + t_{n+1}

Let t_{1} = a and t_{2} = b, so t_{3} = b-a, t_{4} = -a, t_{5} = -b and t_{6} = a-b.

After, this, the terms will repeat i.e. t_{7} = a, t_{8} =b, t_{9} = b-a and so on.

So, looking at this series we can say that terms repeat after a period of 6.

Also, the sum of the first six terms is zero and as the terms repeat after that, so sum of next terms is also zero and this cycle continues.

**Example 9:** In a series T_{n} = T_{n-1} + T_{n+1}. If the sum of first 100 terms of the series is 20 while the sum of first 50 terms is 10. Then, what is the value of the first term and the second term.

**Solution:** Let the first term be a and the second term be b

Sum of first 96 terms is zero. So t_{97} + t_{98} +t_{99} +t_{100} = 2b –a = 20. (1)

Sum of first 48 terms is zero. So, t_{49} + t_{50} = a +b = 10 (2)

Adding (1) and (2) we get 3b = 30. Or b = 10 and a = 0

**So, the first term is 0 while the second term becomes 10.**

**Example 10:** In a series, any term except the first and the last is equal to the product of its neighbour. If the product of the first 20 terms is 20 while the product of 100 terms is 400. What is the product of the first 93 terms?

**Solution:** Let the first term (t_{1}) be a while the second term (t_{2}) be b.

Now t_{3} = b/a, t_{4} = 1/a, t_{5} = 1/b and t_{6} = a/b

Now, again the terms after t_{6} will repeat themselves, i.e. t_{7} = a, t_{8} = b and so on.

So, we can say that the product of first six terms = product of next six terms and so on = 1

Now, given that the product of first 20 terms =24

The product of first 18 terms will be 1. So, product of first 20 terms = ab = 20 (1)

Similalry product of first 96 terms will be 1. So, product of first 100 terms = b^{2}/a =400 (2)

From (1) and (2) we get: b = 20 and a = 1

**The product of first 93 terms = product of first 3 terms = b ^{2} = 100**

** **

__Series that can be expressed as Difference of two terms:__

**Example11: **Find the sum of the series:

Multiply and divide the entire series by 2 (as the difference between the denominators in each term is 2), we get :

As a^{2} –b^{2} = (a+b)(a-b),

The given series can be written as :

**Example 13:** Find the sum of 100 terms of the series:

1 +3 +7 +15+………

**Solution:** Let S = 1s +3 +7 + 15 +. ……… t_{n}

S= .. 1 + 3 + 7 +………….t_{n-1 }+t_{n}

Subtract both we get

0 = 1 +2 +4 +8+…………..-t_{n}

or t_{n} = 1+(2+4+8+….. upto n-1 terms)(This series is in GP)

t_{n} = 1+ 2(2n-1 -1)/(2-1). = 2^{n} -1

S = St_{n} = (2^{1}-1) + (2^{2}-1) + (2^{3}-1) +…….(2^{n} -1)

= 2(2^{n}-1)/(2-1). – (1+1+1+……upto n terms)

= 2^{n+1} – 2 – n

Put n = 100, we get:

**S = 2 ^{101} -2 -100. = 2(2^{100}-1) – 100**