1.8 Progressions of numbers:

Will it not be interesting to solve a puzzle similar to the one given below?

Puzz1e 1:

Suppose you had taken a loan of Rs.10, 000 from a friend and you agree to pay few Rupees every day. You have following options to choose:

1.      You repay the loan at the rate of Rupee a day. Do you think your friend will agree for this? He may not agree. (Because repayment will take nearly 28 years (10000/365))

2.      You agree to pay the amount equal to the day of payment (1st day 1Rs, 2nd day 2Rs, 3rd day 3Rs. 4th day 4Rs..). Will you agree to pay for indefinite number of days?

3.      First day you pay 1Rs and subsequently you pay twice the amount of previous day(1st day 1Rs, 2nd day 2Rs, 3rd day 4Rs, 4th day 8Rs..) Will you agree to pay for indefinite number of days?

In the case of last 2 options don’t you need to know the number of days of repayments?

Puzz1e 2:  You decide to participate in a Cycling race of 70km.  In the first hour of race you cycle at the rate of 16km/hr. If your cycling speed reduces by 1km every hour thereafter, find out how much time you require to complete the race?

Let us see how mathematics can help us in finding solutions to these real life problems.

1.8.1 Sequence:

1.8.1 Example 1: If you are asked to list out all the classes in your school. What will you write?

Will you write the list as 3,10,4,1,12,8,7,5,6,2,9,11?

No, You will probably write  them as

1,2.3,4,5,6,7,8,9,10,11,12

1.8.1 Example 2: Similarly if you are asked to write the dates of Sundays in January 2006 you will write them as?

1, 8,15,22,29

What did you do? Without being aware you applied a rule in both the cases.

In the first case you started with 1st standard and applied the rule of ‘one more than previous number’ to list other classes. You stopped at 12 as it was the last standard in your school.

Similarly in the 2nd case you found that first Sunday of January 2006 is 01 and applied the rule ’add 7 to previous number till the date is less than31’. You stopped at 29 as any month does  not have more than 31 days. But if you were asked to find the Sundays in February of a year, you will apply a different rule, depending upon whether the year is a leap year or not.

Observe the following list

1.8.1 Example 3 : 2,4,6,8,10,12……..

What did you notice? It is a list of even numbers and the list does not end at all

Definition : A sequence is an ordered arrangement of numbers according to a rule.

The individual numbers in the sequence are called ‘terms’ of the sequence.

The terms of a sequence are generally denoted by T1 T2T3T4T5 …. as shown below

 Order number of the term ==à 1st 2nd 3rd 4th ---- nth --- Corresponding notation ===à T1 T2 T3 T4 ---- Tn ---

The sequence is generally denoted (represented) by {Tn  }

A sequence which has definite number of terms is called a ‘finite sequence,

If the sequence which has indefinite number of terms is called a ‘infinite sequence,

In the first example we had 12 terms and in the 2nd example we had 5 terms and hence both of them are examples of finite sequence.

The list of even numbers does not have finite number of terms and hence it is an infinite sequence.

1.8.1 Example 4 : The sequence can also be of fractions like

2/1 , 3/2 , 4/3, 5/4 ,,,,,,,,

What is the general term Tn  here?

We observe

T1 = (1+1)/1

T2  =(2+1)/2

T3  =(3+1)/3

T4    =(4+1)/4

Thus

Tn=(n+1)/n .With this general representation of the nth term we can find any term of the sequence.

Hence the 6th term isT6   =(6+1)/6 =7/6

1.8.1 Problem  1 : If Tn    =2n2+1 find the value of n if Tn=73

We have Tn    =2n2+1 =73; 2n2 =73-1=72: 2n2 =72 : n2 =36: n = =

Since it is a natural number, it has to be positive and hence n=6.

Verify that T6 = 2*62+1 = 2*36+1=73

1.8.2 Series

Definition : The sum of terms of a sequence is called the series of the corresponding sequence and is usually denoted by S  or Sn.

Observe that  sum of the series is meaningful for finite sequence, as summing up terms of an infinite sequence is meaningless. (In later sections you will learn about infinite series whose sums are meaningful).

Sn = T1 + T2+T3.........Tn

Prove that Tn =Sn- Sn-1

We have Sn- Sn-1=( T1 + T2+T3.........Tn-1+ Tn) -( T1 + T2+T3.........Tn-1)= Tn

1.8.2 Problem  1 : If Tn    ={(-1)n} prove that S1 = S3 : S2 = S4

Since Tn    =(-1)n we have

T1= (-1)1 = -1, T2 = (-1)2 =1, T3 = (-1)3 = -1, T4= (-1)4 = 1

Substituting values for series we have

S1 = T1   = -1

S3 = T1 + T2+T3= -1+1-1 = -1 Thus S1 = S3

S2  = T1 + T2  =-1+1 =0

S4 =T1 + T2+T3 +T4= -1+1-1+1 =0 Thus S2 = S4

1.8.3 Arithmetic Progression:

In the Example1.8.1.1 did you notice that, the difference between 2 successive terms of the sequence is 1 ?

Similarly what was the difference between 2 successive terms in Example 1.8.1.2 ?  It is 7.

Definition : A sequence in which the difference between 2 successive(consecutive) terms is constant is called ‘Arithmetic Progression’(AP).  The Common difference which is a constant is denoted by ‘d’

Thus by definition in an AP  Tn+1 Tn  =d  and Tn-1+d =  Tn

The first term of an AP is generally a constant and is denoted by ‘a’ and hence  T1 = a

Its other terms are

T2= a+d

T3=  T2+d  =(a+d)+d = a+2d = a + (3-1)d

T4=  T3+d  =(a+2d)+d =a+3d= a+(4-1)d

….

General term Tn=  Tn-1+d = a+(n-1)d : thus d= (Tn  -a)/(n-1)

{AP}= {a, a+d, a+2d,a+3d …, a+(n-1)d}

1.8.3 Problem  1 :Find the AP in which Sn = 5n2+3n

Solution:

Sn-1 = 5(n-1)2+3(n-1) = 5(n2 -2n+1) +3n-3 = 5n2-10n+5+3n-3 = 5n2-7n+2

We know that Tn= Sn- Sn-1= (5n2+3n) –(5n2-7n+2) = 10n-2

T1 = 8

T2  =18

T3  =28

So {AP} is {8,18,28…..}

Verification: S3 = T1 + T2 + T3 =8+18+28 = 54 = 45+9 = 5*32+3*3=( 5n2+3n with n=3)

1.8.3 Problem  2  : In an AP T10 =20    T20  =10 Find    T30

Solution:

If we can find a and d we can arrive at the solution.

By definition Tn = a+(n-1)d

Thus

T10 = a+(10-1)d = a+9d

But  it is given that T10= 20 so we have

a+9d=20: a=20-9d                              ====à(1)

Be definition T20 = a+(20-1)d = a+19d  ====à(2)

But  it is given that T20= 10

By substituting the value of ‘a’ obtained from (1) in (2) we get

20-9d+19d =10: 20+10d =10: 10d =(10-20)= -10: d = -1

By (1) we get a =20-9d = 20+9 =29

T30  =a+(30-1)d = 29+29*(-1) = 29-29 =0

Verification: Note that T10 =29+9*(-1)=20: T20 =29+19*(-1)=10 which are the given terms.

1.8.3 Problem 3: Find AP whose 5th and 10th terms are in the ratio of 1:2 and T12 =36

Solution:

It is given that T5 : T10 = 1:2 (i.e T5 /T10 =1/2)

2T5 = T10

By substituting generic value for T5 & T10 in the above equation

We get 2(a+4d) = (a+9d) i.e.  2a+8d =a+9d and by transposition of terms we get a=d.

But it is given that T12 =36

Hence a+ 11d = 36: Since a=d we get 12d =36 and hence d=3 and since a=d, a=3

Therefore the {T} = 3,6,9,12…

Verification: Note that T5 = 15 and   T10 =30 which are in the ratio of 1:2 which is the given ratio

1.8.3 Problem 4: Find the three numbers in AP whose sum is 15 and product is 105

Solution:

Let the middle number be a. hence the first term is a-d and the 3rd term is a+d.

Sum of three numbers = (a-d)+a+(a+d ) = 3a which is given to be 15 and hence a =5.

The product of these 3 numbers = (a-d)*a*(a+d) = a*(a2-d2) which is given to be 105

a*(a2-d2) =105

I.e. 5(52-d2) = 105

I.e. (25-d2) = 21

I.e. 25-d2 = 21

I.e. -d2 = 21-25

I.e. -d2= -4

I.e. d2= 4

I.e.  d =

So the series is 3,5,7 or 7,5,3

1.8.3 Problem 5: Find the number of integers between 60 and 600 which are divisible by 9

Solution:

The first number greater than 60 and divisible by 9  is 63

The  last number lesser  than 600 and divisible by 9  is 594

Thus the sequence is 63, 72,81 ….594

Here a= 63, d=9 and Tn = a+(n-1)d = 594

I.e.  63+(n-1)9 = 594

I.e. (n-1)9 = 594-63 = 531

(n-1) = 59

n=60

Thus there are 60 numbers between 60 and 600 which are divisible by 9

1.8.3 Problem 6: If the mth  term of an AP is 1/n and nth term is 1/m show that its (mn)th term is 1

Solution:

Let a be the first term and d be the common difference of AP

Tm = a+(m-1)d = 1/n and Tn = a+(n-1)d = 1/m

By subtracting the Tm  and Tn we get

1/n – 1/m = a+md –d –(a+nd –d)

I.e. (m-n)/mn = (m-n)d

d =  1/mn

Substituting this value  of d in Tm we get

a+(m-1)/mn = 1/n

a = 1/n – (m-1)/mn

= 1/mn

Tmn = a+(mn-1)d =

(1/mn) + (mn-1)/mn =

(1+mn-1)/mn

=1

1.8.4 Summation of arithmetic series

Let us consider the puzzle discussed in the beginning (Section 1.8). for which we wanted to find the solution

1.8.4 Problem 1 : Suppose you had taken a loan for Rs.10,000 from a friend  and you agree to pay him, few Rupees every day.   Assume that you have two options:

Option 1: You repay the loan at the rate of 1 Rupee a day. Do you think your friend will agree to this?  He may not, as repayment will take nearly 28 years = .

Option 2:  You agree to pay the amount equal to the day of payment (1st day 1Rs, 2nd day 2Rs, 3rd day 3Rs. 4th day 4Rs and so on). Will you agree to pay the amount indefinitely?

With the 2nd option let us find out how much money you would have paid totally in 10 days

Total money paid after 10 days = 1+2+3+4+5+6+7+8+9+10 = 55Rs.

How do we find the money paid after 100days?  Does it not take time to find out?

Let us say we are asked to find the sum of 1+2+3+. . . . +99+100. Do not we find it difficult to add

Instead let us pair the numbers as follows:

Then sum   = (1+100)+(2+99)+(3+99) . . +(50+51)  = 101*50 = 5050

Let us apply this logic to find the sum of the first ‘n’ numbers of the series of natural numbers.

{T} = {1,2,3……n}

Sn  = 1     +    2  +   3  ………….+(n-2)+ (n-1) +n(there are n terms)

+ Sn  =  n    +(n-1)+(n-2)              +  3  +   2     +1(repeated  in reverse order)

==================================

­­­­­­­­­­­­­­­­­­­­­­2Sn= (n+1)+(n+1)+(n+1)    …..    .+(n+1)+(n+1)+(n+1) (there are n terms) = n(n+1)

Sn=

Using this formula in the above problem, let us cross check the correctness of the amount paid by you after 10 days:

S10  =10*11/2= 55 Rs which is correct!

Now Let us find the amount paid by you after 100 days: S100 = 100*101/2  =  5050 Rs

Do you think you will require 200 days to clear the loan? : S200 = 200*201/2 =20,100 Rs

In 200 days you would have paid 10,100 Rs extra! This is a trial and error method. (In next sections you will learn to find the value of n satisfying the condition n(n+1) =1000)  For now, note S141 =  =10,011

So you require 141 days to return the loan.

The above sum (Sn ) is also denoted by the symbol  =

Definition: A series whose terms are in AP is called an ‘arithmetic series’ For example:

{2,5,8},  {1,4,7,},  {3,7,11}

Find the first ‘n’ terms of an AP:

{AP}= {a, a+d, a+2d, a+3d ….,a+(n-1)d}

Sn= [a+(a+d)+(a+2d)+(a+3d) …..a+(n-1)d] = [a+a+a ….(n times) +d(1+2+3+     +(n-1)] = na+d[] =

na+   (apply the formula by replacing n by (n-1) in  )

Sn = na+   =  = n*()

= n*()=n*()

Let us use the above formula to arrive at .

= 1+2+3+4+5+6+7+8+ . . . +n   is an AP with a =1, d=1

Sn =

= n*()

= n*(1+{1+(n-1)*1}/2

= n*(n+1)/2

This is the same formula we had arrived earlier in the beginning of 1.8.4.

1.8.4 Problem 2  : Find the sum of arithmetic series which contains 25 terms and whose middle term is 20

Solution:

Given : n=25, T13 =20,  we are required to find S25

But T13 = a+12d

S25  = n*(a+ T25)/2= 25*(a+a+24d)/2 = 25*2*(a+12d)/2 = 25*(a+12d) = 25*20(T13 = a+12d) = 500

1.8.4 Problem 3   : Find the sum of all natural numbers between 101 and 201 which are divisible by 4

Solution:

{AP} = (104,108,112 …200}

Sn  = 104+108+112+……

= 104+(104+4) + (104+8)… (104+96) (104 repeats 25 times)(Note that 1st term =104, last term is 200 and difference = 4

we have 24 =  terms after the first term, Thus in all 25 terms)

= 104*25 +4(1+2+3…..24)

= (104*25) +4*( )

=2600+1200=3800

1.8.4 Problem 4 :

Assume you went on a trip to Shravanabelagola where the statue of Bahubali, carved out of a single stone is installed.  Assume that you climbed 23 steps in the first minute. After that you started climbing 2 steps less than what you had climbed in the previous minute. If you reached a resting place after 7 minutes of climbing, find out how many steps did you climb before reaching  the resting place?

Solution:

Notice that you are climbing  2 steps  less than the previous minute. Hence your steps of climbing are an AP. Since you have taken 7 minutes to reach the resting place, we are required to find S7 of an AP.

{AP} = {23,21,19….) so we have a=23 and d = -2

Since Sn  of an AP = n*( )

S7

= 7* ( )

= 7*[46-12]/2

= 7*17 = 119

Exercise: If you need to climb 1000 steps to reach the statue, find out how much time you will require to reach the statue?

1.8.4 Problem 5:  You decide to participate in a Cycling race of 70km.  In the first hour of the race you cycle at the rate of 16km/hr. If your cycling speed reduces by 1km every hour thereafter, find out how much time you require to complete the race.

Solution:

Notice that your cycling speed is (16,15,14, …) which is an AP. We are required to find n such that Sn =70

In the given AP, note that a =16 and d = -1

Since Sn   = n*( )

= n*( )

= n*()

= n*()

n*()  = 70( total distance =70km)

Thus  we have an equation to solve

(33n-n2 ) = 2*70=140 or

-n2 +33n -140 =0 or

n2 -33n +140 =0 or

(n-5)*(n-28) = 0

Thus  n=5 or n=28

As per this solution the distance is covered in 5 hours or 28hours.

Though mathematically we have 2 answers to the same problem, we have to eliminate one answer in this real life problem.

Let us observe the 28th term of this AP

Tn=  a+(n-1)d  so

T28=  16+(28-1)*(-1) = 16-27 = -11

Since the cyclist can not cycle in a negative speed, n=28 is not a correct answer to the real life problem.

Thus the correct answer is 5 hours.

1.8.4 Problem 6:   To catch a herd of elephants, a king starts the journey starting with 2 ‘Yojana’s( unit of distance) on the first day. O genius, tell me the increase in distance he has to cover every day,  if he takes 7 days to  cover the total distance of  80 Yojana’s ( Lilavati Shloka 126 ).

Solution:

The distance covered by the king is an AP, with a =2, n=7  and Sn =70. We need to find d

Sn   = n*( )

= 7*( )

= 7*()

= 7*(2+3d) = 80

2+3d = 80/7

3d = (80/7)-2 = (66/7)

d = (80/7)-2 = (22/7)

Thus the king needs to increase the distance covered every day by (22/7) Yojanas.

1.8.4 Problem 6:   A person gifts every day 3 Pallas (A unit of measurement of volume) of grains and increases it by 2 Pallas every day, O Lilavati tell me the number of days needed to gift 360 Pallas.

( Lilavati Shloka 124 )

Solution:

The grains gifted by the person is an AP, with a =3, d=2  and Sn =360. We need to find n

Sn   = n*( )

= n*( )

= n*(3n+2n-2) = n(n+2)

n2+2n =360

n2+2n -360 =0

(n+20)*(n-18) =0

n= -20 or n =18

Since number of days can not be negative, the person needs 18 days to gift the entire 360 Pallas.

1.8.5 Geometric Progression (GP):

Let us take some examples

1.  {T}= {2,4,8,16 …….}.  In this series we observe that any term is twice the previous term. I.e. next term = 2* previous term or previous term = ½ of next term. The ratio of the terms = 1:2.

2.  {T}= {27,9,3,1 …….}.  In this series we observe that any term is one third the previous term. I.e.  next term = 1/3* previous term or previous term = 3times the next term. The ratio of terms =3:1

Definition : A sequence whose ratio of term and its preceding or succeeding term is constant is called ‘Geometric Progression(GP)’

Thus by definition, in a GP Tn   /Tn-1 = constant. In the first case T3   /T2==2 and in the 2nd case T3   /T2= = 1/3

In a GP if the first term T1 = a and the ratio is r we have

T2= T1*r= ar(2-1)

T3= T2*r= ar*r =ar2= ar(3-1)

T4= T3*r= ar2*r = ar3= ar(4-1)

In general Tn= ar(n-1)

In a GP we also know that Tn= Tn-1*r

So {a, ar, ar2, ar3 ……….. ar(n-1)}  is the standard form of  GP.

1.8.5 Problem 1 :  In a  GP 7th term is eight times the fourth term and the 5th term is 12 find the GP if  S10:  S5=  33:1 and T6= 32

Solution:

Tn = arn-1

T7=a r6 and  T4=a r3 it is also given that T7= 8T4

a r6= 8a r3

r3= 8

r=2

We know T5=a r4

= a 24=16a =12 (given)

a =  =

Therefore {GP} = {, *2, *22 , *23….} = {3/4, 3/2,3,6…}

Let us find the sum of n terms of a GP = {a, ar, ar2, ar3 ……….. ar(n-1)}(n terms)

(1)        Sn= a +ar+ar2+ ar3 ……….. +ar(n-1)      By multiplying  this equation by r we get

(2)      rSn=       ar+ar2+ ar3 ……       +ar(n-1)+ arn

Subtracting (2) from (1)  we get  Sn- rSn=a- arn

I.e. Sn(1-r) =a(1- rn)

Sn= a (1- rn)    / (1-r)   -----à we use this formula  when r <1

= -a (1- rn)  /-(1-r) (multiply numerator and denominator by -1)

= a   ( rn-1) /  (r-1)      -----à we use this formula  when r >1

What are the possible values of r ? ( r=1, r>1,r<1)

1) If r=1 then GP = {a ,a,a.a,a….}

2) When r<1.

Let us arrive at few terms of the GP when r =  = 0.9 and n is very large.

 r2= 0.81 r4= 0.66 r8= 0.43 r16= 0.19 r64= 0.0012

Thus when n becomes a large number rn  almost becomes zero (we say rn  approaches 0).

This is true even when r is very close to 1(say 999/1000).

Sum of infinite terms of GP when r<1

Sn= a (1- rn) / (1-r)

When n approaches infinity we have

Sinfinity = =

From the above it follows that

1+(1/2)+(1/4)+(1/8)+(1/16)+………. = 1/[1-(1/2)] = 2

In a GP prove that S2n/ Sn = rn+1

S2n/ Sn = [a(1- r2n)/(1-r)]/ [a(1- rn)/(1-r)]

= [a(1- r2n)*(1-r)]/[a (1- rn)*(1-r)]

= (1- r2n)/ (1- rn)

= (1- rn) (1+ rn)/ (1- rn)    ===à apply the formula (a2- b2) = (a-b)*(a+b) and note r2n= (rn)2

=  (1+ rn)

1.8.5 Problem 2 :  Find the sum of the finite series { 1,0.1,0.01,0.001,…. (0.1)9} (Note that the series has 10terms and not 9 terms)

Solution:

In this problem a=1, r=1/10

We know Sn = a (1- rn)    / (1-r)

S10 = 1(1- (1/10)10 )   / (1-1/10)

=  [(1010 -1)/1010]/(9/10)

=  (1010 -1)/(9*109)

1.8.5 Problem 3 : Find the GP if  S10:  S5=  33:1 and T6= 32

Solution:

We know in a GP S10:  S5 = [a(r10-1)/(r-1)]/ [a(r5-1)/(r-1)]

= (r10-1)/ (r5-1)

=  (r5+1)  =====à apply the formula (a2- b2) = (a-b)*(a+b) and note r10= (r5)2

= 33 (given)

r5 =33-1=32    r =2

We know Tn = arn-1

T6 = a25

= 32(given)

a=1

{GP} = (1, 2, 4, 8, 16, 32,…}

1.8.5 Problem 4 :  Suppose you  decide to  celebrate your birthday by distributing sweets to students of few schools. Assume that you distribute sweets   in such a way that packets given to a school is 4 times the packets given in the previous school. To  how many schools can you  distribute packets, if you have 341 sweet packets with you?

Solution:

Let us assume that the 1st school gets 1 packet.

Note that the sequence is a {GP} of {1,4,16,….} and hence a=1, r=4. Sn = 341. We have been asked to find n

Since r >1 We know Sn = [a(rn-1)/(r-1)]

Sn = a(4n-1)/(4-1)

= 1(4n-1)/3

= 341 (given)

(4n-1) = 3Sn = 3*341=1023 or 4n= 1024

n =5

Thus, you can distribute sweets to 5 schools.

1.8.5 Problem 5 : The sum of the first three terms of a GP is 39/10 and their product is 1. Find the terms and the common ratio.

Solution:

Let the first three terms be a/r, a ar

(a/r)*a*ar =1

a3=1

a=1

It is given that a/r+a+ar = 39/10

1/r+1+r = 39/10(a=1)

(1+r+r2)/r =39/10

10(1+r+r2)=39r

10r2-29r+10=0

(2r-5)(5r-2) =0

r =5/2 or r=2/5

The two GPs satisfying the given conditions are

2/5, 1,5/2 Or 5/2,1,2/5

1.8.5 Problem 6: Find a rational number which when expressed as a decimal will have 1. as its expansion.

Solution:

We write 1. = 1.565656…

= 1+ 0.56 +.0056+.00056

= 1+a +ar +ar2+. . .  with a =.56 and r =0.01

= 1+0.56/(1-0.01)

= 1+0.56/0.99

= 1+56/99

=155/99

Thus 155/99 is the required rational number.

1.8.5 Problem 7: If a, b, c are three consecutive terms of an AP, then show that  ka kb and kc are in GP

Solution:

Let b=a+d and c=b+d

Thus d=b-a =c-b

Hence k(b-a)= k(c-b): kb/ ka= kc/ kb

1.8.5 Problem 8: If a, b, c, d are in GP. Prove that

(b-c)2+(c-a)2+(d-b)2 =(a-d)2

Solution:

Since  a, b, c , d are in GP b =ar, c= ar2 and d = ar3

LHS =  (ar –ar2)2 + (ar2-a)2+(ar3-ar)2

= a2{ r(1-r)2+(r2-1)2+ r2(r2-1)2}

= a2{ r6-2r3+1) = a2(r3-1)2 = (ar3-a)2

= (d-a)2 = RHS

1.8.5 Problem 7: If a person gifts 2 Varatakas ( A unit of measurement of money) on the first day and gives out on subsequent days, twice the amount given on previous day, O Lilavati tell me quickly,how much he gives out in a month?( Lilavati Shloka 130)

Solution:

The series given out as gift { 2, 4, 8,16 . . . } is a GP.  If  a=2, r =2 and n=30 we need to find out  Sn

Sn = [a(rn-1)/(r-1)]

Sn = 2(230-1)/(2-1)

= 2(10243-1)    (  230 ={210}3=10243

= 2147483646

Thus the number of Varatakas given out as gift is  214,74,83,646.

1.8.6 Harmonic Progression:

Consider the sequences:

{,  ,  ,…}

{,,…}

By taking reciprocals of the terms of these sequences we get

{ 3, 6, 9 12…} which  is an {AP}(Problem 1.8.3.3)

{8,18,28….} which  is an {AP}(Problem 1.8.3.1)

Definition : A sequence whose terms are reciprocals of terms of an AP  is called ‘Harmonic progression’ and is denoted by ‘{HP}’

We have seen that the general term Tn of an {AP} is a+(n-1)d and hence general term Tn of a {HP} is  ( Reciprocal of  nth term of an AP)

{HP}= {, , , ……. }

Note: There is no formula to find Sn of a HP.

For easy understanding we can say {HP} = {1/AP}. In order to solve a problem on HP, we could take reciprocal of terms of given HP and then solve the problem as if it is a problem on AP.

1.8.6 Problem 1 : In a HP T4=  and T10=  find T19.

Solution:

In a HP Tn=

T4=  =  (given)

T4= =

a+3d =12  ==========à(1)

T10=  = (given)

a+9d =42  ==========à (2)

Subtract (1) from (2) we get

a+9d-(a+3d) =42-12

6d = 30

d =5

Substitute 5 for d in (1) we get

a+3*5 =12

a = (12-15) = -3

Substitute values for a and d in T19  we get

T19=

=

=

1.8.7 Arithmetic, Geometric and Harmonic means (AM,GM and HM)

Definition : if {a, A and b }are in AP then A is called ‘Arithmetic Mean (AM)’ between a and b and is denoted by ‘A’

The derivation for A is shown below.

Since a, A and b are in AP, by definition we have

A-a =b-A( difference between a term and its preceding term is a constant in an AP)

2A = a+b

A =

Definition : if {a, G and b }are in GP then G is called  ‘Geometric Mean (GM)’ between a and b and is denoted by ‘G’

The derivation for G is shown below.

Since a, G and b are in GP, by definition we have

=( the ratio  of a term to its preceding term  is constant in a GP)

G2= ab

G =

Definition : if {a, H and b }are in HP then H is called  ‘Harmonic Mean (HM)’ between a and b and is denoted by ‘H’

The derivation for H is shown below.

Since a, H and b are in HP, by definition we have

(,,) are in AP

Thus - =  - ( difference between a term and its preceding term  is constant in an AP)

= +

=

2ab =H(a+b)

H =

1.8.7 Theorem: If A, G and H are AM, GM and HM of 2 positive numbers respectively, then prove that A,G and H are in GP

We need to prove that G/A =H/G (i.e. the ratio of term to its preceding term is constant)

We are given:

A =

G =

H =

A*H = * = ab= ()2= G2

Or   = G/A : This proves that A,G and H are in GP

Observations : We notice  that AGH for two positive numbers

1.8 Summary of learning

 No Points to remember 1 {AP}= {a, a+d, a+2d,a+3d …..a+(n-1)d} General term  of an AP  is Tn=  a+(n-1)d 2 = 3 Sn  of an AP =  n*[2a+(n-1)*d]/2= n*(a+ Tn)/2 4 {GP} = {a, ar, ar2, ar3 ……….. ar(n-1)}  Tn= Tn-1*r  = ar(n-1) 5 Sn of a  GP = a(1- rn)/(1-r)        If r<1 then  Sinfinity =  a/(1-r) 6 {HP}= {, , , …} General term  of a HP  is Tn= 7 Arithmetic Mean(AM): A= 8 Geometric Mean (GM): G = 9 Harmonic  Mean (HM): H =

1.8.8 Mathematical Induction:

In 1.8.4, we have arrived at the formula for the series 1+2+3…+n as

Sn=

We shall prove the same by principle of mathematical induction:

Mathematical Induction states that :

If f(n) is a statement such that f(n) is true for n=1, then we prove the statement for n=n+1 after assuming it to be true for n.

Note that by using this method, we cannot arrive at a formula/statement which is true for all values of n, however given the formula/statement, We can prove them.

Example : Prove that 1+2+3 ….+n =  by mathematical induction.

Proof :

Let the statement be f(n) =  (1+2+3 ….+n)

We notice that f(1) = 1

And we also notice that 1(1+1)/2=1. Hence f(1) is true

Let the given statement be true for n

i.e. 1+2+3 …+n =

Let us now prove that statement is also true for n+1.

Add the next term, (n+1) to both sides of the above statement.

1+2+3 ….+n +(n+1) = +(n+1)

= (n+1)(n/2+1) = (n+1)(n+2)/2  which is again of the form  m(m+1)/2, where m = n+1.

This proves that the given statement is true.

1.8.8 Problem 1: 1*3 + 3*5 + 5*7 +…….+ (2n-1)*(2n+1) = n(4n2+6n-1)/3

Proof :

Let the statement be f(n) = 1*3 + 3*5 + 5*7 +…….+ (2n-1)*(2n+1)

We notice that f(1) = 1*3

And we also notice that 1(4*12+6*1-1)/3 = 3

Hence f(1) is true.

Let the given statement be true for n

i.e. 1*3 + 3*5 + 5*7 +…….+ (2n-1)*(2n+1) = n(4n2+6n-1)/3

Let us now prove that statement is also true for n+1.

Add the next term (2(n+1)-1)* (2(n+1)+1) =  (2n+1)*(2n+3) to both sides of the above statement.

1*3 + 3*5 + 5*7 +…….+ (2n-1)*(2n+1) + (2n+1)*(2n+3) = n(4n2+6n-1)/3 +(2n+1)*(2n+3)

= (n+1)(n/2+1) = (n+1)(n+2)/2 {After simplification: verify yourself}

which is again of the form  m(m+1)/2 where m = n+1.

This proves that the given statement is true.

Exercises: Using mathematical induction to prove the following:

1. n2  = n(n+1)(2n+1)/6

2. n3  = n2(n+1)2/4 = ()2

3.  1+3+5  . . . . +(2n-1) =  n2 ( Note that this is an AP also)