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 (1^{st}
day 1Rs, 2^{nd} day 2Rs, 3^{rd} day 3Rs. 4^{th} 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(1^{st} day 1Rs, 2^{nd} day 2Rs, 3^{rd}
day 4Rs, 4^{th} 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 1^{st}
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 2^{nd }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 T_{1 }T_{2}T_{3}T_{4}T_{5}
…. as shown below
Order
number of the term ==à |
1st |
2nd |
3rd |
4th |
---- |
nth |
--- |
Corresponding
notation ===à |
T_{1} |
T_{2} |
T_{3} |
T_{4} |
---- |
T_{n} |
--- |
The sequence is generally denoted (represented) by
{T_{n}_{ } }
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 2^{nd}
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 T_{n} here?
We observe
T_{1 }= (1+1)/1
T_{2 } =(2+1)/2
T_{3 } =(3+1)/3
T_{4 }=(4+1)/4
Thus
T_{n}=(n+1)/n .With this
general representation of the nth term we can find any term of the sequence.
Hence the 6^{th} term isT_{6 }=(6+1)/6 =7/6
1.8.1 Problem 1 : If T_{n}_{
}=2n^{2}+1 find the value of n if T_{n}=73
We have T_{n}_{ }=2n^{2}+1 =73; 2n^{2}
=73-1=72: 2n^{2} =72 : n^{2} =36: n = _{}=_{}
Since it is a natural number, it has to be positive
and hence n=6.
Verify that T_{6} = 2*6^{2}+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 S_{n}_{.}
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).
S_{n} = T_{1 }+ T_{2}+T_{3}.........T_{n}_{}
Prove that T_{n} =S_{n}- S_{n-1} _{}
We have S_{n}- S_{n-1}=(_{ }T_{1 }+ T_{2}+T_{3}.........T_{n-1}+ T_{n}) -(_{ }T_{1 }+ T_{2}+T_{3}.........T_{n-1})= T_{n}_{}
_{ }
1.8.2 Problem 1 : If T_{n}_{
}={(-1)^{n}} prove that S_{1} = S_{3} : S_{2}
= S_{4}
Since T_{n}_{ }=(-1)^{n}_{
}we have
T_{1}= (-1)^{1 }= -1, T_{2 }=
(-1)^{2 }=1, T_{3 }= (-1)^{3 }= -1, T_{4}= (-1)^{4
}= 1
Substituting values for series we have
S_{1} = T_{1 } = -1
S_{3} = T_{1 }+ T_{2}+T_{3}=
-1+1-1 = -1 Thus S_{1} = S_{3}
S_{2} = T_{1 }+ T_{2} =-1+1 =0_{}
S_{4} =T_{1 }+ T_{2}+T_{3}
+T_{4}= -1+1-1+1 =0 Thus S_{2} = S_{4}
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 T_{n+1 }– T_{n}_{ }=d and T_{n-1}+d = T_{n}_{ }
The first term of an AP is generally a constant and is denoted by ‘a’ and hence T_{1 }= a_{}
Its other terms are
T_{2}= a+d
T_{3}=
T_{2}+d =(a+d)+d = a+2d = a + (3-1)d
T_{4}=
T_{3}+d =(a+2d)+d =a+3d=
a+(4-1)d
….
General term T_{n}= T_{n-1}+d = a+(n-1)d : thus d= (T_{n}_{ }-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 S_{n}_{ }= 5n^{2}+3n
Solution:
S_{n-1 }= 5(n-1)^{2}+3(n-1) = 5(n^{2
}-2n+1) +3n-3 = 5n^{2}-10n+5+3n-3 = 5n^{2}-7n+2
We know that T_{n}=
S_{n}- S_{n-1}= (5n^{2}+3n)
–(5n^{2}-7n+2) = 10n-2
_{}T_{1 }= 8
_{}T_{2 } =18
_{}T_{3 } =28
So {AP} is {8,18,28…..}
Verification: S_{3 }= T_{1 }+ T_{2 }+ T_{3 }=8+18+28
= 54 = 45+9 = 5*3^{2}+3*3=( 5n^{2}+3n with n=3)
1.8.3 Problem 2 : In an AP T_{10 }=20_{ }T_{20 }=10 Find _{ }T_{30 }
Solution:
If we can find a and d we can arrive at the
solution.
By definition T_{n}_{
}= a+(n-1)d
Thus
T_{10 }= a+(10-1)d = a+9d
But it is
given that T_{10}= 20 so we have
a+9d=20: a=20-9d ====à(1)
Be definition T_{20 }= a+(20-1)d = a+19d ====à(2)
But it is
given that T_{20}= 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
_{} T_{30 }=a+(30-1)d = 29+29*(-1) = 29-29
=0
Verification: Note that T_{10 }=29+9*(-1)=20: T_{20 }=29+19*(-1)=10
which are the given terms.
1.8.3 Problem 3: Find AP whose 5^{th} and 10^{th}
terms are in the ratio of 1:2 and T_{12} =36
Solution:
It is given that T_{5 }: T_{10 }=
1:2 (i.e T_{5 }/T_{10} =1/2)
_{} 2T_{5} = T_{10}
By substituting generic value for T_{5}
& T_{10 }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 T_{12} =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 T_{5 }= 15 and _{ }T_{10} =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 3^{rd} 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*(a^{2}-d^{2}) which is given to be
105
_{}a*(a^{2}-d^{2}) =105
I.e. 5(5^{2}-d^{2}) = 105
I.e. (25-d^{2}) = 21
I.e. 25-d^{2} = 21
I.e. -d^{2} = 21-25
I.e. -d^{2}= -4
I.e. d^{2}= 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 T_{n}
= 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 m^{th}^{
} term of an AP is 1/n and n^{th}
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
_{} T_{m} = a+(m-1)d
= 1/n and T_{n} = a+(n-1)d = 1/m
By
subtracting the T_{m } and T_{n} 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 T_{m} we get
a+(m-1)/mn = 1/n
_{} a = 1/n – (m-1)/mn
= 1/mn
_{} T_{mn}
= 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 (1^{st}
day 1Rs, 2^{nd} day 2Rs, 3^{rd} day 3Rs. 4^{th} day 4Rs
and so on). Will you agree to pay the amount indefinitely?
With the 2^{nd} 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}
S_{n}_{ } = 1 + 2 + 3 ………….+(n-2)+
(n-1) +n(there are n terms)
+ S_{n}_{ } = n +(n-1)+(n-2) … +
3
+ 2 +1(repeated in reverse order)
==================================
2S_{n}= (n+1)+(n+1)+(n+1)
….. .+(n+1)+(n+1)+(n+1)
(there are n terms) = n(n+1)
_{} S_{n}=
_{}
Using this formula in the above problem, let us cross
check the correctness of the amount paid by you after 10 days:
S_{10 } =10*11/2= 55 Rs which is correct!
Now Let us find the amount paid by you after 100
days: S_{100} = 100*101/2 = 5050 Rs
Do you think you will require 200 days to clear the
loan? : S_{200} = 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 S_{141 }= _{} =10,011
So you require 141 days to return the loan.
The above sum (S_{n}
) 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}
S_{n}= [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 _{} )
S_{n}_{ }= 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
_{} S_{n}_{ }=
= 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, T_{13 }=20, we are required to find S_{25}
But T_{13} = a+12d
S_{25 } = n*(a+ T_{25})/2= 25*(a+a+24d)/2 =
25*2*(a+12d)/2 = 25*(a+12d) = 25*20(_{}T_{13} = 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}
S_{n}_{ } = 104+108+112+……
=
104+(104+4) + (104+8)… (104+96) (104 repeats 25 times)(Note that 1^{st}
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 S_{7 }of an AP.
{AP} = {23,21,19….) so we have a=23 and d = -2
Since S_{n}_{ } of an AP = n*(_{} )
_{}S_{7 }
= 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 S_{n}_{
}=70
In the given AP, note that a =16 and d = -1
Since S_{n}_{ } = n*(_{} )
= n*(_{} )
= n*(_{})
= n*(_{})
n*(_{}) = 70(_{} total distance =70km)
Thus we have
an equation to solve
(33n-n^{2}
) = 2*70=140 or
-n^{2} +33n -140 =0 or
n^{2}
-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 28^{th} term of this AP
T_{n}= a+(n-1)d
so
T_{28}=
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 S_{n}_{
}=70. We need to find d
S_{n}_{ } = 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 S_{n}_{
}=360. We need to find n
S_{n}_{ } = n*(_{} )
= n*(_{} )
= n*(3n+2n-2) = n(n+2)
_{} n^{2}+2n =360
_{} n^{2}+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 T_{n}_{ }/T_{n-1 }= constant. In the
first case T_{3 }/T_{2}=_{}=2 and in the 2^{nd} case T_{3 }/T_{2}=_{} = 1/3
In a GP if the first term T_{1} = a and the
ratio is r we have
T_{2}= T_{1}*r= ar^{(2-1)}
T_{3}= T_{2}*r= ar*r
=ar^{2}= ar^{(3-1)}
T_{4}= T_{3}*r= ar^{2}*r =
ar^{3}= ar^{(4-1)}
In general T_{n}=
ar^{(n-1)}
In a GP we also know that T_{n}=
T_{n-1}*r
So {a, ar, ar^{2},
ar^{3} ……….. ar^{(n-1)}} is the standard form of GP.
1.8.5 Problem 1 : In a GP 7^{th} term is eight times the
fourth term and the 5^{th} term is 12 find the GP if S_{10}: S_{5}= 33:1 and T_{6}= 32
Solution:
T_{n}_{ }= ar^{n-1}
_{} T_{7}=a r^{6} and T_{4}=a r^{3} it is also given that T_{7}= 8T_{4}
_{} a r^{6}= 8a r^{3}
_{} r^{3}= 8
_{} r=2
We know T_{5}=a r^{4}
= a 2^{4}=16a
=12 (given)
_{} a = _{} =_{}
Therefore {GP} = {_{}, _{}*2, _{}*2^{2} , _{}*2^{3}….} = {3/4, 3/2,3,6…}
Let us
find the sum of n terms of a GP = {a, ar, ar^{2}, ar^{3} ……….. ar^{(n-1)}}(n terms)
(1) S_{n}=
a +ar+ar^{2}+ ar^{3} ……….. +ar^{(n-1}) By
multiplying this equation by r we get
(2) rS_{n}= ar+ar^{2}+ ar^{3}
…… +ar^{(n-1)}+
ar^{n}
Subtracting (2) from (1) we get S_{n}- rS_{n}=a- ar^{n}
I.e. S_{n}(1-r)
=a(1- r^{n})
S_{n}= a (1- r^{n}) /
(1-r) -----à we use this
formula when r <1
= -a (1- r^{n})
/-(1-r) (multiply numerator and denominator by -1)
= a (
r^{n}-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.
r^{2}= |
0.81 |
r^{4}= |
0.66 |
r^{8}= |
0.43 |
r^{16}= |
0.19 |
r^{64}= |
0.0012 |
Thus when n becomes a large number r^{n}^{ } almost becomes zero (we say r^{n}^{ }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
S_{n}= a (1- r^{n}) / (1-r)
When n approaches infinity we have
S_{infinity}_{ }= _{}= _{}
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 S_{2n}/ S_{n}_{ }= r^{n}+1
S_{2n}/ S_{n}_{
}= [a(1- r^{2n})/(1-r)]/ [a(1- r^{n})/(1-r)]
=
[a(1- r^{2n})*(1-r)]/[a (1- r^{n})*(1-r)]
=
(1- r^{2n})/ (1- r^{n})
=
(1- r^{n}) (1+ r^{n})/
(1- r^{n})
===à apply the formula (a^{2}- b^{2}) = (a-b)*(a+b) and note r^{2n}= (r^{n})^{2}
= (1+ r^{n})
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 S_{n} = a
(1- r^{n})
/ (1-r)
_{}S_{10} = 1(1- (1/10)^{10} ) / (1-1/10)
= [(10^{10 }-1)/10^{10}]/(9/10)
= (10^{10 }-1)/(9*10^{9})
1.8.5 Problem 3 : Find the GP if
S_{10}: S_{5}= 33:1 and T_{6}= 32
Solution:
We know in a GP S_{10}: S_{5 }= [a(r^{10}-1)/(r-1)]/
[a(r^{5}-1)/(r-1)]
=
(r^{10}-1)/ (r^{5}-1)
= (r^{5}+1) =====à apply the formula (a^{2}-
b^{2}) = (a-b)*(a+b) and note r^{10}=
(r^{5})^{2}
=
33 (given)
_{} r^{5 }=33-1=32 _{} r =2
We know T_{n}_{ }= ar^{n-1}
T_{6} = a2^{5}
^{ }= 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 1^{st} school gets 1
packet.
Note that the sequence is a {GP} of {1,4,16,….} and
hence a=1, r=4. S_{n}_{ }= 341. We
have been asked to find n
Since r >1 We know S_{n}_{
}= [a(r^{n}-1)/(r-1)]
_{}S_{n} = a(4^{n}-1)/(4-1)
= 1(4^{n}-1)/3
= 341
(given)
_{} (4^{n}-1) = 3S_{n}
= 3*341=1023 or 4^{n}= 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
_{}a^{3}=1
_{} a=1
It is given that a/r+a+ar
= 39/10
_{}1/r+1+r = 39/10(_{}a=1)
_{} (1+r+r^{2})/r
=39/10
_{} 10(1+r+r^{2})=39r
_{}10r^{2}-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 +ar^{2}+.
. . 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
k^{a} k^{b} and k^{c}
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)}: k^{b}/
k^{a}= k^{c}/ k^{b}
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= ar^{2} and d = ar^{3}
LHS = (ar –ar^{2})^{2} + (ar^{2}-a)^{2}+(ar^{3}-ar)^{2}
= a^{2}{ r(1-r)^{2}+(r^{2}-1)^{2}+
r^{2}(r^{2}-1)^{2}}
= a^{2}{ r^{6}-2r^{3}+1) = a^{2}(r^{3}-1)^{2} = (ar^{3}-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 S_{n}
_{}S_{n}_{ }= [a(r^{n}-1)/(r-1)]
_{}S_{n} = 2(2^{30}-1)/(2-1)
= 2(1024^{3}-1) ( _{} 2^{30 }={2^{10}}^{3}=1024^{3}
=
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 T_{n}_{ }of an {AP} is a+(n-1)d and hence
general term T_{n} of a {HP} is _{} ( Reciprocal of nth term of an AP)
{HP}= {_{}, _{}, _{}, _{}……. _{}}
Note: There is no formula to find S_{n} 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 T_{4}= _{} and T_{10}= _{} find T_{19}.
Solution:
In a HP T_{n}= _{}
_{} T_{4}= _{} = _{} (given)
_{}T_{4}= _{}= _{}
_{} a+3d =12 ==========à(1)
T_{10}= _{} = _{}(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 T_{19 }^{ }we get
T_{19}= _{}
= _{}
= _{}
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)
G^{2}= 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}= G^{2}
Or _{}= G/A : This proves that A,G and H are in GP
Observations : We
notice that A_{}G_{}H 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 T_{n}= a+(n-1)d |
2 |
_{}= _{} |
3 |
S_{n}_{ } of an AP =
n*[2a+(n-1)*d]/2= n*(a+ T_{n})/2 |
4 |
{GP}
= {a, ar, ar^{2}, ar^{3} ……….. ar^{(n-1)}}
T_{n}= T_{n-1}*r = ar^{(n-1)} |
5 |
S_{n} of a GP = a(1- r^{n})/(1-r)
If r<1 then S_{infinity}_{
}= a/(1-r) |
6 |
{HP}=
{_{}, _{}, _{}, _{}…} General term of a
HP is T_{n}= _{} |
7 |
Arithmetic
Mean(AM): A= _{} |
8 |
Geometric
Mean (GM): G =_{} |
9 |
Harmonic Mean (HM): H = _{} |
Additional Points:
1.8.8 Mathematical
Induction:
In 1.8.4, we have arrived at the formula for the
series 1+2+3…+n as
S_{n}= _{}
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(4n^{2}+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*1^{2}+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(4n^{2}+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(4n^{2}+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. _{}n^{2 } =
n(n+1)(2n+1)/6
2. _{}n^{3 } = n^{2}(n+1)^{2}/4 = (_{})^{2}
3.
1+3+5 . . . . +(2n-1) = n^{2 }( Note that this is an AP also)