sequence, series and progressions

7/27/2019 Sequence, Series and Progressions

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Quantitative AbilitySequence, Series and Progressions


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Importance in CAT

One of the major testing areas in CAT

Average weightage over the years has been

10% of the Quantitative Ability section The questions related to sequence and series

can be usually solved logically

The questions related to progressions areusually formulae based.


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Key Concepts

Sequences and Series Sum or difference

Cumulative Sequence

Power Series

Alternate Sequence

Progressions Arithmetic Progressions

Geometric Progressions Harmonic Progressions

Arithmetic/geometric/harmonic mean


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Types of Questions frequently asked

Sequence and Series

Identify the next term in a sequence or series

Find the sum of a given series

Progressions Sum of terms in an Arithmetic/Geometric/Arithmetico-

Geometric progression

Number of terms or value of a certain term in anarithmetic/geometric/harmonic progression

Questions based on the concepts ofarithmetic/geometric/harmonic progressions

Questions based on concepts of arithmetic/geometric mean

Word problems related to concepts of arithmetic/geometricprogression and mean


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ExamplesSequence and Series

Find the nth term of the sequence 1, 2, 4, 7, 11, 16,

Let us observe the given sequence. The nth term is denoted as Tn.

T1 = 1

T2 = (1 + 1) = 2

T3 = (2 + 2) = 4

T4= (4 + 3) = 7

T5 = (7 + 4) = 11 and so on.

Thus we can generalize the given series in the following way to

find out any term.Tn= T(n1) + (n1)

This is an example of a sum/difference sequence.


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Find the general term of the sequence 3, 7, 10, 17, 27, 44 Let us denote the nth term of the given sequence as Tn

We can see that

T1 = 3

T2 = 7

T3 = (T1 + T2) = (3 + 7) = 10

T4 = (T2 + T3) = (7 + 10) = 17

T5 = (T3 + T4) = (10 + 17) = 27 and so on.

Thus as we have found the pattern, we can define Tnwithout loss

of generality asTn = Tn 2 + Tn 1 which is the required general term for the

aforementioned sequence.



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Ifa1 = 1 and an+1 = 2an + 5, n = 1, 2,... , then a100 is equal to

(1) (5 2996) (2) (5 299 + 6)

(3) (6 299 + 5) (4) (6 2995)

Let us find the values of first few terms of the given series by substitutingvalues starting from 1 for n.

a1 = 1 = 6 5 = 6 205

a2 = 7 = 6 2 5 = 6 215

a3 = 19 = 6 4 5 = 6 225

a4 = 43 = 6 8 5 = 6 235and so on.

We can see that the value of a particular term an is of the form

6 2(n 1)5

Thus, a100 = 6 299 5

Hence, option 4.



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Find the next two terms in the sequence 0, 6, 8, 16, 64, 116,216,

Let us analyse and break up the sequence. Assume that thenth term is

denoted by Tn. Let us try and find a pattern in the given sequence.

T1 = 0

T2 = 6

T3 = 8 = 2T4 = 16 = 6 + 10

T5 = 64 = 4

T6 = 116 = 16 + 100

T7 = 216 = 6

Thus the above sequence is a combination of 2 sequences, one being the

cubes of even numbers

0, 8, 64, 216,

and the other being the one with

powers of 10 added to the previous terms

6, 16, 116,

T8 = 10 + 116 = 1116 and T9 = 83 = 512 are the next two terms.



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ProgressionsImportant Properties

and Formulae

Arithmetic Progressions

A sequence in which consecutive terms are a fixed distance

apart i.e. the difference between two consecutive terms is

always constant is said to be in Arithmetic Progression. Ifa is the first term and dis the common difference of the

sequence in Arithmetic Progression, then,

nth term Tn= a + (n 1)d

Sum of terms S = [2 + 1 ]

This can also be written as S =

[ + + 1 ]

S =

[ + T]


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Geometric Progressions A sequence in which the ratio of two successive terms is

always constant is said to be in Geometric Progression.

Ifa is the first term and ris the common ratio of the given

sequence in Geometric Progression, then

The th term T

= Sum of terms S =

|| > 1 or

0 1

Sum of terms S =

for 0 1 Sum of infinite terms doesnt exist when |r| > 1 because the

series is divergent i.e. the sum doesnt tend towards a certain

value.


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Harmonic Progressions

The reciprocals of terms in Arithmetic Progression are said to

be in Harmonic Progression.

Ifa is the first term of an A.P and dis the common difference,

,

:,

:, ,

: are terms in Harmonic Progression

The nth term of a harmonic progression is given as

:

.

Arithmetic, Geometric and Harmonic mean Ifa, b and c are in Arithmetic Progression then ba = cb

Thus 2b = a + c and b = (a + c)/2.

Thus b is the arithmetic mean ofa and c.

We can generalize this result for n terms.

Average ofn terms =

=

:


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Ifa, b and c are in Geometric Progression then

=

Thus b2 = ac and =

b is said to be the geometric mean ofa and c.

If

,

and

are in Arithmetic Progression, then

a, b and c are in Harmonic Progression.

Ifa, b and c are in Harmonic Progression, then = :

b is the harmonic mean ofa and c.

Note: Arithmetic Mean (A.M)of two numbers is always

greater than or equal to the corresponding Geometric Mean(G.M.) which in case is always greater than or equal to the

corresponding Harmonic mean (H.M.).

(G.M)2 = (A.M) (H.M)


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Example - Arithmetic Progression

The 54th and the 4thterms of an A.P. are 61 and 64respectively. Find the nth term.

Solution:

Assume that the first term of the A.P is a and the common difference is d.

We will follow this convention for all A.P problems.The 54th term is represented as T54 = a + 53d= 61 (i)

The 4th term is represented as T4 = a + 3d= 64 (ii)

By subtracting (ii) from (i), we get,

50d= 125

d= 5/2 =2.5By substituting the value ofdin equation (ii), we get

a = 64 3(2.5) = 71.5

Hence, the nth term of this arithmetic progression is

Tn = 71.5 + (n 1)(2.5) = 74 2.5n


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Example - Arithmetic Progression

How many terms of the series 12, 9, 6, must be taken sothat the sum may become 78?

Solution:

Recall that the formula for the sum ofn terms in an arithmetic progression

is given by =

+

Here, values ofa, dand Snare known and we have to find out n.

Let us substitute the known values in the above equation to obtainn.

[24 + ( 1)3] = 78n (n 9) = 52

n2 9n 52 = 0

(n 13)(n + 4) = 0

n = 13

Thus sum of 13 terms will be needed to obtain 78.


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Word ProblemArithmetic Progression A teacher observes that the marks that the students in her class have scored

are all different. She arranges her students in a line in increasing order of their

marks such that difference in marks scored by any two students next to eachother is 4. The lowest marks that any student has scored are 11. The sum ofthe marks that all her students have scored is 585. Find the marks scored bythe student standing in the middle of the line.

Solution:

Recall that the formula for sum ofn terms in arithmetic progression is =

+

It is given that Sn = 585, d= 4 and a = 11 and we need to find out the number

of students present in the line.

585 =

+

On re-arranging the above equation, we will obtain a quadratic equation.One of the roots of this equation is n = 15 and the other one is negative.

Thus there are 15 students present in the line and the 8th student stands in the

middle of the line.

Note that the score of the 8th student is nothing but the average of the scores of all

the students.Thus the score of the 8th student = 585/15 = 39.


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ExampleGeometric Progression

The product of the first three consecutive terms of anincreasing G.P. is 216 and their sum is 21. Find the fourth termof this G.P.

Solution:

Let the three terms of the G.P. bex/r,xandxrrespectively.It is given that the product of the three terms (a/r a ar) = 216

x3 = 216. Thus the second term (x)= 6

Also, sum of the three terms (x/r+x+xr) = 21

6/r+ 6/r= 15

6r2 15r+ 6 = 0 (2r 1)(r 2) = 0

= 2 or =

As the G.P given is an increasing G.P., r= 2 and the first term a = 3

The fourth term of this G.P. is T4 = ar3 = 3 23 = 24


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Find the sum ofn terms of the series 9.01 + 99.001 + 999.0001 +

Solution:

In the current situation, the terms of the given sequence are not in G.P

Let us split the terms such that a combination Geometric sequencescan be formed from the given sequence.

Sn= 9.01 + 99.001 + 999.0001 + upto n terms

Sn= (9 + 99 + 999 + upto nterms) + (0.01 + 0.001 + 0.0001 + upto n terms)

Sn= *(10 1) + (100 1) + (1000 1) + upto n terms]+ (0.01 + 0.001 + 0.0001 + upto n terms)

Thus we have been able to express the given sequence as a sum of 2

different sequences in G.P.

contd.


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Let S1 = (10 1) + (100 1) + (1000 1) + upto n terms

and S2 = 0.01 + 0.001 + 0.0001 + upto n terms

Recall that sum ofn terms of a sequence in G.P is

S =

|| > 1 or

0 1

S1= (10 + 100 + 1000 + ) n

S =

9

S =. ;.

.9

S=

9+

. ;.

.9



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Word ProblemGeometric Progression A man decides to save 10% of his income every year. But from the

second year onwards, he manages to save only 90% of his savingspercentage in the previous year. How much money does he saveat the end of 5 years if his income in the 1st year is Rs. 2,00,000and his income increases every year by 10%?

Solution:Let us calculate the salary and savings of this person for the first 3 years

in order to find a establish a pattern regarding the amount he saves yearly.

Savings in the first year =x= 200000 0.1 = 20000

Salary for the second year = (200000 1.1)

Thus, savings in the 2nd year = (200000 1.1) (0.1 0.9)

= (200000 0.1) (1.1 0.9) = 0.99x

contd.


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Word ProblemGeometric Progression

Salary for the third year = (200000 1.12)Savings in the third year = (200000 1.12) (0.1 0.92) = (0.99)2x

Similarly, savings in the fourth and fifth years are (0.99)3xand (0.99)4x.

His savings at the end of five years

=x(1 + 0.99 + 0.992 + 0.993 + 0.994)

= 20000(1 0.995)/(1 0.99) [ 0 < r< 1 hence Sn =

]

2000000(10.95) 98020

The man saves approximately Rs. 98,020 at the end of five years.


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ExampleArithmetico - Geometric

Progression

What is the approximate sum of 50 terms of the series givenbelow?

+

+

9

+

6+

Solution:

The numerators of the terms in the given series are in A.P., with a

common difference of 2. Thus the numerator of the 50th term is 103.

The denominators are in G.P. with common ratio = 1/2. The denominator

of the 50th term is 250.

S =

+

+9

+ +

(i)

Let us divide the above given equation by 2.

S =

+

+

9

6+ +

+

(ii)

Subtracting (ii) from (i),

contd.


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ExampleArithmetico - Geometric

Progression

S = +;

+ 9;

+ + ;

S =

+ 2

+

+

6+ +

S =

+ 2

;

S = 5

+ 2

;

;

1 and

0

S50 5 + 2 = 7

W d P bl I fi it G t i


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Word ProblemInfinite Geometric

series A ping pong ball is dropped from a 45 metres high multi-

storey building. The ball bounces back three fifth of thedistance each time before coming to rest. The total distancetraversed by the ball is:

Solution:

Every time the ball bounces, it travels 3/5

th

of the distance ithad previously travelled twice (once going upwards and once while

coming down)

The distance traversed by the ball is

4 5 + 2

45 + 2

45 + 2

45 +

= 4 5 + 2 4 5

+

+

+

= 4 5 + 2 4 5

;

= 180 m


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ExampleHarmonic Mean If three positive real numbers a, b and c (c > a) are in

Harmonic Progression, thenlog (a + c) + log (a 2b + c) is equal to:

(1) 2 log (cb) (2) 2 log (ac)

(3) 2 log (ca) (4) log a + log b + log c

Solution:

As a, b, c are in H.P.,

=

:

log (a + c) + log (a 2b + c) = log [(a + c)2 2b(a + c)]

= log [(a + c)2 4ac)]

= log (ac)2

= 2 log(ca) (c > a)

Hence, option 3.


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ExampleRelation amongst A.M,

G.M and H.M.

The arithmetic mean of two numbers exceeds its geometricmean by 18. The geometric mean is 2.125 times the harmonicmean. Find the difference between the two numbers.

Solution:

Letp and q be the two numbers. Let us denote Arithmetic Mean byA,

Geometric Mean by Gand Harmonic mean by H.

The following data has been given to us

A = G + 18

G = 2.125H = 17H/8

But, G2 =A H

9

6 =

+ 18

On simplifying the equation that we have obtained, we get =

.


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ExampleRelation amongst A.M,

G.M and H.M.

A = 34 and G = 16

We have to find the value of (pq).

(pq)2 = (p + q)2 4pq

(pq)2 = (2A)2 4G2

(pq)2 = 682 4 162

(pq)2 = 3600

(pq) = 60

Difference between the two numbers = 60


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Important Properties of A.P and G.P

If we add, subtract, multiply or divide any constant (say k) to

all the terms of an A.P, the resulting sequence is also in A.P.

If each term of a G.P. is multiplied or divided by the some non-

zero quantity, then the resulting sequence is also a G.P. with

the common ratio remaining the same

In a finite G.P., the product of two terms equidistant from the

first and the last terms is same as the product of the first and

the last term.

The reciprocals of the terms of a given G.P. also form a G.P.,

where the common ratio is the reciprocal of that of the earlier

G.P.


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Important Series

Sum of first n natural numbers = :

Sum of squares of first n natural numbers = : :

6

Sum of cubes of first n natural number = :

2 + 2 + 2 + + 2 = 2 1

=

! +

! +

! + +

!


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Example

Ginny is adding the numbers from 1 to 50 for her homework. She makes amistake and wrongly adds the reverse of a two-digit number instead of thenumber. As a result, her sum is 1347. Which is the number that she addsthe reverse of?

Solution:

Sum of numbers from 1 to 50 = 50 51/2 = 1275

Ginnys sum = 1347

Let the number that Ginny wrongly adds be 10x+ y

Ginnys sum = 1275 (10x+ y) + (10y+x)

1347 = 1275 9(xy)

9(yx) = 72yx= 8

y= 9 andx= 1

Ginny wrongly adds the reverse of 19.

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