Progressions concepts for CAT

This is one of the chapters that contributes extremely logical problems that can be readily solved by innovative methods as opposed to conventional methods (such as formula application, etc.). The CAT is a test where logic predominates over formal mathematics.

Therefore, students who use creativity to solve these problems receive higher grades than those who rely on formulas. In the beginning of the chapter, I presented conventional mathematical methods, and then, as the chapter progressed, I introduced novel techniques for quicker and more accurate calculation. Remember that 1–2 questions must be posed on the CAT, and this chapter is also essential for other entrance exams, such as XAT, FMS, IIFT, CET, and MAT.


Sequence

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A sequence is a collection of numbers occurring in a specific order or according to a rule. Thus, any set of numbers a1, a2, a3,..., an such that for each positive integer, there is a number n, i.e., a sequence, is sequential. The numerals a1, a2, a3,..., an are referred to as elements (or members) of the sequence.

The number of its elements determines whether a sequence is finite or infinite. If there are a finite number of elements, the sequence is finite; if there are an infinite number of elements, the sequence is infinite.


Series

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Let u1,u2,u3,...,un be provided as a sequence of n integers, where n is a specified positive integer. The sum of these numbers, u1 + u2 + u3 +... + un, is referred to as a finite series, while the numbers u1, u2, u3,..., un are referred to as terms of the series.

If Sn = u1 + u2 + u3 +... + un, then Sn is the sum of n terms (or the sum of the firstn terms) of the series.


Progressions

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Progression is the name for a certain kind of sequence. When the nth term "an" is given, the series "a1,a2,...,an" is known.

(i) If an =2n, then the sequence is 2, 4, 6, 8, ...
(ii) If an = 2n −1, then the sequence is 1, 3, 5, 7, ....

Based on the type of sequence , progressions can be of three types.

  • Arithmetic Progression – AP
  • Geometric Progression – GP
  • Harmonic Progression – HP

Arithmetic Progression - AP

A series of numbers a1, a2, a3,..., an is said to be in arithmetic progression when a2 – a1 = a3 – a2 =..., which means that the numbers keep going up or down by the same amount. This constant amount is called the common difference of the AP.

Nth term of an AP is denoted by Tn and is given by

T= a + (n-1)d

Last term of an AP is denoted by l and is given by

l = a + (n-1)d

Sum of First n Terms of an A.P.

Let if a be the first term of the progression and d be the common difference, then the

First term T1= a

Second term T2 = a+d

Third term T3 = a+2d

Fourth term T4 = a+3d

Fifth term T5 = a+4d and so on.

S= T1+T2+T3+T4+….+Tn

 S= a+( a+d) +(a+2d) +(a+3d)+….+a+(n-1)d

 Sn\(\frac{n}{2}\) [2a + (n-1)d]

Arithmetic mean(AM)

When there are three quantities in AP, the middle one is referred to as the arithmetic mean (A.M.) of the other two.

a, b, c are in AP, then A.M. of a and c i.e. \(b=\frac{a+c}{2}\) = AM (Arithmetic Mean)

How to Insert n Arithmetic Means Between a and b

If a and b are the first and last terms between which we want to insert n arithmetic means m1, m2, m3,..., mn, then m1, m2,..., mn.with common difference d = \(\frac{b-a}{n+1}\)

Properties of Arithmetic Progressions (AP)

  1. When a specific number is added or subtracted from each term of a given AP, the resulting sequence is also an AP and has the same common difference as the given AP.
  2. When each term of an AP is multiplied by a constant (or divided by a nonzero constant), the resulting sequence is also an AP.
  3. If m1, m2, m3,..., mk and n1, n2, n3,..., nk are in AP, then (m1 +n1),(m2 +n2),(m3 +n3),...,(mk +nk ) are also in AP.
  4. If m1, m2, m3,..., mk and n1, n2, n3,..., nk are in AP, then (m1n1),(m2n2),(m3n3),...,(mk nk ) are also in AP
  5. If there are an odd number of terms in an AP as a1, a2, a3,..., an, then the term in the middle is the A.M.
  6. If there are even number of terms in an AP as a1 , a2 , a3 ,..., an , then a1 +an a2 +an–1a3 +an–2a4 +an–3 etc.
  7. The sum of an AP with n terms is given by Sn n(A.M.)
  8. If (m - k), m, and (m + k) are consecutive elements of an AP, and k is the common difference. Similarly, five AP terms can be represented as (m - 2k), (m - k), m, (m + k), and (m + 2k).
  9. If we must take 4 AP terms, then the terms are(m - 3k), (m - k), (m + k), (m + 3k), where (m - 3k), (m - k), (m + k), (m + 3k) are consecutive AP terms and (2k) is the common difference. Similarly, six AP terms are as (m–5k), (m–3k), (m–k), (m+k), (m+3k), and (m+5k).
  10. The sum of first n natural numbers

\( \Sigma n = 1+ 2+3+4+......+n= \frac{n(n+1)}{2}\)

  1. The sum of squares of first ‘n’ natural numbers

\(\Sigma n^2= 1^2+ 2^2+ 3^2+4^2+.......+n^2 = \frac{n(n+1)(2n+1)}{6}\)

  1. The sum of cubes of first ‘n’ natural numbers

\(\Sigma n^3= 1^3+ 2^3+ 3^3+ 4^3+........+ n^3= [ \frac{n(n+1)}{6}]^2= [\Sigma n]^2\)

  1. The sum of first ‘n’ odd numbers i.e., 1+3 +5 + 7 +... + (2n –1) = n2
  2. The sum of first n even numbers i. e. , 2 + 4 + 6 + ... + 2 n = n ( n + 1)

Geometric Progression-GP

A sequence a1, a2, a3,..., an is said to be in Geometric Progression if

 \(\frac{a_2}{a_1}= \frac{a_3}{a_2}= \frac{a_4}{a_3}=.........r\)

And a1,a2,a3,...,an are all non-zero numbers and r is said to be the common ratio of the G.P

Nth term of Geometric Progression (GP)

First term T1 = a

Second term T2 = ar2

Third term T3 = ar3

Fourth term T4 = ar4

Fifth term T5 = ar5 and so on.

So the nth term of a GP is given by. Tn = a.r(n-1) it is also the last term (l) of GP.

Sn=T1+T2+T3+T4+….+Tn

S\(\frac{a(r^n-1)}{(r-1)}\) , if r>1

S\(\frac{a(1-r^n)}{(1-r)}\) , if r<1 also S\(\frac{lr-a}{r-1}\) , if r>1

Sum of an Infinite G.P.

S\(\frac{a}{1-r}\) ; |r|<1

a is the first term and r is the common ratio of the G.P. of infinite terms.

Geometric Mean (G.M.)

If three quantities are in G.P., the middle one is called the

geometric mean (G.M.) of the other two quantities. If a, m, b are in G.P., then m is the G.M. of a and b.

\(\frac{m}{a}= \frac{b}{m} \)

m2 = ab

m = \(\pm \sqrt{ab}\)

if a1, a2, a3,..., an are in G.P., then

G.M. = (a1 .a2 .a3 .....an )\(\frac{1}{n}\)

Properties of Geometric Progression

  1. If each term of a G.P. is multiplied or divided by a certain non-zero constant, then the resulting sequence is also a G.P.
  2. If a1 ,a2 ,a3 ,... and b1 ,b2 ,b3 ,... are two geometric progressions, then the sequence a1 b1 , a2 b2 , a3 b3 ,... is also a G.P.
  3. If there are odd number of terms in a G.P. as a1, a2, a3,..., an then the middle most term is the G.M. of the given sequence.
  4. If there are even number of terms in a G.P. as a1 , a2 , a3 ,..., an , then

\(\sqrt{a_1.a_n} = \sqrt{a_2.a_{n-2}} = \sqrt{a_3.a_{n-3}}= ......GM\)

  1. If we have to take three terms in G.P., it is convenient to take them as (m/k) , m , mk; where (m/k), m and mk are consecutive terms of G.P. and k is the common ratio. Similarly we can take five terms of G.P. as (m/k2),(m/k), m, mk, mk2
  2. If a1, a2, a3,..., an are in G.P. then a1m, a2m, a3m, a4m,……..anm are also in G.P.
  3. If a1,a2,a3,...,an are in G.P. (for every ai >0; iI+ then log a1, log a2, log a3,...are in AP

Harmonic Progression-HP

A series of quantities a1, a2, a3, a4 ..., an is said to be in H.P. when their reciprocals \(\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3},......., \frac{1}{a_n}\) are in AP

Harmonic Mean (H.M.)

When three quantities are in H.P., the middle one is called the Harmonic Mean (H.M.) between the other two.

If a, H, b are in H.P.

then H is the H.M.of a, b.

 \(\frac{1}{H} - \frac{1}{a} = \frac{1}{b}- \frac{1}{H}\)

 \(\frac{2}{H}= \frac{1}{a} + \frac{1}{b}\)

 \(\frac{2}{H}= \frac{a+b}{ab}\)

 H = \(\frac{2ab}{a+b}\)


Previous Year CAT Questions

Ques 1: Let X be the set of integers {9, 15, 21, 27, ...,375}. Y denotes a subset of X, such that the sum of no two elements of Y is 384. Find the maximum number of elements in Y. (CAT 2014)

  1. 29
  2. 30
  3. 31
  4. 32

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Ans: (C)

X is a set of integers whose elements, when arranged in ascending order, form an arithmetic progression with 9 as the first term and 6 as the common difference. Let us say it has n constituents.

 375 = 9 + (n − 1)6

 n=62

In an arithmetic progression with an even number of terms, the sums of the kth term from the beginning and the kth term from the end are identical.

The utmost number of elements in y is therefore 31.

Ques 2: Along a road lie an odd number of stones placed at intervals of 10 m. These stones have to be assembled around the middle stone. A person can carry only one stone at a time. A man carried out the job starting with the stone in the middle, carrying stones in succession, thereby covering a distance of 4.8 km. Then, the number of stones is __  (CAT 1994)

  1. 35
  2. 15
  3. 29
  4. 31

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Ans: (D)

It is stated that a man traveled 4,800 meters to position stones on both sides of the central stone. It indicates that 2400 meters will be traveled to assemble the stones of one side.

Now, the distance he traveled to bring the first stone to the center (he begins from the center) is 10 + 10 = 20 meters.

Distance traveled to deliver second stone to center = 20 + 20 = 40

If n stones are kept on one side, then the total distance traveled to assemble n stones is n.

=20+40+60+80+…

2400 = \(\frac{n}{2}\) [2 x 20 + (n-1)20]

n=15

Total number of stones = (2n + 1) = 31

Ques 3: Let x, y, z be three positive real numbers in a geometric progression such that x < y < z. If 5x, 16y and 12z are in an arithmetic progression, then the common ratio of the geometric progression is (CAT 2018)

  1. \(\frac{5}{2}\)
  2. \(\frac{1}{6}\)
  3. \(\frac{3}{2}\)
  4. \(\frac{3}{6}\)

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Ans: (A)

Given three positive real numbers = x, y, z

⇒ The given numbers are in geometric progression that means in

the form of a, ar, ar2 .........

Solving using options:

Take option (i) ⇒ 5/2

5x,16y,12z

Put (x = 1), 5(1) = 5

 16y= 16 x \(\frac{5}{2}\)= 40

 12y = 12 x \(\frac{25}{4}\)

The common ratio is GP is \(\frac{5}{2}\)

Ques 4: a, b, c , d and e are 5 distinct numbers that are from an arithmetic progression. They are not necessarily consecutive terms but from the first 5 terms of the AP. It is known that c is the arithmetic mean of a and b, d is the arithmetic mean of b and c. Which of the following statements are true?

  1. Average of all 5 terms put together is c.
  2. Average of d and e is not greater than average of a and b.
  3. Average of b and c is greater than average of a and d. (CAT 2017)
  1. A and B
  2. B and C
  3. A, B and C
  4. A and C

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Ans: 

  1. c, is the arithmetic mean of a and b; therefore, c falls between a and b. And it precisely falls between the two categories. As in, the number of terms between a and c and b and c should be equal. a, c, and b could be the first, second, and third terms, or the first, third, and fifth terms, or the second, third, and fourth terms, or the third, fourth, and fifth terms. The terms could also be used in reverse order. Thus, b, c, and a could be the first, second, and third terms, or the first, third, and fifth terms, and so on. This is a very basic but extremely potent concept.

d is the arithmetic mean of b and c; therefore, d rests between b and c.

Using statements I and II, we can conclude that a, c, and b must be either first, third, and fifth or fifth, third, and first, as there is also an element between b and c.

Thus, term c is the third. a and b are, in some order, first and fifth. Statement I: The weighted average of all five terms is c. The middle term is c. Therefore, this is true.

The average of d and e is not higher than the average of a and b. Average of a, b is c, and d and e are the second and fourth terms in this series (in some order). Therefore, their average should also equal c. Therefore, both are equivalent. Therefore, this assertion is also true.

Statement III: The mean of b and c exceeds the mean of a and d. d is the average of b and c. The average of a and d may be higher or lower than d. Thus, this need not be the case.

Ques 5: If the sum of the first 11 terms of an arithmetic progression equals that of the first 19 terms, then what is the sum of the first 30 terms? (CAT 2004)

  1. 0
  2. −1
  3. 1
  4. Not unique

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Ans: (A)

a is the first term and d is the common difference.

 S11= S19

 \(\frac{11}{2}\) (2a+10d)\(\frac{19}{2}\) (2a+ 18d)

 16a+232d =0

 2a+29d=0

 S30 = \(\frac{30}{2}\)(2a+29d) = 15 x 0 = 0

Ques 6: The number of common terms in the two sequences 17, 21, 25, ...,417 and 16, 21, 26, ...466 is __ . (CAT 2008)

  1. 78 
  2. 19
  3. 20
  4. 77
  5. 22

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Ans: (C)

Both the sequences (17,21,25, ... ) and (16,21,26,...) are arithmetic progression with a common difference of 4 and 5, respectively.

In both the sequences first common term is 21. Hence, a new arithmetic sequence containing the common terms of both the series can be formed with a common difference of LCM of (4, 5), i.e. 20.

∴New sequence will be 21, 41, 61, ..., 401.

nth term = a + (n −1)d

401=21+(n-1)20

(n-1)= 401- 21/20 = 19

n=20

Ques 7: On January 1, 2004 two new societies, S1 and S2 are formed, each with n members. On the first day of each subsequent month, S1 adds b members while S2 multiple its current number of members by a constant factor r. Both the societies have the same number of members on July 2, 2004. If b = 10. 5n, what is the value of r? (CAT 2004)

  1. 2.0
  2. 1.9
  3. 1.8
  4. 1.7

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Ans: (A)

Let the number of members as on January 1, 2004 in S1 and S2 be n each. At the beginning of every month, members of society S1 are in arithmetic progression while those of S2 are in geometric progression. Hence, on July 1, 2004 the number of members in S1 and S2 are

n + (7 − 1)b and nr7−1, i.e.

n + 6b and nr6, respectively.

Hence, n + 6b = n + 6(10.5n) = 64n = nr6

 r6 = 64

 r = 2.

Ques 8: Let C be a circle with centre P0 and AB be a diameter of C. Suppose P1 is the mid-point of the line segment P0 B, P2 is the mid-point of the line segment P0 B and so on. Let C1 ,C2 ,C3 , ... be circles with diameters P0 P1 , P1 P2 , P2 P3 , ......, respectively. Suppose the circles C1, C2 , C3,... are all shaded.The ratio of the area of the unshaded portion of C to that of the original circle C is __ . (CAT 2004)

  1. 8 : 9
  2. 9 : 10
  3. 10 : 11
  4. 11 : 12

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Ans: (D)

Let the radius and area of the big circle (C) be R and A, respectively. The diameter of the biggest of the smaller circles P0 P1 is of R /2, i.e. its radius in R /4 and its area A0 is A /16.

The areas of successive circles form a geometric progression with common ratio 1/4.

The Shaded area = \(\frac{A/16}{1- (1/4)} = \frac{A}{12}\)

Unshaded portion is 1-(A/12) =(11A/12)

Ratio of unshaded portion to the total area is 11:12.

Ques 9: Let u and v be the roots of the equation x2 − 2x + p = 0 and let y and z be the roots of the equation x2 − 18x + q = 0. If u < v < y < z are in arithmetic progression, then p, q respectively equal to (CAT 2013)

  1. 8,17 
  2. 3,7
  3. -3,11
  4. None of these

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Ans: (D)

Let u=a−3d,v=a−d,y=a+dandz=a+3d.

sum of producers in first equation,

(a − 3b) + (a − d) = 2 2a − 4 d = 2

Sum of products in second equation,

(a + d) + (a + 3d) = 18

2a+4d=18

solving Eqs. (i) and (ii),

a=5,d=2

u=5-2x3=-1

v=5-2=3

y=5+2=7

z=5+2x3=11

p=u.v =-1 x 3 = -3

q= y.z = 7 x11 =77

Ques 10: If (a2 + b2 ), (b2 + c2 ) and (a2 + c2 ) are in geometric progression, which of the following holds true? (CAT 2012)

  1. b2 – c2\(\frac{a^4-c^4}{b^2+ a^2}\)
  2. b2 – a2\(\frac{a^4- c^4}{b^2+ c^2}\)
  3. b2 – c2\(\frac{b^4-c^4}{b^2+a^2}\)
  4. b2 – a2\(\frac{b^4-c^4}{b^2+a^2}\)

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Ans: (B)

(b2+c2)2=(a2+b2)(a2+c2)

b4 + c4 +2b2c2 =a4 + a2b2 + a2c2 + b2c2

b4 + c4 + b2c2 = a4 + a2b2 + a2c2

b2 (b2 + c2)+ c4 =a2 (b2 + c2)+ a4

⇒(b2a2)(b2 + c2)=a4c4

 b = b2- a2 = \(\frac{a^4-c^4}{b^2+ c^2}\)

Ques 11: If ax2 + bx + c = 0 and 2a, b and 2c are in arithmetic progression, which of the following are the roots of the equation (CAT 2012)

  1. a,c
  2. −a,− c
  3. \(-\frac {a}{2}, -\frac{c}{2}\)
  4. \(-\frac{c}{a}, -1\)

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Ans: (D)

As 2a, b and 2c are in arithmetic progression,

2b = 2a + 2c

b=a+c

Putting the value in the quadratic equation for finding the value of x

 \(x = {-b \pm \sqrt{(a+c)^2-4ac} \over 2a}\)

 \(x = \frac{-b \pm (a-c)}{2a}\)

 \(x= \frac{-(a+c) \pm (a-c)}{2a}\)

 \(x =- \frac{c}{a} \ or\ -1\)

Ques 12: The sum of 3rd and 15th elements of an arithmetic progression is equal to the sum of 6th, 11th and 13th elements of the same progression. Then, which element of the series should necessarily be equal to zero? (CAT 2010)

  1. 1st
  2. 9th
  3. 12th
  4. None of these

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Ans: (C)

Considering third term to be x

15th term will be (x+12d)

6th term will be (x+3d)

11th term will be (x+8d)

13th term will be (x+10d)

Based on the given condition

2x+12d=3x+21d

x+9d = 0

So (x+9d) is the 12th term.

Ques 13: If x, y and z are in harmonic progression, which of the following statement(s) is/are true? (CAT 2009)

  1. x = \(\frac{y(x+z)}{2z}\)
  2. x =\(\frac{z(x-y)}{y-z}\)
  3. x = \(\frac{y-z}{x-z}\)
  1. only I
  2. I and II
  3. only II
  4. II and III

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Ans: (B)

x, y and z are in Harmonic progression.

(1/x),(1/ y) and (1/z) are in Arithmetic progression

(1/z)-(1/y) = (1/y)-(1/x)

(y-z)/yz=(x-y)/xy

Multiplying xyz both the sides

x(y-z)=z(x-y)

x= z(x-y)/(y-z)

 II is true

(1/z)-(1/y)=(1/y)-(1/x)

(2/y)=(1/z)+(1/x)

x=(y(x+z))/2z

I is true

Ques 14: Consider the set S = {1,2,3,…..,1000}. How many arithmetic progressions can be formed from the elements of S that start with 1 and end with 1000 and have atleast 3 elements? (CAT 2006)

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Ans: Let number of terms in an arithmetic progression be n, then

1000 = 1 + (n − 1)d

(n−1)d = 999=33 ×37

Possible values of (n − 1) are 3, 37, 9, 111, 27, 333, 999.

Therefore, the number of possible values of n will also be 7, hence required progressions can be made.

Ques 15: The sum of 3rd and 15th elements of an arithmetic progression is equal to the sum of 6th, 11th and 13th elements of the progression. Then, which element of the series should necessarily be equal to zero (CAT 2003)

  1. 1st
  2. 9th
  3. 12th
  4. None of these

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Ans: (C)

Tn = a + (n − 1)d. Hence, we get 3rd + 5th term = (a + 2d) + (a + 4d) = 2a + 6d.

So, 6, 11 and 13th terms.

= (a + 5d) + (a + 10d) + (a + 12d) = 3a + 27d.

Now, 2a +6d =3a +27d,

a +11d =10.

Therefore 12th term is zero.


How to approach Progression questions on CAT

  • Go through all the formulae of AP, GP and HP especially the condition of finding Sum of all the terms of sequence.
  • Practice questions on numbers and algebra to get more exposed to CAT level questions.

CAT Related Questions

1.
A shopkeeper sells half of the grains plus \(3 \, \text{kg}\) of grains to Customer 1, and then sells another half of the remaining grains plus \(3 \, \text{kg}\) to Customer 2. When the 3rd customer arrives, there are no grains left. Find the total grains that were initially present.

    • 10
    • 36
    • 42
    • 18

    2.
    \(\sqrt{x + 6\sqrt{2}} - \sqrt{x - 6\sqrt{2}} = 2\sqrt{2}.\)
    Find \( x \).

        3.
        For any natural Number 'n', let an be the largest number not exceeding \(\sqrt{n}\) , then a1 + a2 + a3... +a50 =

            4.
            X is a +ve real no, 4 log10 (x) + 4log 100 (x) + 8 log1000 (x) = 13, then the greatest integer not exceeding 'x'

                5.
                A fruit seller has a stock of mangoes, bananas, and apples with at least one fruit of each type. At the beginning of a day, the number of mangoes makes up 40% of his stock. That day, he sells half of the mangoes,96 bananas, and 40% of the apples. At the end of the day, he ends up selling 50% of the fruits. What is the smallest possible total number of fruits in the stock at the beginning of the day?

                    6.
                    \(3^{3333}\) divided by 11, then the remainder would be?

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