How To Solve Easily Progressions For CAT Concepts

Here you can learn the basics of progressions for CAT and understand the concepts with practice questions for CAT- by IIM Skills – online CAT coaching

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The progression and series is one of those sections that are frequently asked or used in questions in CAT. Usually, 4-5 questions are asked from progressions for CAT each year. This is why getting your basics right and practicing a lot of questions is important in progressions and series for CAT.

 

Table of contents:

  • Theory
  • Practice questions

 

There are three types of progressions:

  1. Arithmetic progression
  2. Geometric progression
  3. Harmonic progression

 

Find the list of Top CAT Coaching Institutes in Delhi here

 

Arithmetic progression

 

Arithmetic Progression is one major topic for the CAT quantitative aptitude section. 

 

Whenever a sequence of numbers falls in a pattern so that the difference between the consecutive numbers is constant, the numbers are said to be in an Arithmetic Progression or an AP.

 

For example {3, 5, 7, 9} are part of an arithmetic progression (AP) as the difference between all the consecutive numbers is 3. This first term of this AP is 3 and the common difference is 2.

 

Similarly, {11, 7, 3,  -1, -5} form an arithmetic progression (AP). Here, the first term of AP is 11 and the common difference of the AP is -4.

 

The general form of an AP

 

Lets us assume an AP that has the first term as ‘a’ and the common difference is ‘d’.

 

If the First-term = a

 

Then, Second term = a + d

 

And, Third term = a + 2d

 

Similarly, Fourth term = a + 3d

 

So the nth term = a + (n – 1) d

 

So, sum of arithmetic progression to first ‘n’ terms = a + a + d + a + 2d + a + 3d + a + 4d + … + a + (n – 1) d

 

=> na + d + 2d + 3d + 4d + … + (n – 1) d

 

=> na + d( 1 + 2 + 3 + … + (n – 1))

 

=> na + d n(n-1)/2

 

=> (Taking n/2 common)= (n/2) ( 2a + (n-1)d)

 

=>(n/2) ( a + a + (n – 1)d)

 

=>(n/2) (first term + last term).

 

 

 

For any AP that has the 

first term= ‘a’ 

and 

common difference = ‘d’

 

nth term of an AP : Tn = a + (n – 1)d

 

Sum of arithmetic progression for first ‘n’ terms will be : 

Sn= n/2 (2a+(n-1)d) = (n/2) (first term + last term)

 

 

Average Of “n” Terms in an AP

 

If you look at the formula for the sum of first ‘n’ terms of an AP= (n/2)(first term + last term)

 

You can write it as 

Sum= n/2(first term + last term)

 

Now, it is known that sum of ‘n’ terms = average × number of terms 

Here, number of terms = ‘n’

 

Average of ‘n’ terms of an AP will be = (first term + last term)/2

 

The middle point of an AP, ie. (first term + last term)/2 is the average of all the terms of the AP upto that term.

If we look at any 3 consecutive terms in an AP, we can take them as a-d, a and a + d

Any Four consecutive terms in an AP; we can take them as (a – 3d), (a – d), (a + d), (a + 3d)

Any Five consecutive terms in an AP we can take them as (a – 2d),(a – d), a, (a + d), (a + 2d).

 

 

Geometric progression

 

The geometric progression is another sub-topic of progression and series. This topic also has considerable importance in CAT quantitative section. 

A geometric progression “GP” is a sequence of numbers such that the ratio of any (n+1)th term to its preceding nth term is always the same. 

This ratio is called the common ratio(r).

 

For example, take a progression of numbers like: 2, 4, 8, 16.

Now the ratio of all terms to their preceding terms 4/2=8/4=16/8= 2. 

So this progression of numbers forms a GP 

And has a common ratio = 2.

 

The general form of a GP

 

Consider a GP that has

first term = ‘a’ 

and 

common ratio = ‘r’.

 

Then, 1st term = a

 

And, 2nd term = ar

 

Similarly, 3rd term = ar2

 

And, 4th term = ar3

 

In this way,

 

nth term of a GP= Tn= arn-1

 

Sum of a GP: 

The nth term of a geometric progression is given by Tn=arn-1

 

So sum to ‘n’ terms,

 

Sn = a + ar + ar2 + ar3 + … + arn-1

 

= a (1 + r + r2 + r3 + … + rn-1)

 

= a (1- rn ) / (1-r) ……………….if r < 1 

or 

a (rn-1) / (r-1)……………………….. if r > 1

 

 

The Harmonic Progression

 

The Harmonic Progression is another type of progressions and is a significant section of the progression topic in CAT quantitative aptitude. 

Let us learn the basics of harmonic progressions before we try and attempt practice questions. 

 

Any sequence of numbers a, b, c are said to be in Harmonic Progression or HP, if the reciprocals of all these numbers (1/a,1/b,1/c) are in an Arithmetic Progression. 

 

It is important to note that no term in a Harmonic progression can be 0. 

 

Let us consider an example of an HP: 1, 1/3 and 1/5.

 

Now you can see that the reciprocals of each of the numbers are 1, 3, 5 are in AP. 

Hence, it can be said that 1, 1/3, 1/5 are in an HP.

 

The general form of a Harmonic Progression

 

The nth term of an HP = 1/((nth term of corresponding AP))

 

For example, take the HP: 1/2, 1/5, 1/8

 

The corresponding AP of the given HP is 2, 5, 8 

That has the first term= 2 and common difference = 3.

 

So given the sequence, the 6th term in this AP = 2 + (5)(3) = 17.

 

Hence, the 6th term in the corresponding HP = 1/17.

 

 Example Question:

The first two terms of a given HP are 24 and 12, what is the 4th term?

 

Solution:

 

Let 3rd term of the HP be = x

 

Since the reciprocals of the terms of an HP are in AP, 

 

=> 1/24,1/12,1/x are in AP

 

=> 1/x-1/12=1/12-1/24 

=> 1/x=1/6–1/24

=1/8

 

Let the 4th term of the HP be = y 

Then, 

1/y-1/8=1/8-1/12

=>1/y=1/6

=>y =6

 

Therefore the 4th term of the HP = 6

 

Alternatively, we can also solve by using the percentage equivalents,

 

it becomes like this,

 

1/24= 4.16

 

1/12= 8.33. 

So, 4.16 & 8.33 are the first two terms of the corresponding AP,

 

Similarly, 3rd term = 12.5, 

And, the 4th term = 16.66 = 1/6,

 

Finding the reciprocal of the 4th term of the AP would give us the 4th term of the HP ie. 6.

 

 

Harmonic Mean (HM):

 

Let a and b be two given numbers. Subsequently, we insert n harmonic means h1, h2, h3,….hn between the quantities a and b

 

=> 1/a,1/h1,1/h2,…..1/hn,1/b are in A.P

 

Harmonic mean of n numbers a1, a2,……an is given by: n/[1/a1 + 1/a2 + 1/a3 + …+ 1/an ]

 

Harmonic Mean Formula:

If a, b, c are three numbers in Harmonic Progression, then

 

“HM, b =2/[1/a + 1/c ] = 2ac/((a + c) )”

 

“Harmonic mean of 4 numbers a, b, c, d = 4/((1/a + 1/b + 1/c +1/d) )”

 

This shows that between any two terms of an HP, the middle term is the harmonic mean of the first and the third term.

 

 

 

There are many applications of Harmonic Mean. One of them is to find the average speed within a distance covered. When the intervals of the distance covered are equal, then the average speed is equal to the harmonic mean of all the speeds would be the average speed.

Harmonic mean helps to find out the average speed of a journey. The average speed when the 

 

 

Progressions for CAT practice questions

Below are some practice questions for progressions for CAT basics to clear your concepts.

 

Question 1: In a Geometric progression, the sum of the first 12 terms is equal to the sum of the first 14 terms in the same GP. The sum of the first 17 terms is 92. Then the third term in the GP is…

 

  1. 92
  2. -92
  3. 46
  4. 231

Solution:

The solution to this progressions for CAT problem is very simple.

Since the addition of two extra terms to the sum of 12 terms does not change the sum. 

Sum of the first 12 terms is equal to the sum of the first 14 terms.

 

Sum of 14 terms = Sum of 12 terms + 13th term + 14th term

=> 13th term + 14th term = 0

 

Let us assume 

13th term = k, and 

common ratio = r. 

In this way, the 14th term = kr.

=> k + kr = 0

=> k (1 + r) = 0

Since k cannot be 0; because then all terms would be 0.

=> r = -1 

 

So, common ratio, r = -1.

 

Now, if the first term of this GP = a, 

then 

second term = -a 

And the third = a 

and so on

Therefore the GP would be a, -a, a, -a, a, -a,…

 

Sum to any even number of terms of the GP = 0

Sum to any odd number of terms of the GP = a

 

Sum to 17 terms is 92 => a = 92

Third term = a = 92

 

Hence the correct answer is A. 92

 

Question 2: In an AP, the sum of the first 25 terms is 525, and the sum of the subsequent 25 terms is 725, find out the common difference?

 

  1. 825
  2. 425
  3. 625
  4. 125

Solution:

We first need to formulate a relationship between the 

Let the sum of the first 25 terms, S25 = 525

And the sum of the next 25 terms, K25 = 725

a26 = a1 + 25d

a27 = a2 + 25d

 

K25 = a26 + a27 + …… + a50

=> K25 = a1 + 25d + a2 + 25d + …… + a25 + 25d

=> K25 = a1 + a2 + …… + a25 + 25(25d)

 

Or K25 = S25 + 25(25d)

i.e., 725 = 525 + 25 * 25d

=>200 = 25 * 25d

=>8 = 25d

=>d = 825

 

This gives us the common difference as 825

Hence the answer is A

 

Question 3: a, b, c, d, and e form an arithmetic progression. But they may or may not be consecutive terms of the AP. c is the arithmetic mean of numbers a and b. Similarly, d is the arithmetic mean of b and c. which of the following is true?

  1. Average of all 5 terms put together is c.
  2. Average of d and e is not greater than the average of a and b.

iii. Average of b and c is greater than the average of a and d.

  1. i and ii only
  2. ii and iii only
  3. all three statements are true
  4. i and iii only

 

Solution:

The solution to this progressions for CAT problem is very simple.

Let us first arrange the terms in order

If c is the arithmetic mean of a and b, and d is the arithmetic mean of b and c then the correct sequence of numbers would be

a,c,b

And

b,d,c  

 

This way,

a, c, b could be either 

1st 2nd and 3rd terms or

1st, 3rd and 5th, or

2nd, 3rd, 4th, or

3rd, 4th, 5th.

 

The terms could also be in the reverse order. That means b, c, a could also be the 1st 2nd and 3rd terms, or the 1st, 3rd and 5th, and so on. 

 

Again, given d is the arithmetic mean of b and c. 

=> d lies between b and c. 

Using the above information we can say that a, c, b either have to be 1st, 3rd and 5th 

or 5th, 3rd, and 1st 

Since there is a number between b and c also.

 

So, c is the third term in the AP. a and b are 1st and 5th in one of these orders.

 

b ? c ? a 

or 

a ? c ? b

 

Since d is the arithmetic mean of numbers b and c.

 

Possible order of the AP are:

 

b   d c   e a    

or   

a   e c    d b

 

Hence the correct answer would be (A). “i and ii only”

 

Question 4: Consider numbers a, b, c are in a G.P. such that |a + b + c| = 15. The median of these three numbers of the GP is a, and b = 10. If a > c, what is the product of the first 4 terms of this G.P.?

 

  1. 40000
  2. 32000
  3. 8000
  4. 48000

 

Solution:

This progressions practice question for CAT is going to clear a few concepts for you. 

Does the median need to be the middle term? It is not necessary.

 

If the Median is the first term 

=> Common ratio must be negative

Explanation:

When a > 0,

If  r > 1, this would be an increasing G.P.

If  0<r<1, this would be a decreasing G.P.

 

In both cases, the middle term will be the median. If a < 0, the order will be the other way around, but the middle term would still be the median.

If the middle term is not the median, we can say that r < 0. Now, let us go the solution

b = 10, a and c should be negative. Solution, a + b + c cannot be 15.

a + b + c = –15

10r + 10 + 10r = -15 ⇒ 2r + 2r = -5

Solving the above quadratic equation, 

Possible solutions of r = −12 or -2.

The sequence is one of these: 

– 5, 10, – 20 or 

– 20, 10, – 5.

If a > c 

=> the sequence must be – 5, 10, – 20.

The product of the first 4 numbers = – 5 * 10 * –20 * 40 = 40000.

 

Hence the correct answer is (A) “40,000”

 

Question 5:A Series P is stated as pn = pn-1 + 3, p1 = 11, Sequence Q is stated as qn = qn-1 – 4, q3 = 103. If pk > qk+2, what is the smallest possible value of k?

 

  1. 6
  2. 11
  3. 14
  4. 15

Solution:

 

Series  P is an A.P. with a = 11, and common difference 3.

So, Pk = 11 + (k – 1)3.

Series Q is an A.P. with third term = 103 and with common difference = – 4.

=>t3 = a + 2d

=>103 = a + 2 (– 4) 

or a = 111

=>qk+2 = 111 + (k +1) (– 4)

=>qk + 2 = 111 – 4k – 4 = 107 – 4k

 

=>pk > qk + 2

=>11 + (k–1)3 > 107 – 4k

=>8 + 3k > 107 – 4k

 

=>7k > 99

=>k > 99/7

k must be an integer, so the smallest value of k is 15.

The question is “what is the smallest value k can take?”

 

Hence the correct answer is (D) “15”

 

Question 6: a, b, c and d are ofur numbers in A.P., What is true about terms bcd, acd, abd and abc?

 

  1. They are also in A.P.
  2. They are also in H.P.
  3. They are also in G.P.
  4. They are not in an A.P., G.P. or H.P.

 

Solution:

The solution to this progressions for CAT practice question is very simple

 

a, b, c and d are in A.P.

Dividing all the terms of this A.P. by abcd, we get

=>a/abcd , b/abcd , c/abcd , d/abcd  in A.P.

 

1/bcd , 1/acd , 1/abd , 1/abc are in A.P.

This implies that bcd, acd, abd and abc are in H.P.

 

Hence the correct answer to this is B “They are in H.P.”

 

Question 7: the Second term in an Arithmetic Progression is 8 and the 8th term of the AP is 2 more than three times the second term. What is the sum up to 8 terms of this AP?

 

  1. 124
  2. 108
  3. 96
  4. 110

Solution:

 

It is given that the Second term of the AP is 8 

=> a+d =8, 

 

And that the 8th term is 2 more than thrice the second term 

=> a+7d = 2 +3(a+d) = 2+3*8 =26 .

=>a+d =8 ————– (1

=>a+7d = 26 ————– (2

Solving for d in the two equations, we get d = 3 and a =5.

Sum of n terms in an AP = n2 * [2a +(n-1)d].

=> Sum upto 8 terms in this AP = 82 * [2*5 + (8-1)3] => 4*[10+21] = 4*31 = 124.

The question is “Second term in an AP is 8 and the 8th term is 2 more than thrice the second term. Find the sum up to 8 terms of this AP.”

 

Hence the correct answer is (A) “124”

 

Question 8: Hari invests a total amount of rupees 2000 on governmental bonds in a span of 4 years. If these investments are a sequence in an A.P and the sum of squares of the investments is 1200000. Find the investment made by Hari each year. He always invested more than the previous year.

 

  1. 200,400,600,800
  2. 875,625,375,125
  3. 125,375,625,875
  4. 50,350,650,950

 

Solution:

 

A tip for solving these types of practice questions for progressions in CAT

Take the numbers like this for convenience

(a-3d),(a-d),(a+d),and (a+3d)

So, Let the investments made by Hari each year be (a-3d),(a-d),(a+d),(a+3d)

Then,

(a-3d) + (a-d) + (a+d) +(a+3d) = 2000

=> 4a = 2000        (dividing both sides by 4)

=> a = 500

 

Now, again AQ

(a-3d)2 +(a-d)2 +(a+d)2 +(a+3d)2 = 1200000

=> 4(a2 + 5 d2) = 1200000

=> (a2 + 5 d2) = 300000

=> 250000 + 5d2 = 300000

=> 5d2 = 50000

=> d2 = 10000

=> d = ±100

 

∴ d = 100 

As it is given that the investment increases every year 

so d = -100 is not a solution

Therefore the investments made are 200,400,600,800

 

Note – If there are 3 figures in an arithmetic series then take the numbers as (a-d), a,(a+d)

If there are 5 figures in an arithmetic series then take the numbers as (a-2d),(a-d), a,(a+d),(a+2d)

…….. so on

 

While If there are 4 quantity in the arithmetic series then take the quantity as (a-3d),(a-d), ,(a+d),(a+3d)

If there are 6 quantity in the arithmetic series then take the quantity as (a-5d),(a-3d),(a-d), ,(a+d),(a+3d), (a+5d)

……………. so on

 

Hence the correct answer for this progressions for CAT practice question is (A) “200,400,600,800”

 

You can find similar helpful blog posts for your CAT preparation by IIM Skills- online CAT coaching. Visit the CAT blog.

 

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Gaurav is a Content Writer at IIM Skills. He has a B.Tech. degree but then he switched to the creative side by doing his master's in advertising and public relations. Gaurav is also a part-time blogger and graphic designer currently living in Mumbai

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