- Sequential Output Tracing (Input Output)
- Syllogism
- Critical Reasoning: Assumptions & Conclusions
- Relationships
- Cause & Effect
- Circular Arrangement
- Coding-Decoding
- Complex Arrangements
- Critical Reasoning: Course of Action
- Critical Path
- Direction and Word Puzzle
- Games And Tournaments
- Linear Arrangement
- Team Selection
- Venn Diagram
- Bar Graphs
- Data Sufficiency
- Line Graph
- Pie Chart
- Tables and Caselets

We’re going to learn about one particular branch of sequence and series, i.e. Progression and its principles in relation. Sequence and Series Questions carry a large weightage when it comes to the Quant Section. And a big part of Sequence and Series is the Arithmetic Progression Questions, along with the other progression questions. The questions that come from this particular topic are relatively easy. Hence, if one has practiced sequence and series problems thoroughly, they can definitely score full in this topic.

In general, from the topic of progression and series in CAT, XAT, GMAT, SNAP, NMAT, and other exams report, approximately 2-3 were questions asked each year. Hence, to be able to easily ace these entrance tests, it is therefore very important for the candidates to be well prepared in this topic. Some of the Progression and Series questions can be very difficult and time-consuming, while others can be very easy. This section’s trick is to quickly work out whether or not a topic is solvable and not to waste time on very complicated problems.

Year | No of Questions | Good attempt | Difficulty | |

2019 | Slot 1 | 3 | 2 | Moderate |

Slot 2 | – | – | – | |

2018 | Slot 1 | 2 | 2 | Easy |

Slot 2 | 4 | 2 or 3 | Difficult |

__XAT :__

Year | No of Questions | Good attempt | Difficulty |

2020 | 2 | 2 | Easy |

2019 | 2 | 1 | Moderate |

__SNAP : __

Year | No of Questions | Good attempt | Difficulty |

2019 | 3 | 3 | Easy |

2018 | 3 | 3 | Easy |

Here we will look at important concepts of sequence and progression, types of progression in maths along with ap and gp formulas:

• Arithmetic Progressions (A.P.) – If the difference in a sequence of two consecutive terms is constant, then the sequence is called an arithmetic sequence or a progression of arithmetic

• Geometric Progression (G.P.) – If the elements of a sequence are such that they can be produced by multiplying a certain number of the previous term, it is called a geometric sequence

• Harmonic Progression (H.P.) – If we have a series and if the reciprocals of the series’ terms form an arithmetic progression, the sequence is called Harmonic Progression

• We will now look at formula of AP and GP:

For AP: d=an−an−1

an=a+(n−1)d

For GP: r=anan−1

an=arn−1

• Geometric Mean = nth root of product of n terms in the GP

• Sum of infinite terms of a GP (r < 1) = (a) / (1 – r)

For two numbers, if A, G and H are respectively the arithmetic, geometric and harmonic means, then

1. A ≥ G ≥ H

2. A H = G2, i.e., A, G, H are in GP

• Arithmetic Progression Sum:

Sum of ‘n’ terms of an AP = 0.5 n (first term + last term) = 0.5 n [ 2a + (n-1) d ]

We will now look at some previous year questions on Harmonic Progressions, Arithmetic Progression and Geometric Progression:

1. [CAT 2020]

Let the m-th and n-th terms of a geometric progression be 3/4 and 12, respectively, where m < n. If the common ratio of the progression is an integer r, then the smallest possible value of r + n – m is

A) -2 B) 6 C) 2 D) -4

3/4 12

mth nth

m > n

Given r is an integer, So rk = 12341234 = 16

rk = 24 or 42

Since asked for minimum possible value, Taking rk = {(-2)}2 or {(-4)}2

k = n – m (From m how many terms we have to jump to reach n)

We have two cases r = – 2 and n – m = 4 —-> r + n – m = 2

r = – 4 and n – m = 2 —-> r + n – m = – 2

-2 is the smallest possible value

2. [CAT 2019]

If a(base1), a(base2), … are in A.P., then, 1/√a(base1)+√a(base2) + 1/√a(base2)+√a(base3) + … + 1/√a(base n)+√a(base n + 1) is equal to?

A. n-1/√a(base1)+√a(base n)

B. n/√a(base1)+√a(base n + 1)

C. n-1/√a(base1)+√a(base n – 1)

D. n/√a(base1)-√a(base n + 1)

The best method to this problem is to use hit and trial method , for this put n =1 in the question you will get the first term only option a) is equal to that so option a) is correct.

(Alternative method)

As given a(base1) , a(base2) ,a(base3) , a(base4) , ………..,a(base n) are in AP.

So a(base2) – a(base1) = a(base3) – a(base2)= a(base4) – a(base3) = …….= a(base n) – a(base n-1) = d (common difference)

Now 1/(√a1+√a2) + 1/(√a2+√a3) + 1/(√a3+√a4) + 1/(√a4+√a5)+⋯.

= 1/(√a1+√a2)×(√a1-√a2)/(√a1-√a2)+1/(√a2+√a3)×(√a2-√a3)/(√a2-√a3)+1/(√a3+√a4)×(√a3-√a4)/(√a3-√a4)+⋯.+⋯.1/(√(a(n-1))+√an)×(√an-√(a(n+1)))/(√an-√(a(n+1)))

= (√a1-√a2)/(a1-a2 )+(√a2-√a3)/(a2-a3 )+(√a1-√a2)/(a1-a2 )+⋯..+(√an-√(a(n+1)))/(an-a(n+1) )

= (√a1-√a2)/(-d )+(√a2-√a3)/(-d )+(√a3-√a4)/(-d )+⋯..+(√an-√(a(n+1)))/(-d )

= (√a1-√an)/(-d )

= (√a1-√(a(n+1)))/(-d )×(√a1+√(a(n+1)))/(√a1+√(a(n+1)))

= 1/(√a1+√an)× (a(n+1) –a1)/d ——1)

Now as we know a(n+1) = a1 + nd

So a(n+1) – a1= nd

Putting this value of a(n+1) – a1in eq 1)

We get required sum = n/(√a1+√(a(n+1)))

3. [CAT 2018]

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

A)3/6 B) 3/2 C) 5/2 D) 1/6

Since x, y ,and z are in G.P. and x<y<z, let x = a, y=ar and z=ar2, where a>0 and r>1. It is also given that, 15x, 16y and 12z are in A.P.

Therefore, 2×16y=5x+12z

Substituting the values of x, y and z we get,

32ar = 5a +12ar2

32r = 5 +12r2

⇒ 32r = 5 +12r2

⇒12r2 − 32r + 5 = 0

On solving the above quadratic equation we get r=1/6 or 5/2. Since r>1, therefore r=5/2.

2r2 − 32r + 5 = 0

On solving the above quadratic equation we get r=1/6 or 5/2. Since r>1, therefore r=5/2.

4. [CAT 2018]

Let a1, a2, … , a52 be positive integers such that a1 < a2 < … < a52. Suppose, their arithmetic mean is one less than the arithmetic mean of a2, a3, …, a52. If a52 = 100, then the largest possible value of a1 is

A) 20 B) 23 C) 48 D) 45

We want to maximize the value of a1, subject to the condition that a1 is the least of the 52 numbers and that the average of 51 numbers (excluding a1) is 1 less than the average of all the 52 numbers. Since a52 is 100 and all the numbers are positive integers, maximizing a1 entails maximizing a2, a3, ….a51.

The only way to do this is to assume that a2, a3…. a52 are in an AP with a common difference of 1.

Let the average of a2, a3…. a52 i.e. a27 be A.

(Note: The average of an odd number of terms in an Arithmetic Progression is equal to the value of the middle-most term)

Since a52 = a27 + 25 and a52 = 100

A = 100 – 25 = 75

a2 + a3 + … + a52 = 75×51 = 3825

Given a1 + a2 +… + a52 = 52(A – 1) = 3848 Hence a1 = 3848 – 3825 = 23

It will help you minimize the time taken to solve relevant questions in CAT by keeping these Tricks & Tips in mind for Progression and Sequence. First learn the basic concepts and tricks in for calculation so that you can apply them to questions quickly. For this, we recommend using the MBAP CAT E-Book (Concept Theory) study material.

There is a chance that the questions that will come in the exam will be direct questions based on the concept and hence it is advised to have clarity on basic concepts of Series and Progression. If you find difficulty understanding the concepts, use the MBAP Live Lecture Recording (Basic) on Basic Concepts.

For problems on arithmetic progression, use MBAP CAT E-Book (Practice Questions) and solve all the relevant practice questions. Also, solve the MBAP lecture assignment in order to complete the basics.

Write down the formulas on a sheet of paper and put them on a wall so that you can see them regularly. It’ll help memorize the formulas.

Questions based on the combination of A.P and G.P, etc., may also include comprehension of other math principles such as trigonometry, logarithmic, geometry, etc. If you do not prepare on these topics extensively, you will be clueless on how to proceed in the exam. Therefore, to build and learn the application of progression, you need to practice several questions.

It will be way easier for you to attempt these particular questions by solving previous year papers. They are all readily available if you use MBAP Topic Wise Previous Year CAT Questions.

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