Progression and Series Questions for CAT

Progression and Series is an important topics in the CAT Quant section. If you find these questions a bit tough, make sure you solve more CAT Progression and Series questions. Learn all the important formulas and tricks on how to answer questions on Progression and Series. You can check out these CAT progressions and series questions from the CAT Previous year papers. Practice a good number of sums in the CAT Progression and Series so that you can answer these questions with ease in the exam. In this post, we will look into some important Progression and Series Questions for CAT quants. These are a good source of practice for CAT 2022 preparation; If you want to practice these questions, you can download this Important CAT Progression and Series Questions and Answers PDF along with the video solutions below, which is completely Free.

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Question 1: The number of common terms in the two sequences 17, 21, 25,…, 417 and 16, 21, 26,…, 466 is

a) 78

b) 19

c) 20

d) 77

e) 22

1) Answer (C)

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Solution:

The terms of the first sequence are of the form 4p + 13

The terms of the second sequence are of the form 5q + 11

If a term is common to both the sequences, it is of the form 4p+13 and 5q+11

or 4p = 5q -2. LHS = 4p is always even, so, q is also even.

or 2p = 5r – 1 where q = 2r.

Notice that LHS is again even, hence r should be odd. Let r = 2m+1 for some m.

Hence, p = 5m + 2.

So, the number = 4p+13 = 20m + 21.

Hence, all numbers of the form 20m + 21 will be the common terms. i.e 21,41,61,…,401 = 20.

Question 2: 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?

a) 1st

b) 9th

c) 12th

d) None of the above

2) Answer (C)

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Solution:

The sum of the 3rd and 15th terms is a+2d+a+14d = 2a+16d
The sum of the 6th, 11th and 13th terms is a+5d+a+10d+a+12d = 3a+27d
Since the two are equal, 2a+16d = 3a+27d => a+11d = 0
So, the 12th term is 0
Question 3: The 288th term of the series a,b,b,c,c,c,d,d,d,d,e,e,e,e,e,f,f,f,f,f,f… is

a) u

b) v

c) w

d) x

3) Answer (D)

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Solution:

1, 2, 3, 4,….n such that the sum is greater than 288
If n = 24, n(n+1)/2 = 12*25 = 300
So, n = 24, i.e. the 24th letter in the alphabet is the letter at position 288 in the series
​So, answer = x

Question 4: If the product of n positive real numbers is unity, then their sum is necessarily

a) a multiple of n

b) equal to n + 1/n

c) never less than n

d) a positive integer

4) Answer (C)

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Solution:

Let the numbers be $a_1,a_2….a_n.$

Since the numbers are positive,

$AM\geq GM$

$\frac{a_1+a_2+a_3….+a_n}{n}\geq (a_1*a_2….*a_n)^{1/n}$

$a_1+a_2+a_3….+a_n \geq n$

Question 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?

a) 0

b) -1

c) 1

d) Not unique

5) Answer (A)

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Solution:

Sum of the first 11 terms = 11/2 ( 2a+10d)

Sum of the first 19 terms = 19/2 (2a+18d)

=> 22a+110d = 38a+342d => 16a = -232d

=> 2a = -232/8 d = -29d

Sum of the first 30 terms = 15(2a+29d) = 0

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Question 6: Consider the sequence of numbers $a_1, a_2, a_3$……. to infinity where $a_1 = 81.33$ and $a_2 = -19$ and $a_j = a_{j-1} – a_{j-2}$ for $j\ge3$. What is the sum of the first 6002 terms of this sequence?

a) -100.33

b) -30.00

c) 62.33

d) 119.33

6) Answer (C)

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Solution:

According to given conditions the terms are 81.33, -19, -100.33, -81.33, 19, 100.33, 81.33,-19,.. Hence the series repeats after every 6 terms . Also summation of these 6 terms is 0. Hence summation is 60002 terms will we sum of first 2 terms which is 62.33.

Question 7: Consider a sequence where the $n^{th}$ term, $t_n = n/(n+2), n =1, 2, ….$ The value of $t_3 * t_4 * t_5 * …..* t_{53}$ equals.

a) 2/495

b) 2/477

c) 12/55

d) 1/1485

e) 1/2970

7) Answer (A)

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Solution:

substituting 3,4…53 in the given function, we get
$t_3 = \frac{3}{5}$
$t_4 = \frac{4}{6}$
$t_5 = \frac{5}{7}$
$t_6 = \frac{6}{8}$

Multiplying the values, we get $\frac{3}{5}*\frac{4}{6}*\frac{5}{7}*….\frac{52}{54}*\frac{53}{55} $ which ultimately after cancellations give $\frac{3*4}{54*55}=\frac{2}{495}$

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Question 8: A group of 630 children is arranged in rows for a group photograph session. Each row contains three fewer children than the row in front of it. What number of rows is not possible?

a) 3

b) 4

c) 5

d) 6

e) 7

8) Answer (D)

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Solution:

Let x be in the front row.

So no. of children in next rows will be x-3,x-6,x-9,x-12,x-15,x-18,x-21….

Suppose there are 6 rows, then the sum is equal to x + x-3 + x-6 + x-9 + x-12 + x-15 = 6x – 45

This sum is equal to 630.

=> 6x – 45 = 630 => 6x = 585

Here, x is not an integer.

Hence, there cannot be 6 rows.

Question 9: For a Fibonacci sequence, from the third term onwards, each term in the sequence is the sum of the previous two terms in that sequence. If the difference in squares of 7th and 6th terms of this sequence is 517, what is the 10th term of this sequence?

a) 147

b) 76

c) 123

d) Cannot be determined

9) Answer (C)

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Solution:

It is given that in a Fibonacci sequence, from the third term on wards, each term in the sequence is the sum of the previous two terms in that sequence.

Let x and y be the 1st and 2nd term respectively.

3rd term = x+y

4th term = x+2y

5th term = 2x+3y

6th term = 3x+5y

7th term = 5x+8y

We know that difference of the squares of 6th and 7th terms is 517 = 47*11 .

And $a^2-b^2=(a+b)(a-b)$.

Applying above formula we get (8x+13y)(2x+3y) = 47*11.

So only possible way is (8x+13y)=47 and

2x+3y=11 .

Solving we get x=1 and y=3 .

Using the concept that every term is the sum of the previous two terms, as used in the beginning of the solution, we get 10th term as 21x+34y, which gives 10th term as 123.

Question 10: If $a_1 = 1$ and $a_{n+1} = 2a_n +5$, n=1,2,….,then $a_{100}$ is equal to:

a) $(5*2^{99}-6)$

b) $(5*2^{99}+6)$

c) $(6*2^{99}+5)$

d) $(6*2^{99}-5)$

10) Answer (D)

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Solution:

$a_2 = 2*1 + 5$
$a_3 = 2*(2 + 5) + 5 = 2^2 + 5*2 + 5$
$a_4 = 2^3 + 5*2^2 + 5*2 + 5$
…
$a_{100} = 2^{99} + 5*(2^{98} + 2^{97} + … + 1)$
$= 2^{99} + 5*1*(2^{99} – 1)/(2-1) = 2^{99} + 5*2^{99} – 5 = 6*2^{99} – 5$