# Progressions, Sequences and Series Questions for CAT PDF

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Progressions, Sequences and Series for CAT PDF consists of important questions on progressions for CAT. Every year 3-5 questions will be asked from this sequences and series topic.

Here we are giving some important questions to practice on this topic.

Hope you find this Progressions for CAT PDF useful. You can also download the Progression Tricks/formulas for CAT PDF.

You can download the Progressions, Sequences and Series Questions for CAT PDF or you can go through the details below.

Profit and Loss Questions for CAT:

Question 1:

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

Question 2:

A child was asked to add first few natural numbers (i.e. 1 + 2 + 3 + …) so long his patience permitted. As he stopped, he gave the sum as 575. When the teacher declared the result wrong, the child discovered he had missed one number in the sequence during addition. The number he missed was

A. less than 10

B. 10

C. 15

D. more than 15

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

Question 4:

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

You can download the CAT Maths formulas PDF for other Quant formulas for CAT.

Question 5:

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

Answers and Solutions for Progressions Questions for CAT PDF:

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.

If the child adds all the numbers from 1 to 34, the sum of the numbers would be 1+2+3+…+34 = 34*35/2 = 595

Since the child got the sum as 575, he would have missed the number 20.

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

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