CAT Progressions and Series Questions PDF [Most Important]

Progressions and Series is one of the most important topics in the Quantitative Ability section of CAT. It is an easy topic and so one must not avoid this topic. You can check out these Progressions and Series CAT Previous year questions. Practice a good number of questions on CAT Progressions and Series questions. In this article, we will look into some important Progressions and a Series of Questions for CAT. These are a good source for practice; If you want to practice these questions, you can download this CAT progression and Series Questions PDF 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 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

2) 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 3: 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

3) 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 4: 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) $(6*2^{99}-5)$

4) Answer (D)

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Solution:

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

Question 5: What is the value of the following expression?
$(1/(2^2-1))+(1/(4^2-1))+(1/(6^2-1))+…+(1/(20^2-1)$

a) 9/19

b) 10/19

c) 10/21

d) 11/21

5) Answer (C)

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Solution:

$(1/(2^2-1))+(1/(4^2-1))+(1/(6^2-1))+…+(1/(20^2-1)$ = 1/[(2+1)*(2-1)] + 1/[(4+1)*(4-1)] + … + 1/[(20+1)*(20-1)]

= 1/(1*3) + 1/(3*5) + 1/(5*7) + … + 1/(19*21)

=1/2 * ( 1/1 – 1/3 + 1/3 – 1/5 + 1/5 – 1/7 + … +1/19 – 1/21)

=1/2 * (1 – 1/21) = 10/21

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Question 6: The value of $\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}$

a) $\frac{8}{1-x^8}$

b) $\frac{4x}{1+x^2}$

c) $\frac{4}{1-x^6}$

d) $\frac{4}{1+x^4}$

6) Answer (A)

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Solution:

$\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}$
or $\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}$
or $\frac{4}{1-x^4}+\frac{4}{1+x^4}$
or $\frac{8}{1-x^8}$

Question 7: N the set of natural numbers is partitioned into subsets $S_{1}$ = $(1)$, $S_{2}$ = $(2,3)$, $S_{3}$ =$(4,5,6)$, $S_{4}$ = $(7,8,9,10)$ and so on. The sum of the elements of the subset $S_{50}$ is

a) 61250

b) 65525

c) 42455

d) 62525

7) Answer (D)

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Solution:

According to given question $S_{50}$ will have 50 terms
And its first term will be 50th number in the series 1,2,4,7,………$T_{50}$
$T_1 = 1$
$T_2 = 1+1$
$T_3 = 1+1+2$
$T_4 = 1+1+2+3$
$T_n = 1+(1+2+3+4+5….(n-1))$
= $1+\frac{n(n-1)}{2}$
So $T_{50} = 1+1225 = 1226$
Hence $S_{50} = (1226,1227,1228,1229……..)$
And summation will be = $\frac{50}{2} (2\times 1226 + 49 \times 1 ) = 62525$

Question 8: If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is

a) 2:3

b) 3:2

c) 3:4

d) 4:3

8) Answer (A)

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Solution:

The seventh term of an AP = a + 6d. Third term will be a + 2d and second term will be a + 16d. We are given that
$ (a + 6d)^2 = (a + 2d)(a + 16d)$
=> $ a^2 $ + $36d^2$ + 12ad = $ a^2 + 18ad + 32d^2 $
=> $4d^2 = 6ad$
=> $ d:a = 3:2$
We have been asked about a:d. Hence, it would be 2:3

Question 9: Let $a_1$, $a_2$,………….,  $a_{3n}$ be an arithmetic progression with $a_1$ = 3 and $a_{2}$ = 7. If $a_1$+ $a_{2}$ +…+ $a_{3n}$= 1830, then what is the smallest positive integer m such that m($a_1$+ $a_{2}$ +…+ $a_n$) > 1830?

a) 8

b) 9

c) 10

d) 11

9) Answer (B)

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Solution:

$a_{1}$ = 3 and $a_{2}$ = 7. Hence, the common difference of the AP is 4.
We have been given that the sum up to 3n terms of this AP is 1830. Hence, $1830 = \frac{m}{2}[2*3 + (m – 1)*4$
=> 1830*2 = m(6 + 4m – 4)
=> 3660 = 2m + 4m$^2$
=> $2m^2 + m – 1830 = 0$
=> (m – 30)(2m + 61) = 0
=> m = 30 or m = -61/2
Since m is the number of terms so m cannot be negative. Hence, must be 30
So, 3n = 30
n = 10
Sum of the first ’10’ terms of the given AP = 5*(6 + 9*4) = 42*5 = 210
m($a_1$+ $a_{2}$ +…+ $a_n$) > 1830
=> 210m > 1830
=> m > 8.71
Hence, smallest integral value of ‘m’ is 9.

Question 10: Let $a_{1},a_{2},a_{3},a_{4},a_{5}$ be a sequence of five consecutive odd numbers. Consider a new sequence of five consecutive even numbers ending with $2a_{3}$
If the sum of the numbers in the new sequence is 450, then $a_{5}$ is

10) Answer: 51

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Solution:

Sum of the sequence of even numbers is $2a_{3} + (2a_{3} – 2) + (2a_{3} – 4)$ $+ (2a_{3} – 6) + (2a_{3} – 8) = 450$
=> $10a_{3} – 20 = 450$
=> $a_{3} = 47$
Hence $a_{5} = 47 + 4 = 51$

Question 11: An infinite geometric progression $a_1,a_2,…$ has the property that $a_n= 3(a_{n+1}+ a_{n+2} + …)$ for every n $\geq$ 1. If the sum $a_1+a_2+a_3…+=32$, then $a_5$ is

a) 1/32

b) 2/32

c) 3/32

d) 4/32

11) Answer (C)

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Solution:

Let the common ratio of the G.P. be r.
Hence we have $a_n= 3(a_{n+1}+ a_{n+2} + …)$

The sum up to infinity of GP is given by $\frac{a}{1-r}$ where a here is $a_{n+1}$

=> $a_n= 3(\frac{a_{n+1}}{1-r})$
=> $a_n= 3(\frac{a_{n}\times r}{1-r})$
=> $ r = \frac{1}{4}$
Now, $a_1+a_2+a_2…+=32$
=> $\frac{a_1}{1-r} = 32$
=> $\frac{a_1}{3/4} = 32$
=> $a_1 = 24$

$a_5 = a_1 \times r^4$
$a_5 = 24 \times (1/4)^4 = \frac{3}{32}$

Question 12: If $a_{1}=\frac{1}{2\times5},a_{2}=\frac{1}{5\times8},a_{3}=\frac{1}{8\times11},…,$ then $a_{1}+a_{2}+a_{3}+…+a_{100}$ is

a) $\frac{25}{151}$

b) $\frac{1}{2}$

c) $\frac{1}{4}$

d) $\frac{111}{55}$

12) Answer (A)

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Solution:

$a_{100} = \frac{1}{ (3\times100 -1) \times (3\times100 + 2)}= \frac{1}{ 299 \times 302}$

$\frac{1}{2\times5} = \frac{1}{3} \times (\frac{1}{2} – \frac{1}{5})$

$\frac{1}{5\times8} = \frac{1}{3} \times (\frac{1}{5} – \frac{1}{8})$

$\frac{1}{8\times11} = \frac{1}{3} \times (\frac{1}{8} – \frac{1}{11})$
….

$\frac{1}{299\times302} = \frac{1}{3} \times (\frac{1}{299} – \frac{1}{302})$

Hence $a_{1}+a_{2}+a_{3}+…+a_{100}$ = $\frac{1}{3} \times (\frac{1}{2} – \frac{1}{5})$ + $\frac{1}{3} \times (\frac{1}{5} – \frac{1}{8})$ + $\frac{1}{3} \times (\frac{1}{8} – \frac{1}{11})$ + … + $\frac{1}{3} \times (\frac{1}{299} – \frac{1}{302})$

= $\frac{1}{3} \times (\frac{1}{2} – \frac{1}{302})$

= $\frac{25}{151}$

Question 13: 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) $\frac{3}{6}$

b) $\frac{1}{6}$

c) $\frac{5}{2}$

d) $\frac{3}{2}$

13) Answer (C)

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Solution:

Let x = $a$, y = $ar$ and z = $ar^2$
It is given that, 5x, 16y and 12z are in AP.
so, 5x + 12z = 32y
On replacing the values of x, y and z, we get
$5a + 12ar^2 = 32ar$
or, $12r^2 – 32r + 5$ = 0
On solving, $r$ = $\frac{5}{2}$ or $\frac{1}{6}$

For $r$ = $\frac{1}{6}$, x < y < z is not satisfied.

So, $r$ = $\frac{5}{2}$

Hence, option C is the correct answer.

Question 14: Let $\ a_{1},a_{2}…a_{52}\ $ be positive integers such that $\ a_{1}$ < $a_{2}$ < … < $a_{52}\ $. Suppose, their arithmetic mean is one less than arithmetic mean of $a_{2}$, $a_{3}$, ….$a_{52}$. If $a_{52}$= 100, then the largest possible value of $a_{1}$is

a) 48

b) 20

c) 23

d) 45

14) Answer (C)

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Solution:

Let ‘x’ be the average of all 52 positive integers $\ a_{1},a_{2}…a_{52}\ $.

$a_{1}+a_{2}+a_{3}+…+a_{52}$ = 52x … (1)

Therefore, average of $a_{2}$, $a_{3}$, ….$a_{52}$ = x+1

$a_{2}+a_{3}+a_{4}+…+a_{52}$ = 51(x+1) … (2)

From equation (1) and (2), we can say that

$a_{1}+51(x+1)$ = 52x

$a_{1}$ = x – 51.

We have to find out the largest possible value of $a_{1}$. $a_{1}$ will be maximum when ‘x’ is maximum.

(x+1) is the average of terms $a_{2}$, $a_{3}$, ….$a_{52}$. We know that $a_{2}$ < $a_{3}$ < … < $a_{52}\ $ and $a_{52}$ = 100.

Therefore, (x+1) will be maximum when each term is maximum possible. If $a_{52}$ = 100, then $a_{52}$ = 99, $a_{50}$ = 98 ends so on.

$a_{2}$ = 100 + (51-1)*(-1) = 50.

Hence,  $a_{2}+a_{3}+a_{4}+…+a_{52}$ = 50+51+…+99+100 = 51(x+1)

$\Rightarrow$ $\dfrac{51*(50+100)}{2} = 51(x+1)$

$\Rightarrow$ $x = 74$

Therefore, the largest possible value of $a_{1}$ = x – 51 = 74 – 51 = 23.

Question 15: The value of the sum 7 x 11 + 11 x 15 + 15 x 19 + …+ 95 x 99 is

15) Answer: 80707

View Video Solution

Solution:

S = 7 x 11 + 11 x 15 + 15 x 19 + …+ 95 x 99

Nth term of the series can be written as $T_{n} = (4n+3)*(4n+7)$

Last term, (4n+3) = 95 i.e. n = 23

$\sum_{n=1}^{n=23} (4n+3)*(4n+7)$

$\Rightarrow$ $\sum_{n=1}^{n=23}16n^2+40n+21$

$\Rightarrow$ $16*\dfrac{23*24*47}{6}+40*\dfrac{23*24}{2}+21*23$

$\Rightarrow$ $80707$

Progressions & Series Questions | CAT Preparation Videos

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