# Progression and Series Questions for CAT

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

a)Â 78

b)Â 19

c)Â 20

d)Â 77

e)Â 22

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:Â 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,fâ€¦ is

a)Â u

b)Â v

c)Â w

d)Â x

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

Question 3:Â 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)Â 3

b)Â 4

c)Â 5

d)Â 6

e)Â 7

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

a)Â $(5*2^{99}-6)$

b)Â $(5*2^{99}+6)$

c)Â $(6*2^{99}+5)$

d)Â $(6*2^{99}-5)$

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:Â 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)Â 9/19

b)Â 10/19

c)Â 10/21

d)Â 11/21

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

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

a)Â $\frac{8}{1-x^8}$

b)Â $\frac{4x}{1+x^2}$

c)Â $\frac{4}{1-x^6}$

d)Â $\frac{4}{1+x^4}$

Solution:

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

Question 7:Â 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)Â 61250

b)Â 65525

c)Â 42455

d)Â 62525

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:Â 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)Â 2:3

b)Â 3:2

c)Â 3:4

d)Â 4:3

Solution:

The seventh term of an AP = a + 6d. Third term will be aÂ + 2d and second term will be aÂ + 16d. We are given that
$(a + 6d)^2 = (a + 2d)(a + 16d)$
=> $a^2$ + $36d^2$ +Â 12ad = $a^2 + 18ad + 32d^2$
=> $4d^2 = 6ad$
=> $d:a = 3:2$

Question 9:Â Let $a_1$, $a_2$,………….,Â  $a_{3n}$ 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)Â 8

b)Â 9

c)Â 10

d)Â 11

Solution:

$a_{1}$ = 3 andÂ $a_{2}$ = 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($a_1$+Â $a_{2}$Â +…+ $a_n$) > 1830
=> 210m >Â 1830
=> m > 8.71
Hence, smallest integral value of ‘m’ is 9.

Question 10:Â 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

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:Â 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)Â 1/32

b)Â 2/32

c)Â 3/32

d)Â 4/32

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 byÂ $\frac{a}{1-r}$ where a here isÂ $a_{n+1}$

=> $a_n= 3(\frac{a_{n+1}}{1-r})$
=> $a_n= 3(\frac{a_{n}\times r}{1-r})$
=> $r = \frac{1}{4}$
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:Â 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)Â $\frac{25}{151}$

b)Â $\frac{1}{2}$

c)Â $\frac{1}{4}$

d)Â $\frac{111}{55}$

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:Â 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)Â $\frac{3}{6}$

b)Â $\frac{1}{6}$

c)Â $\frac{5}{2}$

d)Â $\frac{3}{2}$

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

So,Â $r$ =Â $\frac{5}{2}$

Hence, option C is the correct answer.

Question 14:Â Let $\ a_{1},a_{2}…a_{52}\$ be positive integers such that $\ a_{1}$ < $a_{2}$ < … <Â $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)Â 48

b)Â 20

c)Â 23

d)Â 45

Solution:

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

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

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

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

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

$a_{1}+51(x+1)$ =Â 52x

$a_{1}$ = x – 51.

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

(x+1) is the average of termsÂ $a_{2}$, $a_{3}$, ….$a_{52}$. We know thatÂ $a_{2}$ < $a_{3}$ < … <Â $a_{52}\$ andÂ $a_{52}$ = 100.

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

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

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

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

$\Rightarrow$ $x =Â 74$

Therefore,Â theÂ largest possible value ofÂ $a_{1}$ = x – 51 = 74 – 51 = 23.

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

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,Â (4n+3) = 95 i.e. n = 23

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

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

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

$\Rightarrow$Â $80707$