CAT Number Series Questions & Solutions

A sample of real CAT Number Series past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 77 Number Series questions in all — sign up free to practise them timed.

1. Q1.CAT 2025

   For any natural number k , let $a_{k}=3^{k}$. The smallest natural number m for which \left\{(a_{1})^{1}\times(a_{2})^{2}\times...\times(a_{20})^{20}\right\}<\left\{a_{21}\times a_{22}\times...\times a_{20+m}\right\}, is

   • 58

   • 59

   • 56

   • 57

   Show solution

   Given expression is  \left\{(a_{1})^{1}\times(a_{2})^{2}\times...\times(a_{20})^{20}\right\}<\left\{a_{21}\times a_{22}\times...\times a_{20+m}\right\} ,

   $\left\{(a_{1})^{1}\times(a_{2})^{2}\times...\times(a_{20})^{20}\right\}$ =  $\left\{3^1\times3^4\times3^9...\times3^{400}\right\}$

   Sum of square of n natural numbers is  $\frac{n\cdot\left(n+1\right)\cdot\left(2n+1\right)}{6}$

   =  $3^{\dfrac{\left(20\cdot21\cdot41\right)}{6}}$ =  $3^{2870}$

   On right hand side of inequlaity we have $\left\{a_{21}\times a_{22}\times...\times a_{20+m}\right\}$

    = $3^{21}\times3^{22}\times...\times3^{20+m}$ =  $3^{21+22+...+20+m}$

   Using the sum of the first (n) natural numbers,

   $1+2+\cdots+n = \frac{n(n+1)}{2}$

   $21 + 22 + \cdots + (20+m)$

   = $1+2+\cdots+(20+m) - (1+2+\cdots+20)$

   $1+2+\cdots+(20+m)=\frac{(20+m)(21+m)}{2}$

   $1+2+\cdots+20 = \frac{20\cdot21}{2} = 210$
   

   So, $21+22+\cdots+(20+m)$

   = $\frac{(20+m)(21+m)}{2} - 210$
   

   Expanding, $(20+m)(21+m)=m^2+41m+420$
   

   Thus, $\frac{m^2+41m+420}{2}-210$

   $= \frac{m^2+41m}{2}$

   Since the bases are equal, we must compare the powers.

   2870<\frac{m^2+41m}{2} \Rightarrow 5740<m^2+41m

   Here, we can put in the option to check the minimum value that satisfies the inequality.

   56: We get 5740<5264. This is false

   57: We get 5740<5586. This is false

   58: We get 5740<5742. This is the minimum possible value.

2. Q2.CAT 2024

   Suppose $x_{1},x_{2},x_{3},...,x_{100}$ are in arithmetic progression such that $x_{5}=-4$ and $2x_{6}+2x_{9}=x_{11}+x_{13}$, Then,$x_{100}$ equals

   • -194

   • -196

   • 204

   • 206

   Show solution

   Using the arithmetic progression formula for the nth term, where
   $x_n=a+\left(n-1\right)d$
   Substituting the value for n and using that in the equation that is given, 
   $2x_{6}+2x_{9}=x_{11}+x_{13}$ , Then, $x_{100}$ equals

   We get,  $2\left(a+5d\right)+2\left(a+8d\right)=a+10d+a+12d$
   $4a+26d=2a+22d$
   $2a=-4d$
   $a=-2d$

   We are given,  $x_5=-4$
   $a+4d=-4$
   Substituting the value for a in terms of d, 
   $2d=-4$
   $d=-2$
   $a=4$

   $x_{100}=a+99d$
   $x_{100}=4-198=-194$

3. Q3.CAT 2023

   The value of $1 + \left(1 + \frac{1}{3}\right)\frac{1}{4} + \left(1 + \frac{1}{3} + \frac{1}{9}\right)\frac{1}{16} + \left(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}\right)\frac{1}{64} + -------$ is

   • $\frac{15}{13}$

   • $\frac{27}{12}$

   • $\frac{15}{8}$

   • $\frac{16}{11}$

   Show solution

   The given sequence can be written as:

   $1\left(1+\ \frac{1}{4}+\frac{1}{16}+\frac{1}{64}+...\right)+\frac{1}{3}\left(\frac{1}{4}+\frac{1}{16}+...\right)+\frac{1}{9}\left(\frac{1}{16}+\frac{1}{64}+...\right)+..$

   We know that the sum of an infinite G.P. is $\dfrac{a}{1-r}$ , where a is the first term and r is the common ratio.

   => The first term = $\frac{1}{1-\frac{1}{4}}=\dfrac{4}{3}$

   => The second term = $\frac{1}{3}\left(\frac{\left(\frac{1}{4}\right)}{1-\left(\frac{1}{4}\right)}\right)=\dfrac{1}{9}$

   => The third term = $\frac{1}{9}\left(\frac{\left(\frac{1}{16}\right)}{1-\left(\frac{1}{4}\right)}\right)=\dfrac{1}{108}$

   Observing these three terms, we see that they are in G.P. with a common ratio of $\dfrac{1}{12}$

   => Sum of this infinite G.P. = $\dfrac{\left(\dfrac{4}{3}\right)}{1-\left(\dfrac{1}{12}\right)}=\dfrac{16}{11}$

4. Q4.CAT 2022

   Consider the arithmetic progression 3, 7, 11, ... and let $A_n$ denote the sum of the first n terms of this progression. Then the value of $\frac{1}{25} \sum_{n=1}^{25} A_{n}$ is

   • 455

   • 442

   • 415

   • 404

   Show solution

   Sum of n terms in an A.P =  $\dfrac{n}{2}\left(2a+\left(n-1\right)d\right)$

   $A_n=\dfrac{n}{2}\left(6+\left(n-1\right)4\right)=n\left(2n+1\right)$

   $\Sigma\ A_n=\Sigma\ n\left(2n+1\right)=2\Sigma\ n^2+\Sigma\ n=\ \dfrac{\ 2n\left(n+1\right)\left(2n+1\right)}{6}+\ \dfrac{\ n\left(n+1\right)}{2}$

   Substituting n = 25, we get

   $\dfrac{1}{25} \sum_{n=1}^{25} A_{n}$ =  $\dfrac{1}{25}\left(\ \dfrac{\ 2\left(25\right)\left(25+1\right)\left(50+1\right)}{6}+\ \dfrac{\ 25\left(25+1\right)}{2}\right)$

   $\dfrac{1}{25} \sum_{n=1}^{25} A_{n}$ =  $\ 26\left(17\right)+13$ = 455

   The answer is option A. 

5. Q5.CAT 2021

   Consider a sequence of real numbers, $x_{1},x_{2},x_{3},...$ such that $x_{n+1}=x_{n}+n-1$ for all $n\geq1$. If $x_{1}=-1$ then $x_{100}$ is equal to

   • 4849

   • 4949

   • 4950

   • 4850

   Show solution

   Given  $x_{n+1}\ =\ x_n\ +\ n\ -1$ and x1 = -1.

   Considering 

   x1   = -1.      (1)

   x2   = x1+1-1 = x1 + 0    (2)

   x3   = x2 + 2 - 1  =x2 + 1     (3)

   x4   = x3 + 3 - 1 = x3 + 2        (4)

   x100 = x99 + 98      (100)

   Adding the LHS and RHS for the hundred equations we have:

   (x1+x2+......................x100) = (-1+0+.........98) + (x1+x2+...............x99)

   Subtracting this we have :

   (x1+...........x100) - (x1+............. x 99) = $\frac{\left(98\cdot99\right)}{2}$ - 1.

   x100 = 4851 - 1 = 4850
   
   Alternatively
   

   $x_1=-1$

   $x_2=x_1+1-1=x_1=-1$

   $x_3=x_2+2-1=x_2+1=-1+1=0$

   $x_4=x_3+3-1=x_3+2=0+2=2$

   $x_5=x_4+4-1=x_4+3=2+3=5$

   ......

   If we observe the series, it is a series that has a difference between the consecutive terms in an AP.

   Such series are represented as  $t\left(n\right)=a+bn+cn^2$

   We need to find t(100).

   t(1) = -1

   a + b + c = -1

   t(2) = -1

   a + 2b + 4c = -1

   t(3) = 0

   a + 3b + 9c = 0

   Solving we get,

   b + 3c = 0

   b + 5c = 1

   c = 0.5

   b = -1.5

   a = 0

   Now, 

   $t\left(100\right)=\left(-1.5\right)100+\left(0.5\right)100^2=-150+5000=4850$

6. Q6.CAT 2019

   If the population of a town is p in the beginning of any year then it becomes 3 + 2p in the beginning of the next year. If the population in the beginning of 2019 is 1000, then the population in the beginning of 2034 will be

   • $(1003)^{15} + 6$

   • $(997)^{15} - 3$

   • $(997)2^{14} + 3$

   • $(1003)2^{15} - 3$

   Show solution

   The population of town at the beginning of 1st year = p

   The population of town at the beginning of 2nd year = 3+2p

   The population of town at the beginning of 3rd year = 2(3+2p)+3 = 2*2p+2*3+3 =4p+3(1+2)

   The population of town at the beginning of 4th year = 2(2*2p+2*3+3)+3 = 8p+3(1+2+4)

   Similarly population at the beginning of the nth year =  $2^{n-1}$ p+3( $2^{n-1}-1$ ) =  $2^{n-1}\left(p+3\right)$ -3 

   The population in the beginning of 2019 is 1000, then the population in the beginning of 2034 will be  $(2^{2034-2019})\left(1000+3\right)$ -3 =  $2^{15}\left(1003\right)$ -3

7. Q7.CAT 2018

   Let $t_{1},t_{2}$,... be real numbers such that $t_{1}+t_{2}+…+t_{n} = 2n^{2}+9n+13$, for every positive integer $n \geq 2$. If $t_{k}=103$, then k equals

   Answer: 24

   Show solution

   It is given that $t_{1}+t_{2}+…+t_{n} = 2n^{2}+9n+13$ , for every positive integer $n \geq 2$ . 

   We can say that $t_{1}+t_{2}+…+t_{k} = 2k^{2}+9k+13$    ... (1) 

   Replacing k by (k-1) we can say that 

     t_{1}+t_{2}+…+t_{k-1} =&nbsp;2(k-1)^{2}+9(k-1)+13    ... (2)

   On subtracting equation (2) from equation (1)

   $\Rightarrow$ t_{k} =&nbsp;2k^{2}+9k+13 -&nbsp;2(k-1)^{2}+9(k-1)+13

   $\Rightarrow$ 103 =&nbsp;4k+7
   

   $\Rightarrow$ k =&nbsp;24

8. Q8.CAT 2008

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

   • 78

   • 19

   • 20

   • 77

   • 22

   Show 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.

9. Q9.CAT 2003

   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?

   • 1st

   • 9th

   • 12th

   • None of the above

   Show 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
    

10. Q10.CAT 2002

    Let S denotes the infinite sum $2 + 5x + 9x^2 + 14x^3 + 20x^4 + ...$ , where |x| < 1 and the coefficient of $x^{n - 1}$ is n( n + 3 )/2 , ( n = 1, 2 , . . . ) . Then S equals:

    • $(2-x)/(1-x)^3$

    • $(2-x)/(1+x)^3$

    • $(2+x)/(1-x)^3$

    • $(2+x)/(1+x)^3$

    Show solution

    Let  $S = 2+5x+9x^2+....$
    $S*x = 2x+5x^2+9x^3+...$
    $S(1-x) = 2+3x+4x^2+...$
    $S(1-x)*x = 2x+3x^2+4x^3+...$
    $S(1-x)(1-x) = 2+x+x^2+x^3+... = 2+x/(1-x)$
    So, S = [2(1-x) + x]/(1-x)^3 =&gt; S = (2-x)/(1-x)^3

11. Passage

    DIRECTIONS for the following questions: These questions are based on the situation given below: There are fifty integers $a_1, a_2,...,a_{50}$, not all of them necessarily different. Let the greatest integer of these fifty integers be referred to as $G$, and the smallest integer be referred to as $L$. The integers $a_1$ through $a_{24}$ form sequence $S1$, and the rest form sequence $S2$. Each member of $S1$ is less than or equal to each member of $S2$.

    Q11.CAT 1999

    Elements of $S1$ are in ascending order, and those of $S2$ are in descending order. $a_{24}$ and $a_{25}$ are interchanged. Then, which of the following statements is true?

    • S1 continues to be in ascending order

    • S2 continues to be in descending order

    • S1 continues to be in ascending order and S2 in descending order.

    • None of the above

    Show solution

    We know that $a_{24}$ is less than $a_{25}$ .

    So, even if $a_{25}$ replaces $a_{24}$ , the ascending order still exists in S1.

    But, $a_{25}$ is less than $a_{26}$ . Hence, the descending order does not exist in S2 anymore.

12. Passage

    Answer the questions based on the following information. A series $S_{1}$ of five positive integers is such that the third term is half the first term and the fifth term is 20 more than the first term. In series $S_{2}$, the nth term is defined as the difference between the (n+1)th term and the nth term of series $S_{1}$, $S_{2}$ is an arithmetic progression with a common difference of 30.

    Q12.CAT 1996

    What is the sum of series $S_{2}$?

    • 10

    • 20

    • 30

    • 40

    Show solution

    Assume the first series as a,b,a/2,c,a+20
    and second series as x1,x2,x3,x4
    x1=b-a, x2= a/2-b, x3=c-a/2, and x4=a+20-c
    x2-x1=30 => 3a-4b=60
    and x4-x3=30 => 3a-4c=20
    and x4-x2=60 => a-2c+2b=80
    Solving we get, a=100, b=60, and c=70
    S1= 100,60,50,70,120
    

    S2 = -40, -10, 20, 50
    Sum = 20

CAT Number Series previous year questions with solutions — AthenaPrep