# Progressions and Series Questions for CAT

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Progression and Series is one of the most important topics in the CAT Quants section. If you’re weak in Progressions and Series questions for CAT, make sure you learn the basic concepts well. Here, you can learn all the important formulas from CAT Progressions and Series. You can check out these CAT Progression and Series Questions PDF from the CAT Previous year’s papers

This post will look at the important Progressions and Series of questions in the CAT Quant section. These are a good source of practice for CAT preparation; If you want to practice these questions, you can download this CAT Progressions and Series Questions PDF (most important) along with the detailed solutions below, which is completely Free.

Question 1: The arithmetic mean of x, y and z is 80, and that of x, y, z, u and v is 75, where u=(x+y)/2 and v=(y+z)/2. If x ≥ z, then the minimum possible value of x is

Solution:

Given that the arithmetic mean of x, y and z is 80.

$\Rightarrow$ $\dfrac{x+y+z}{3} = 80$

$\Rightarrow$ $x+y+z = 240$  … (1)

Also,  $\dfrac{x+y+z+v+u}{5} = 75$

$\Rightarrow$ $\dfrac{x+y+z+v+u}{5} = 75$

$\Rightarrow$ $x+y+z+v+u = 375$

Substituting values from equation (1),

$\Rightarrow$ $v+u = 135$

It is given that u=(x+y)/2 and v=(y+z)/2.

$\Rightarrow$ $(x+y)/2+(y+z)/2 = 135$

$\Rightarrow$ $x+2y+z = 270$

$\Rightarrow$ $y = 30$   (Since $x+y+z = 240$)

Therefore, we can say that $x+z = 240 – y = 210$. We are also given that x ≥ z,

Hence, $x_{min}$ = 210/2 = 105.

Question 2: 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

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

$t_{1}+t_{2}+…+t_{k-1} = 2(k-1)^{2}+9(k-1)+13$   … (2)

On subtracting equation (2) from equation (1)

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

$\Rightarrow$ $103 = 4k+7$

$\Rightarrow$ $k = 24$

Question 3: Let $a_1, a_2, …$ be integers such that
$a_1 – a_2 + a_3 – a_4 + …. + (-1)^{n – 1} a_n = n,$ for all $n \geq 1.$
Then $a_{51} + a_{52} + …. + a_{1023}$ equals

a) 0

b) 1

c) 10

d) -1

Solution:

$a_1 – a_2 + a_3 – a_4 + …. + (-1)^{n – 1} a_n = n$

It is clear from the above equation that when n is odd, the co-efficient of a is positive otherwise negative.

$a_1 – a_2 = 2$

$a_1 = a_2 + 2$

$a_1 – a_2 + a_3 = 3$

On substituting the value of $a_1$ in the above equation, we get

$a_3$ = 1

$a_1 – a_2 + a_3 – a_4 = 4$

On substituting the values of $a_1, a_3$ in the above equation, we get

$a_4$ = -1

$a_1 – a_2 + a_3 – a_4 +a_5 = 5$

On substituting the values of $a_1, a_3, a_4$ in the above equation, we get

$a_5$ = 1

So we can conclude that $a_3, a_5, a_7….a_{n+1}$ = 1 and $a_2, a_4, a_6….a_{2n}$ = -1

Now we have to find the value of $a_{51} + a_{52} + …. + a_{1023}$

Number of terms = 1023=51+(n-1)1

n=973

There will be 486 even and 487 odd terms, so the value of $a_{51} + a_{52} + …. + a_{1023}$ = 486*-1+487*1=1

Question 4: If $(2n + 1) + (2n + 3) + (2n + 5) + … + (2n + 47) = 5280$, then whatis the value of $1 + 2 + 3 + .. + n?$

Solution:

Let us first find the number of terms

47=1+(n-1)2

n=24

24*2n+1+3+5+….47=5280

48n+576=5280

48n=4704

n=98

Sum of first 98 terms = 98*99/2

=4851

Question 5: The number of common terms in the two sequences: 15, 19, 23, 27, . . . . , 415 and 14, 19, 24, 29, . . . , 464 is

a) 21

b) 20

c) 18

d) 19

Solution:

A: 15, 19, 23, 27, . . . . , 415

B: 14, 19, 24, 29, . . . , 464

Here the first common term = 19

Common difference = LCM of 5, 4=20

19+(n-1)20 $\le\$ 415

(n-1)20 $\le\$ 396

(n-1) $\le\$ 19.8

n=20

Question 6: If $a_1, a_2, ……$ are in A.P., then, $\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + ……. + \frac{1}{\sqrt{a_n} + \sqrt{a_{n + 1}}}$ is equal to

a) $\frac{n}{\sqrt{a_1} + \sqrt{a_{n + 1}}}$

b) $\frac{n – 1}{\sqrt{a_1} + \sqrt{a_{n – 1}}}$

c) $\frac{n – 1}{\sqrt{a_1} + \sqrt{a_n}}$

d) $\frac{n}{\sqrt{a_1} – \sqrt{a_{n + 1}}}$

Solution:

We have, $\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + ……. + \frac{1}{\sqrt{a_n} + \sqrt{a_{n + 1}}}$

Now, $\frac{1}{\sqrt{a_1} + \sqrt{a_2}}$ = $\frac{\sqrt{a_2} – \sqrt{a_1}}{(\sqrt{a_2} + \sqrt{a_1})(\sqrt{a_2} – \sqrt{a_1})}$   (Multiplying numerator and denominator by $\sqrt{a_2} – \sqrt{a_1}$)

= $\frac{\sqrt{a_2} – \sqrt{a_1}}{({a_2} – {a_1}}$

=$\frac{\sqrt{a_2} – \sqrt{a_1}}{d}$   (where d is the common difference)

Similarly, $\frac{1}{\sqrt{a_2} + \sqrt{a_3}}$ = $\frac{\sqrt{a_3} – \sqrt{a_2}}{d}$ and so on.

Then the expression $\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + ……. + \frac{1}{\sqrt{a_n} + \sqrt{a_{n + 1}}}$

can be written as $\ \frac{\ 1}{d}(\sqrt{a_2}-\sqrt{a_1}+\sqrt{a_3}-\sqrt{a_3}+……………………..\sqrt{a_{n+1}} – \sqrt{a_{n}}$

= $\ \frac{\ n}{nd}(\sqrt{a_{n+1}}-\sqrt{a_1})$ (Multiplying both numerator and denominator by n)

= $\ \frac{n(\sqrt{a_{n+1}}-\sqrt{a_1})}{{a_{n+1}} – {a_1}}$     $(a_{n+1} – {a_1} =nd)$

= $\frac{n}{\sqrt{a_1} + \sqrt{a_{n + 1}}}$

Question 7: 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

a) $(1003)^{15} + 6$

b) $(997)^{15} – 3$

c) $(997)2^{14} + 3$

d) $(1003)2^{15} – 3$

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

Question 8: If $a_1 + a_2 + a_3 + …. + a_n = 3(2^{n + 1} – 2)$, for every $n \geq 1$, then $a_{11}$ equals

Solution:

11th term of series = $a_{11}$ = Sum of 11 terms – Sum of 10 terms = $3(2^{11 + 1} – 2)$-3$(2^{10 + 1} – 2)$

= 3$(2^{12} – 2-2^{11} +2)$=3$(2^{11})(2-1)$= 3*$2^{11}$ = 6144

Question 9: If $x_1=-1$ and $x_m=x_{m+1}+(m+1)$ for every positive integer m, then $X_{100}$ equals

a) -5050

b) -5151

c) -5051

d) -5150

Solution:

$x_1=-1$

$x_1=x_2+2$ => $x_2=x_1-2$ = -3

Similarly,

$x_3=x_1-5=-6$

$x_4=-10$

.

.

The series is -1, -3, -6, -10, -15……

When the differences are in AP, then the nth term is $-\frac{n\left(n+1\right)}{2}$

$x_{100}=-\frac{100\left(100+1\right)}{2}=-5050$

Question 10: Let the m-th and n-th terms of a geometric progression be $\frac{3}{4}$ and 12. respectively, where $m < n$. If the common ratio of the progression is an integer r, then the smallest possible value of $r + n – m$ is

a) 6

b) 2

c) -4

d) -2

Solution:

Let the first term of the GP be “a” . Now from the question we can show that

$ar^{m-1}=\frac{3}{4}$    $ar^{n-1}=12$

Dividing both the equations we get $r^{m-1-n+1}=\frac{1}{16}\ or\ r^{m-n}=16^{-1\ }or\ r^{n-m}=16$

So for the minimum possible value we take Now give minimum possible value to “r” i.e -4 and n-m=2

Hence minimum possible value of r+n-m=-4+2=-2

Question 11: If $x_0 = 1, x_1 = 2$, and $x_{n + 2} = \frac{1 + x_{n + 1}}{x_n}, n = 0, 1, 2, 3, ……,$ then $x_{2021}$ is equal to

a) 4

b) 1

c) 3

d) 2

Solution:

$x_0=1$

$x_1=2$

$x_2=\frac{\left(1+x_1\right)}{x_0}=\frac{\left(1+2\right)}{1}=3$

$x_3=\frac{\left(1+x_2\right)}{x_1}=\frac{\left(1+3\right)}{2}=2$

$x_4=\frac{\left(1+x_3\right)}{x_2}=\frac{\left(1+2\right)}{3}=1$

$x_5=\frac{\left(1+x_4\right)}{x_3}=\frac{\left(1+1\right)}{2}=1$

$x_6=\frac{\left(1+x_5\right)}{x_4}=\frac{\left(1+1\right)}{1}=2$

Hence, the series begins to repeat itself after every 5 terms. Terms whose number is of the form 5n are 1, 5n+1 are 2… and so on, where n=0,1,2,3,….

2021 is of the form 5n+1. Hence, its value will be 2.

Question 12: The natural numbers are divided into groups as (1), (2, 3, 4), (5, 6, 7, 8, 9), ….. and so on. Then, the sum of the numbers in the 15th group is equal to

a) 6119

b) 6090

c) 4941

d) 7471

Solution:

The first number in each group: 1, 2, 5, 10, 17…..

Their common difference is in Arithmetic Progression. Hence, the general term of the series can be expressed as a quadratic equation.

Let the quadratic equation of the general term be $ax^2+bx+c$

1st term = a+b+c=1

2nd term = 4a+2b+c=2

3rd term = 9a+3b+c=5

Solving the equations, we get a=1, b=-2, c=2.

Hence, the first term in the 15th group will be $\left(15\right)^2-2\left(15\right)+2=197$

We can see that the number of terms in each group is 1, 3, 5, 7…. and so on. These are of the form 2n-1. Hence, the number of terms in 15th group will be 29. Hence, the last term in the 15th group will be 197+29-1 = 225.

Sum of terms in group 15= $\frac{29}{2}\left(197+225\right)\ =\ 6119$

Alternatively,

The final term in each group is the square of the group number.

In the first group 1, second group 4, …………

The final element of the 14th group is $\left(14\right)^2=\ 196$, similarly for the 15th group this is : $\left(15\right)^2=\ 225$

Each group contains all the consecutive elements in this range.

Hence the 15th group the elements are:

(197, 198, …………………………..225).

This is an Arithmetic Progression with a common difference of 1 and the number of element 29.

Hence the sum is given by :  $\frac{n}{2}\cdot\left(first\ term\ +last\ term\right)$

$\frac{29}{2}\cdot\left(197+225\right)$

6119.

Question 13: Three positive integers x, y and z are in arithmetic progression. If $y-x>2$ and $xyz=5(x+y+z)$, then z-x equals

a) 8

b) 12

c) 14

d) 10

Solution:

Given x, y, z are three terms in an arithmetic progression.

Considering x = a, y = a+d, z = a+2*d.

Using the given equation x*y*z = 5*(x+y+z)

a*(a+d)*(a+2*d) = 5*(a+a+d+a+2*d)

=a*(a+d)*(a+2*d)  = 5*(3*a+3*d) = 15*(a+d).

= a*(a+2*d) = 15.

Since all x, y, z are positive integers and y-x > 2. a, a+d, a+2*d are integers.

The common difference is positive and greater than 2.

Among the different possibilities are : (a=1, a+2d = 5), (a, =3, a+2d = 5), (a = 5, a+2d = 3), (a=15, a+2d = 1)

Hence the only possible case satisfying the condition is :

a = 1, a+2*d = 15.

x = 1, z = 15.

z-x = 14.

Question 14: For a sequence of real numbers $x_{1},x_{2},…x_{n}$, If $x_{1}-x_{2}+x_{3}-….+(-1)^{n+1}x_{n}=n^{2}+2n$ for all natural numbers n, then the sum $x_{49}+x_{50}$ equals

a) 200

b) 2

c) -200

d) -2

Solution:

Now as per the given series :
we get $x_1=1+2\ =3$
Now $x_1-x_2=\ 8$
so$x_2=-5$
Now $x_1-x_2+x_3\ =\ 15$
so $x_3\ =7$
so we get $x_n\ =\left(-1\right)^{n+1}\left(2n+1\right)$
so $x_{49}\ =\ 99$ and $x_{50}\ =-101$
Therefore $x_{49\ }+x_{50}\ =-2$

Question 15: 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

a) 4849

b) 4949

c) 4950

d) 4850

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,

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