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# SNAP Progression Questions PDF [Most Important]

The Progression is an important topic in the Quant section of the SNAP Exam. Quant is a scoring section in SNAP, so it is advised to practice as much as questions from quant. This article provides some of the most important Progression Questions for SNAP. One can also download this Free Progression Questions for SNAP PDF with detailed answers by Cracku. These questions will help you practice and solve the Progression questions in the SNAP exam. Utilize this PDF practice set, which is one of the best sources for practising.

Question 1:Â The average of all 3-digit terms in the arithmetic progression 38, 55, 72, …, is

Solution:

General term = 38 + (n-1)17 = 17n + 21 = 17(n+1) + 4 = 17k + 4

Each term is in the form of 17k + 4

Least 3-digit number in the form of 17k + 4 is at k = 6, i.e. 106

Highest 3-digit number in the form of 17k + 4 is at k = 58, i.e. 990

106, 123, 140,……….., 990

990 = 106 + 17(n-1)

n = 53

Sum =Â $\frac{53}{2}\left(106+990\right)=53\times548$

Average =Â $53\times\frac{548}{53}=548$

Question 2:Â For any natural number n, suppose the sum of the first n terms of an arithmetic progression is $(n + 2n^2)$. If the $n^{th}$ term of the progression is divisible by 9, then the smallest possible value of n is

a)Â 9

b)Â 4

c)Â 7

d)Â 8

Solution:

It is given,

$S_n=2n^2+n$

$S_{n-1}=2\left(n-1\right)^2+\left(n-1\right)$

$S_{n-1}=2n^2-3n+1$

$T_n=S_n-S_{n-1}=2n^2+n-2n^2+3n-1=4n-1$

$T_n=4n-1$

$T_n$ is divisible by 9 when 4n is in the form of 9k + 1

4n = 9k + 1

Least possible value of n is 7

Question 3:Â How many two-digit numbers are divisible by 3?

a)Â 30

b)Â 20

c)Â 29

d)Â 19

Solution:

The least 2-digit number divisible by 3 = 12
The highest 2-digit number divisible by 3 = 99
99 = 12 + (n-1)3
(n-1)3 = 87
n = 30
There are 30 2-digit numbers divisible by 3.

Question 4:Â 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)Â 4849

b)Â 4949

c)Â 4950

d)Â 4850

Solution:

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

Considering

x1Â  Â = -1.Â  Â  Â  (1)

x2Â  Â = x1+1-1 = x1 + 0Â  Â  (2)

x3Â  Â = x2 + 2 – 1Â  =x2 + 1Â  Â  Â (3)

x4Â Â  = x3 + 3 – 1 = x3 + 2Â  Â  Â  Â  (4)

x100 = x99 + 98Â  Â  Â  (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 hasÂ a difference between the consecutive terms in an AP.

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

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,

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

Question 5:Â 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)Â 200

b)Â 2

c)Â -200

d)Â -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 $x_{49\ }+x_{50}\ =-2$

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Question 6:Â 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)Â 8

b)Â 12

c)Â 14

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

a)Â 6119

b)Â 6090

c)Â 4941

d)Â 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 beÂ $ax^2+bx+c$

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 beÂ $\left(15\right)^2-2\left(15\right)+2=197$

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=Â $\frac{29}{2}\left(197+225\right)\ =\ 6119$

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 isÂ $\left(14\right)^2=\ 196$, similarly for the 15th group this is :Â $\left(15\right)^2=\ 225$

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 :Â Â $\frac{n}{2}\cdot\left(first\ term\ +last\ term\right)$

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

6119.

Question 8:Â 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)Â 4

b)Â 1

c)Â 3

d)Â 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 9:Â Find the sum ofthe following series (with infinite terms):
$2\sqrt{2},\frac{4}{\sqrt{3}},\frac{4\sqrt{2}}{3},…..$

a)Â $2\sqrt{3}(\sqrt{3}+\sqrt{2})$

b)Â $2\sqrt{3}(\sqrt{3}-\sqrt{2})$

c)Â $2\sqrt{6}(\sqrt{3}+\sqrt{2})$

d)Â $2\sqrt{6}(\sqrt{3}-\sqrt{2})$

Solution:

The series is an infinite Geometric progression with first term asÂ $2\sqrt{\ 2}$ and common ratioÂ $\sqrt{\ \frac{2}{3}}$
Now we know that sum of an infinite geometric progression = $\frac{a}{1-r}$ where r is the common ratio and a is first term
so we get sum =$\frac{2\sqrt{\ 2}}{1-\sqrt{\ \frac{2}{3}}}$
=$=\ \frac{2\sqrt{\ 6}}{\sqrt{\ 3}-\sqrt{\ 2}}$
Rationalizing the denominator
we get :$=\frac{2\sqrt{\ 6}\left(\sqrt{\ 3}+\sqrt{\ 2}\right)}{3-2}$
= $2\sqrt{\ 6}\left(\sqrt{\ 3}+\sqrt{\ 2}\right)$

Question 10:Â 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)Â 6

b)Â 2

c)Â -4

d)Â -2

Solution:

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

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

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

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

a)Â -5050

b)Â -5151

c)Â -5051

d)Â -5150

Solution:

$x_1=-1$

$x_1=x_2+2$ =>Â $x_2=x_1-2$ = -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 isÂ $-\frac{n\left(n+1\right)}{2}$

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

Question 12:Â A new sequence is obtained from the sequence of positive integers {1,2,3,—} by deleting all the perfect squares. The $2018^{th}$ term of the new sequence is

a)Â 2061

b)Â 2062

c)Â 2063

d)Â 2065

Solution:

Let us write all the number the numbers from 1 to 2018. Here we find that there are 44 numbers that are perfect squares which are 1,4,9… 1936($44^2$). Thus after removing the perfect square it has total of 2018-44 = 1974 terms.

Taking next 44 terms from 2019, it will be (2019,2020,2021….. 2062) in which there is 1 perfect square (2025 which isÂ $45^2$) hence it is removed and now there is total of 2017 terms. Next term has to be 2063 which is also $2018^{th}$ term

Question 13:Â A sequence $\left\{x_n \right\}$ of real numbers is defined as follows:
$x_0 = 1, x_1 = 2,$ and $x_n = \frac{1 + x_{n – 1}}{x_{n – 2}}$ for n = 2, 3, 4 …
It follows that $x_{2018}$ is

a)Â 1

b)Â 2

c)Â 3

d)Â 4

Solution:

Given thatÂ $x_0=1\ and\ x_1=2\$

From above formula we getÂ $x_2=3,\ x_3=2,\ x_4=1,\ x_5\ =\ 1\ and\ x_6=2$

We see that the pattern repeats itself in the period of five withÂ $x_{5k}=1,\ x_{5k+1}=2,\ x_{5k+2}=3,\ x_{5k+3}\ =\ 2\ and\ x_{5k+4}=1$

2018 is of form 5k+3 henceÂ $x_{2018}=2$

Question 14:Â The sum of an infinite geometric series of real numbers is 14, and the sum of the cubes of the terms of this series is 392. The first term of the series is

a)Â -14

b)Â 10

c)Â 7

d)Â -5

Solution:

Let the first term of the infinite G.P. be a and the common ratio be r.

So, Sum of infinite GP=Â $\ \frac{\ a}{1-r}=14$ ………….. (1)

The cubes of the terms of original GP=Â $a^3,\ \left(ar\right)^3,\ \left(ar^2\right)^3…$= $a^3,\ a^3r^3,\ a^3r^6…$

So, in the new infinite GP, first term isÂ $a^3$ and the common ratio isÂ $r^3$

Therefore, sum to infinite terms=Â $\frac{\ a^3\ }{1-r^3}$ ……………(2)

Cubing the eqn 1 and dividing by eqn 2, we get,

$\frac{\ \frac{\ a^3}{\left(1-r\right)^3}\ }{\ \frac{\ a^3}{1-r^3}}=\ \ \frac{\ 14\times\ 14\times\ 14}{392}$

=>$\ \frac{1-r^3\ }{\left(1-r\right)^3}\ =\ \ \frac{\ 14\times\ 14\times\ 14}{392}$

=>Â $\ \frac{\left(1-r\right)\left(1+r^2+r\right)}{\left(1-r\right)^3}\ =\ \ \ 7$

=>$\ \frac{\left(1+r^2+r\right)}{\left(1-r\right)^2}\ =\ \ \ 7$

=>Â $\ \left(1+r^2+r\right)\ =\ \ 7\left(1+r^2-2r\right)$

=>$\ 1+r^2+r\ =\ \ 7+7r^2-14r$

=>$\ -6-6r^2+15r\ =\ 0$

=>$2r^2-5r+2\ =\ 0$

=>$2r^2-4r-r+2\ =\ 0$

=>$2r\left(r-2\right)-1\left(r-2\right)=0$

=>(2r-1)(r-2)=0

.’.r= 1/2 or 2.

In v\case of infinite GP, -1<r<1.

So, r= 1/2

Putting this value in eqn 1, we get,

$\ \frac{\ a}{1-\ \frac{\ 1}{2}}=14$

=>2a=14

.’. a=7- Option C.

Question 15:Â A man is laying stones, from start to end, along the two sides of a 200-meterwalkway. The stones are to be laid 5 meters apart from each other. When he begins, all the stones are present at the start of the walkway. He places the first stone on each side at the walkwayâ€™s start. For all the other stones, the man lays the stones first along one of the walkwayâ€™s sides, then along the other side in an exactly similar fashion. However, he can carry only one stone at a time. To lay each stone, the man walks to the spot, lays the stone, and then walks back to pick another. After laying all the stones, the man walks back to the start, which marks the end of his work. What is the total distance that the man walks in executing this work? Assume that the width of the walkway is negligible.

a)Â 16400 metres

b)Â 4100 metres

c)Â 8050 metres

d)Â 16200 metres

e)Â 8200 metres

Solution:

On one side, to place 1st rock, he had to travel 10m. For 2nd rock he had to travel 20m..similarly, till last rock he had to travel 400.

Total sum would be 10+20+30+…+400 = $\frac{40}{2}\left(410\right)=8200$.

Similarly, on the other side it will be 8200.

Total 8200+8200=16400.

Question 16:Â The 288th term of the sequence a, b, b, c, c, c, d, d, d, dâ€¦ is

a)Â u

b)Â v

c)Â w

d)Â x

Solution:

Here, $\ \frac{\ n\left(n-1\right)}{2}$(not inclusive) term toÂ $\ \frac{\ n\left(n+1\right)}{2}$(inclusive) term is theÂ nth place alphabet.Â  Â (1st place alphabet is a, 2nd place alphabet is b, and so on.)

=> $\ \frac{\ n\left(n-1\right)}{2}$ < 288 <Â $\ \frac{\ n\left(n+1\right)}{2}$

=> n(n-1) < 576 < n(n+1)

n = 24

At 24th place, x is the alphabet.

Question 17:Â From a book, a number of consecutive pages are missing. The sum of the page numbers of these pages is 9808. Which pages are missing?

a)Â The page 9808 is missing

b)Â The pages 291 up to 322 are missing

c)Â The pages 291 up to and including 322 are missing

d)Â Either a or c

Solution:

Let n pages be missing starting from the p th page.

So, p+(p+1)+(p+2)â€¦..+[p+(nâˆ’2)]+[p+(nâˆ’1)]=9808

9808=n/2[2Ã—p+(nâˆ’1)Ã—1]
nÂ²+(2p-1)nâˆ’19616=0
Now both p and n are natural numbers
Now possible cases
n=19616,p=âˆ’9807
n=âˆ’19616,p=9808
n=613,p=âˆ’29
n
=âˆ’613,p=29
n=1,p=9808
n=âˆ’1,p=âˆ’9807
n=32,p=291
n=âˆ’32,p=âˆ’290

But since p and n has to be natural numbers, so all solutions except 5 and 7 are discarded.
Hence we can say A and C are correct

Question 18:Â A man earns Rs 20 on the first day and spends Rs 15 on the next day. He again
earns Rs 20 on the third day and spends Rs 15 on the fourth day. If he continues to save like
this, how soon will he have Rs 60 in hand?

a)Â On $17^{th}$ day

b)Â On $27^{th}$ day

c)Â On $24^{th}$ day

d)Â On $30^{th}$ day

Solution:

For every cycle of 2 days his overall earnings will be 5 Rs ,

In 8 cycles i.e in 16 days he will earn 40 Rs, on 17th day he will earn 20 more rupees and his overall savings will reach to 60 Rs.

Question 19:Â Each of the series S1 = 2 + 4 + 6 + ……… and S2 = 3 + 6 + 9 + ……… is continued
to 100 terms. Find how many terms are identical.

a)Â 34

b)Â 33

c)Â 32

d)Â None of these

Solution:

The identical terms will be of the form k*LCM(d1,d2) +smallest common number .

Where d1 , d2 are common differences in two series and smallest common term in two series = 6

So N= k*LCM ( 2,3) + 6

N= 6k+6Â  Â (where k is a whole number)

Since the last number in 1st series is 200 and last number in 2nd series is 300, it is sufficient to check multiples of till 200

Therefore total number of multiples of 6 below 200 = 33

Question 20:Â Study the series carefully ‘B 8 4 C R M 9 P D K W F A 2 E J 7 X U Q H L T Y 6 G S’. If it is possible to make a meaningful word with the ninth, the sixteenth, the twenty-fourth and the twenty-seventh letters from the left in the above series, which of the following will be the first letter of the word? If no such word can be made, give ‘X’ as an answer. If more than one such word can be made, give ‘M’ as an answer.

a)Â X

b)Â M

c)Â J

d)Â Y

$9^{th}$ term = D
$16^{th}$ term = J
$24^{th}$ term = Y
$27^{th}$ term = S