0
845

# SNAP Progression Questions PDF [Important]

Progression is an important topic in the Progression section of the SNAP Exam. You can also download this free Progression Questions PDF (with answers) for SNAP 2022 by Cracku. These questions will help you to 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 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 2: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 3:Â The first two terms of a geometric progression add up to 12. The sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternately positive and negative, then the first term is

a)Â -2

b)Â -4

c)Â -12

d)Â 8

Solution:

Let the first term of the geometric progression be a and common ratio be r
Now as per given condition :
a+ar= a(1+r) =12Â  (1)
ar^2+ar^3 = 48 (2)
we get ar^2(1+r) =48Â  (3)
Dividing (1) and (3)
we get :
1/r^2 = 1/4
we get 1/r = 1/2 or -1/2
but since terms are alternatively positive and negative
we get r = -2
Substituting
we get a(-1) =12
we get a=-12

Question 4:Â What is the maximum sum of the terms in the arithmetic progression $25, 24\frac{1}{2}, 24, …………..?$

a)Â $637\frac{1}{2}$

b)Â 625

c)Â $662\frac{1}{2}$

d)Â 650

Solution:

It is a decreasing AP 25,24.5,24,23.5,…

The sum will be largest,Â if nth term is zero then

25 + (n-1)*(-.5) = 0

=> n = 51

after 51 terms AP contains -ve terms

so, max sum will be obtained upto 50/51th term

S = $\frac{51}{2}$ *(25+0) = 637.5

Question 5:Â Consider an arithmetic progression of positive terms with the first term as $\alpha$. Let $S_n$ denote the sum of the first n terms of this arithmetic progression and let $\frac{S_m}{S_n} = \frac{m^2}{n^2}$ for m â‰  n. Then the $50^{th}$ term is

a)Â 50 $\alpha$

b)Â 99Â $\alpha$

c)Â 100Â $\alpha$

d)Â 250Â $\alpha$

Solution:

$\frac{m^2}{n^2}$ =Â $\frac{\frac{m*(2\alpha+ (m-1)d)}{2}}{\frac{n*(2\alpha+ (n-1)d)}{2}}$

$\frac{m}{n}$ = $\frac{(2\alpha+ (m-1)d)}{(2\alpha+ (n-1)d)}$

$m(2\alpha+ (n-1)d) = n(2\alpha+ (m-1)d))$

$(n-m)d=(n-m)2\alpha\$

d = 2$\alpha$

$50^{th}$ term =Â $\alpha$+49*2$\alpha$Â = 99$\alpha$

Question 6: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 7: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 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 If the positive real numbers a, b and c are in Arithmetic Progression, such that abc = 4, then minimum possible value of b is:

a)Â $2^{\frac{3}{2}}$

b)Â $2^{\frac{2}{3}}$

c)Â $2^{\frac{1}{3}}$

d)Â None of the above

Solution:

It has been given that a, b, and c are in an arithmetic progression.
Let a = x-p, b = x, and c = x+p
We know that a, b, and c are real numbers.
Therefore, the arithmetic mean of a,b,c should be greater than or equal to the geometric mean.
$\frac{a+b+c}{3} \geq \sqrt[3]{abc}$
$\frac{a+b+c}{3} \geq \sqrt[3]{4}$
$\frac{3x}{3}\geq\sqrt[3]{4}$

$x\geq\sqrt[3]{4}$
We know that $x$ = $b$.
Therefore,$b\geq\sqrt[3]{4}$orÂ $b\geq 2^{\frac{2}{3}}$
Therefore, optionÂ B is the right answer.

Question 10:Â The interior angles of a polygon are in Arithmetic Progression. If the smallest angle is 120Â° and common difference is 5Â°, then number of sides in the polygon is:

a)Â 7

b)Â 8

c)Â 9

d)Â None of the above

Solution:

It has been given that the interior angles in a polygon are in an arithmetic progression.
We know that the sum of all exterior angles of a polygon is 360Â°.
Exterior angle = 180Â° – interior angle.
Since we are subtracting the interior angles from a constant, the exterior angles will also be in an AP.
The starting term of the AP formed by the exterior angles will be 180Â°-120Â° = 60Â° and the common difference will be -5Â°.

Let the number of sides in the polygon be ‘n’.
=> The number of terms in the series will also be ‘n’.

We know that the sum of an AP is equal to 0.5*n*(2a + (n-1)d), where ‘a’ is the starting term and ‘d’ is the common difference.
0.5*n*(2*60Â° + (n-1)*(-5Â°)) = 360Â°
120$n$ – 5$n^2$ + 5$n$ = 720
5$n^2$ – 125$n$ + 720 = 0
$n^2$ – 25$n$ + 144 =0.
$(n-9)(n-16) = 0$

Therefore, $n$ can be 9 or 16.
If the number of sides is 16, then the largest external angle will be 60 – 15*5 = -15Â°. Therefore, we can eliminate this case.
The number of sides in the polygon must be 9. Therefore, option C is the right answer.

TakeÂ  SNAP mock tests here

Question 11:Â If three positive real numbers a, b and c (c > a) are in Harmonic Progression, then log (a + c) + log (a – 2b + c) is equal to:

a)Â 2 log (c – b)

b)Â 2 log (a – c)

c)Â 2 log (c – a)

d)Â log a + log b + log c

Solution:

It has been given that the terms $a, b,$ and $c$ are in harmonic progression.
Therefore, $\frac{1}{b} – \frac{1}{a} = \frac{1}{c} – \frac{1}{b}$
$\frac{2}{b}$ = $\frac{1}{a}$ + $\frac{1}{c}$
$\frac{2}{b} = \frac{a+c}{ac}$
$b = \frac{2ac}{(a+c)}$————–(1)
The given expression is log $(a+c)$ + log $(a-2b+c)$.
log $(a+c) + log (a – 2b + c)$ = log $((a+c)(a-2b + c))$
Substituting (1), we get,
log $(a+c)$ + log $(a – 2b + c)$ = log$((a+c)(a – \frac{4ac}{(a+c)} +c))$
= log ($a^2 + ac – 4ac + c^2 + ac$)
= log $(a^2 + c^2 – 2ac)$
= log $(c-a)^2$ [Since c is greater than a]
= 2 log $(c-a)$
Therefore, optionÂ C is the right answer.

Question 12:Â Suppose a, b and c are in Arithmetic Progression and $a^{2}, b^{2}$ and $c^{2}$ are in Geometric Progression. If $a<b<c$ and a+b+c=$\frac{3}{2}$,, then the value of a=

a)Â $\dfrac{1}{2\sqrt{2}}$

b)Â $\dfrac{1}{2\sqrt{3}}$

c)Â $\dfrac{1}{2} – \dfrac{1}{\sqrt{3}}$

d)Â $\dfrac{1}{2} – \dfrac{1}{\sqrt{2}}$

Solution:

Let us assume that the common difference of the A.P. is ‘d’.

Then, we can say that a = b – d, c = b + d

It is given that a + b + c = 3/2. i.e. b = 1/2.

It is given thatÂ $a^{2}, b^{2}$ and $c^{2}$ are in Geometric Progression. Hence, we can say that

$b^4 = a^2*c^2$

$b^4 = (b – d)^2*(b + d)^2$

$b^4 = (b^2 – d^2)^2$

$\Rightarrow$ $(b^2+d^2-b^2)(b^2-d^2+b^2)=0$

Therefore, $(b^2 – d^2 + b^2) = 0$. i.e. $d = \dfrac{1}{\sqrt{2}}$

Hence, a = b – d = $\dfrac{1}{2}$ – $\dfrac{1}{\sqrt{2}}$.

Question 13:Â If 2, a, b, c, d, e, f and 65 form an arithmetic progression, find out the value of â€˜eâ€™.

a)Â 48

b)Â 47

c)Â 41

d)Â None of the above

Solution:

Given thatÂ 2, a, b, c, d, e, f and 65 are in an AP.

65 = 2 + (8-1)d

d = 9.

Therefore, e = 2+(6-1)*9 = 2+45 = 47. Therefore, option B is the correct answer.

Question 14:Â In a list of 7 integers, one integer, denoted as x is unknown. The other six integers are 20, 4, 10, 4,8, and 4. If the mean, median, and mode of these seven integers are arranged in increasing order, they form an arithmetic progression. The sum of all possible values of x is

a)Â 26

b)Â 32

c)Â 34

d)Â 38

e)Â 40

Solution:

Integers = $4,4,4,8,10,20,x$

Clearly, irrespective of the value of $x$, Mode = $4$

Sum of above integers = $4 + 4 + 4 + 8 + 10 + 20 + x$

= $50 + x$

Mean = $\frac{50 + x}{7}$

Case 1 : If $x < 4$

Median of $x,4,4,4,8,10,20$ = 4

Mode = 4

If these are in A.P. => Mean = 4

=> $\frac{50 + x}{7} = 4$

=> $50 + x = 28$

=> $x = 28 – 50 = -22$

=> It is not possible

Case 2 : If $4 < x < 8$

Median of $4,4,4,x,8,10,20$ = $x$

Mode = 4

Mean = $\frac{50 + x}{7}$

=> $\frac{54}{7} < Mean < \frac{58}{7}$

As these are in AP => $x = 6$ and Mean = $8$

Case 3 : If $x > 8$

Mean = $\frac{50 + x}{7} > \frac{58}{7}$

Median of $4,4,4,8,x,10,20$ = $8$

Mode = $4$

As these are in AP, => Mean = $12$

=> $\frac{50 + x}{7} = 12$

=> $50 + x = 84$

=> $x = 84 – 50 = 34$

$\therefore$ Sum of all possible values of x is = $6 + 34 = 40$

Question 15:Â a,Â  b,Â  c,Â  d andÂ  e are integers such that 1 â‰¤ a < b < c < d < e. If a, b, c, d and e are geometric progression and lcm (m , n) is the least common multiple of m and n, then the maximum value of $\frac{1}{lcm(a,b)}+\frac{1}{lcm(b,c)}+\frac{1}{lcm(c,d)}+\frac{1}{lcm(d,e)}$ is

a)Â 1

b)Â 15/16

c)Â 78/81

d)Â 7/8

e)Â None of these

Solution:

Given that the numbers are in G.P.

Let the common ratio be â€˜râ€™, hence the series a,b,c,d,e can also be expressed as:

$a , ar , ar^2 , ar^3 , ar^4$

lcm(a,b) = lcm$(a,ar) = ar$

lcm(b,c) = lcm$(ar,ar^2) = ar^2$

lcm(c,d) = lcm$(ar^2,ar^3) = ar^3$

lcm(d,e) = lcm$(ar^3,ar^4) = ar^4$

$\therefore \frac{1}{lcm(a,b)}+\frac{1}{lcm(b,c)}+\frac{1}{lcm(c,d)}+\frac{1}{lcm(d,e)}$

= $\frac{1}{ar} + \frac{1}{ar^2} + \frac{1}{ar^3} + \frac{1}{ar^4}$

= $\frac{1}{a} (\frac{1}{r} + \frac{1}{r^2} + \frac{1}{r^3} + \frac{1}{r^4})$

To get max value of this, â€˜aâ€™ and â€˜râ€™ should be minimum.

It is given that $1 \leq a$ => Minimum value of â€˜aâ€™ = 1

For the values in the series to be integers, the minimum common ratio, r = 2Â Â  ($r \leq 1$ wonâ€™t work here as it is an increasing GP)

Substituting values of ‘a’ and ‘r’ in the expression, we get :

Max value = $\frac{1}{1} (\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4})$

= $\frac{8 + 4 + 2 + 1}{16} = \frac{15}{16}$

Question 16:Â If a, b, c, d, e and f are in arithmetic progression, then e-c is equal to

a)Â 2(b-a)

b)Â c-b

c)Â 2(b-f)

d)Â 2(d-b)

Solution:

a, b, c, d, e, f are in AP => They can be written as a, a+x, a+2x, a+3x, a+4x and a+5x.

e – c = a + 4x – a – 2x = 2x

b – a = a + x – a = x

=> e – c = 2(b – a)

Question 17:Â If the sum of the first 11 terms of an arithmetic progression equals that of the first 19 terms, then what is the sum of the first 30 terms?

a)Â 0

b)Â -1

c)Â 1

d)Â Not unique

Solution:

Sum of the first 11 terms = 11/2 ( 2a+10d)

Sum of the first 19 terms = 19/2 (2a+18d)

=> 22a+110d = 38a+342d => 16a = -232d

=> 2a = -232/8 d = -29d

Sum of the first 30 terms = 15(2a+29d) = 0

Question 18:Â 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?

a)Â 1st

b)Â 9th

c)Â 12th

d)Â None of the above

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
Question 19:Â If 7 times the seventh term of an Arithmetic Progression (AP) is equal to 11 times its eleventh term, then the 18th term of the AP will be

a)Â 1

b)Â 0

c)Â 2

d)Â -1

Solution:

Let the first term of the AP be $a$ and the common difference = $d$

7th term = $A_7=a+6d$

11th term = $A_{11}=a+10d$

According to ques,

=> $7 \times (a+6d)=11 \times (a+10d)$

=> $7a+42d=11a+110d$

=> $11a-7a=42d-110d$

=> $4a=-68d$

=> $a=-17d$

=> $a+17d = 0 = A_{18}$

=> Ans – (B)

Question 20:Â What is the sum of the first 13 terms of an arithmetic progression if the first term is -10 and last term is 26?

a)Â 104

b)Â 140

c)Â 84

d)Â 98

$S_{n}=\frac{n}{2}[a+l]$
$S_{13}=\frac{13}{2}[-10+26]$
$S_{13}=\frac{13}{2}[16]=104$