Progressions CAT Previous Year Questions with Answer PDF

x is the Arithmetic Mean of z & y, therefore, z, x, y form an A.P.
It goes without saying that y, x, z also forms an A.P.

Let the common differences of the series a₁, a₂, a₃, … and b₁, b₂, b₃ be p and q respectively.
We are told that p and q are prime numbers.

We have a_n = 46 + 8n and b_m = 98 + 4m where n, m are positive integers ≤ 100.

Let's find the smallest possible values of m, n where a_n = b_m

46 + 8n = 98 + 4m

8n - 4m = 52

2n - m = 13

n = 7, m = 1 are smallest values of m, n where a_n = b_m.

a₇ = b₁ = 102

Therefore, the common elements in the series will be of the form 102+8k .

a₁₀₀ = 846; b₁₀₀ = 498

The largest natural number less than or equal to 498 and of the form 102 + 8k is 494.

So the common terms of both the series are

102 , 110, 118, ..., 494

102, 102 + 8, 102 + 8(2), ..., 102 + 8(49)

So, there are 50 terms in this A.P

Sum of all the terms = 

Let the meeting point of the two trains between stations X and Y be M.
Let the time taken to reach M from X by train A be ‘t’ minutes. Since the two trains start simultaneously, the time taken by train B to reach M from Y will also be ‘t’ minutes.
We know that train A completes the entire journey in 10 minutes, so the time taken by train A to travel from M to Y will be ‘10 - t’ minutes. We are told that train B takes 9 minutes to reach X after it meets train A, which means, train B takes 9 minutes to travel from M to X.
When two bodies travel equal distances at constant speeds, the ratio of time taken by them to travel those distances will be the same.
The time taken by A and B to travel the distance between A and M is ‘t’ minutes and 9 minutes respectively. Similarly, the time taken by them to travel the distance between M and B is ‘10 - t’ and ‘t’ minutes.
The ratio of the times taken should be equal.

t = - 15 or t = 6
t can’t be negative, therefore, t is 6.
The total time taken by train B to reach station X from station Y is 6 + 9 =
15 minutes.

Let S_n stand for the sum of the first n terms of an arithmetic progression.

n, S_n = (n + 2n²)

Give Let T_n stand for the n^th term of the same arithmetic progression.

The smallest value of n for which 4n - 1 is a multiple of 9 is 7.
n = 7

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 ax² + 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 (15)² − 2(15) + 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 15^th group will be 197 + 29 - 1 = 225.
Sum of terms in group 15
= 

Given x, y, z are three terms in an arithmetic progression.
Considering x = a, y = a + d, z = a + 2 x d.
Using the given equation x . y x z = 5 x (x + y + z)
 a(a+d) x (a+2 x d) = 5(a + a + d + a + 2 x d)
= a(a + d) x (a + 2 x d)  = 5(3 x a + 3 x d) = 15(a+d).
= a(a + 2 x d) = 15.
Since all x, y, z are positive integers and y - x > 2. a, a + d, a + 2d 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 x d = 15.
x = 1, z = 15.
z - x = 14.

x₁ = −1
 
......
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(n) = a + bn + cn²
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(100) = (−1.5)100 + (0.5)100²  = −150 + 5000 = 4850 

T_n = 12
Tm = 3 / 4


To get the minimum value for r + n - m, r should be minimum. 
∴ r = - 4
n - m = 2
∴ Required answer =-2 

Let the number of large shirts be l and the number of small shirts be s.
Let the price of a small shirt be x and that of a large shirt be x + 50.
Now, s + l = 64l (x + 50) = 5000
sx = 1800
Adding them, we get,
lx + sx + 50l = 6800
64x + 50l = 6800
Substituting l = (6800 - 64x) / 50, in the original equation, we get

So, x = 75
x + 50 = 125
Answer = 75 + 125 = 200.

Initially number of matches = 40 
Now matches won = 12 
Now let remaining matches be x 
Now number of matches won = 0.6x
Now as per the condition :

24 +1.2x = 40 + x
0.2x = 16
x = 80
Now when they won 90% of remaining = 80(0.9) =72
So total won = 84

Let us solve this question by assuming values(multiples of 100) and not variables(x).
Since we know that the female population was twice the male population in 1990, let us assume their respective values as 200 and 100.
Note that while assuming numbers, some of the population values might come out as a fraction(which is not possible, since the population needs to be a natural number). However, this would not affect our answer, since the calculations are in ratios and percentages and not real values of the population in any given year.

Now, we know that the female population became 1.25 times itself in 1990 from what it was in 1980.
Hence, the female population in 1980 = 200/1.25 = 160
Also, the female population became 1.2 times itself in 1980 from what it was in 1970.
Hence, the female population in 1970 = 160/1.2 = 1600/12 = 400/3
Let the male population in 1970 be x. Hence, the male population in 1980 is 1.4x.
Now, the total population in 1980 = 1.25 times the total population in 1970.
Hence, 1.25 (x + 400/3) = 1.4x + 160
Hence, x = 400/9.
Population change = 300 - 400/9 - 400/3 = 300 - 1600/9 = 1100/9
Percentage change
= 

Let the alloy contain x Kg silver and y kg copper 
Now when mixed with 3Kg Pure silver 
we get 
we get 10x + 30 = 9x + 9y + 27
9y - x = 3    (1)
Now as per condition 2
silver in 2^nd alloy = 2(0.9) = 1.8
So we get 
we get 21y -4x =3   (2)
solving (1) and (2) we get y= 0.6 and x = 2.4
so x + y = 3

From the given data,

Simplifying,

P increases in the series of 2⁰P ,2¹P, 2²P, 2³P,... and so on
2034 is 15 years after 2019. So, the first part of the value would be 2¹⁵ p
Now, let us observe the second pattern.
In 2021, the value increases to 2(2p + 3) + 3
On expanding, (2 x 2) +(2 x 3) + 3 = 2 x 2 + 3(1 + 2)
In 2022, the value increases to 2[2(2p+3) +3]+3
On expanding, 2((2 x 2p) + (2 x 3) + 3) + 3
= (2 x 2 x 2 x p) + (2x 2 x 3) + (2 x 3) + 3
= (2 x 2 x 2 x p) + 3(2⁰ + 2¹ + 2²)
In 2034, the value of the second part would be 3 (2⁰ + 2¹ + 2² +.... +2¹⁴ )
On simplifying using Geometric progression we get, 3 () = 3 (2¹⁵ - 1)
Total population = 2¹⁵p +3 (2¹⁵ - 1)
Substituting p = 1000,
Total population = (2¹⁵ x 1003) - 3

Consider the first term 
Multiply numerator and denominator by - 
We get. We know, a2 - a1 = d
Similarly, repeat the same for the entire series.
We get+ ..... + = (- )
We do not know the value of d.
We know a_n+1 = a₁ + nd
a_n+1 - a₁ = nd
d =  
 =  
So, our expression = = 

Given that the square of the 7^th term of an arithmetic progression with positive common difference equals the product of the 3^rd and 17^th terms
(a + 6d)² = (a + 2d)(a + 16d)
We have to find the ratio of the first term to the common difference
⇒ (a + 6d)² = (a + 2d)(a + 16d)
⇒ a² + 12d + 36 = a² + 2ad + 16ad + 32d²
⇒  4d² + 12 ad – 18 ad = 0
⇒ 4d² = 6ad
⇒ 4d = 6a

The ratio of first term to the common difference is 2 : 3.

Let us assume 3n = k
⇒ k/2 (2a + (k – 1)d) = 1830 we know that a = 3, d = 4
⇒ k/2 (2(3) + (k – 1)4) = 1830
⇒ k/2 (6 + 4k – 4) = 1830
⇒ k (2k + 1) = 1830
⇒ 2k² + k = 1830
⇒ 2k² + k – 1830 = 0
By factorizing we can find that k = 30, n = 10
⇒ 10/2 (2(3) + 4(9)) = 5(6 + 36)
⇒ 5(42) = 210
⇒ m (a₁ + a₂ + ..... + a_n) > 1830
⇒ 210 × m > 1830
⇒ m = 9, since 210 × 9 = 1890

The sum up to infinity, a₁ + a₂ + a₃ +...... = 32.
It can also be written as 
We have been given that, a_n = 3(a_n+1 + a_n+2 +....) if n ≥ 1.
When n = 1, we get,
a = 3(a₂ + a₃ + .....)

⇒ 1 – r = 3r
⇒ 1 = 4r
⇒  r = 1/4
Now finding the value of a is easy.
a = 32 × (1 - (1/4))
⇒ a = 32 × (3/4)
⇒ a = 24
The value of a₅ = ar⁴ , where a = 24 and r⁴ = (1/4)⁴

Given that 
We have to find a₁ + a₂ + a₃ + ...... + a₁₀₀
This can be done by partial fractions

a₁ = 3 - 1 = 2 , a₂ = 6 - 1 = 5 , a₃ = 9 – 1 = 8.

But these are not of the form  Hence these are written as,

Hence this can be written as,