Algebra - Progressions - Previous Year CAT/MBA Questions

Answer: Option D

Explanation :

S_n = n + 2n² 

T_n = S_n – S_n-1 
⇒ T_n = n + 2n² – [(n-1) + 2(n-1)²]
⇒ T_n = n + 2n² – [n-1 + 2n² – 4n + 2]
⇒ T_n = n + 2n² – n  + 1 - 2n² + 4n – 2
⇒ T_n = 4n – 1

T_n is divisible by 9 when n = 7.

Hence, option (d).

Answer: Option C

Explanation :

A_n = 3 + 7 + 11 + ...

⇒ A_n = $\frac{n}{2}$[2 × 3 + (n - 1) × 4] = $\frac{n}{2}$[4n + 2] = 2n² + n

Now, $\frac{1}{25}$$\sum _{n=1}^{25}$ [2n² + n]

= $\frac{1}{25}$[2 × $\left(\frac{25\times 26\times 51}{6}\right)$ + $\left(\frac{25\times 26}{2}\right)$]

= $\frac{1}{25}$[25 × 26 × 17 + 25 × 13]

= 26 × 17 + 13 = 455

Hence, option (c).

Answer: Option C

Explanation :

On nth day ‘n’ particles produce 1 extra particle.
⇒ For every n particles on previous day, their will be (n + 1) particles next day.

∴ On nth day, the number of particles will become $\frac{n+1}{n}$ times the number of particles of previous day.

⇒ Number of particles after day 2 = 100 × $\left(\frac{3}{2}\right)$
⇒ Number of particles after day 3 = 100 × $\left(\frac{3}{2}\right)$ × $\left(\frac{4}{3}\right)$
⇒ Number of particles after day 4 = 100 × $\left(\frac{3}{2}\right)$ × $\left(\frac{4}{3}\right)$ × $\left(\frac{5}{4}\right)$

...

⇒ Number of particles after day m = 100 × $\left(\frac{3}{2}\right)$ × $\left(\frac{4}{3}\right)$ × $\left(\frac{5}{4}\right)$ × ... × $\left(\frac{m+1}{m}\right)$ = 100 × $\left(\frac{m+1}{2}\right)$

⇒ 100 × $\left(\frac{m+1}{2}\right)$ = 1000

⇒ $\left(\frac{m+1}{2}\right)$ = 10

⇒ m = 19

Hence, option (c).

Answer: 548

Explanation :

38, 55, 72, … forms an AP whose first term is 38 and common difference is 17.
∴ T_n = 38 + (n - 1) × 17

To find the average we need to find the highest and lowest 3-digit numbers of this sequence.

Lowest: 38 + (n - 1) × 17 > 99
⇒ 17n - 17 > 61
⇒ 17n > 78
⇒ n > 4
∴ Least possible value of n = 5
⇒ Least such number = 38 + 4 × 17 = 106

Highest: 38 + (n - 1) × 17 < 999
⇒ 17n - 17 < 961
⇒ 17n < 978
⇒ n < 56.5
∴ Highest possible value of n = 56
⇒ Highest such number = 38 + 56 × 17 = 990

∴ The average of the sequence (AP) is same as the average of lowest and highest terms = $\frac{106+990}{2}$ = 548

Hence, 548.

Answer: Option D

Explanation :

Given, x_n+2 = $\frac{1+x_{n+1}}{x_n}$
x₀ = 1, x₁ = 2

Substituting n = 0 we get

x₂ = $\frac{1+2}{1}$ = 3

Similarly, 
x₃ = 2
x₄ = 1

x₅ = 1
x₆ = 2
x₇ = 3
x₈ = 2
x₉ = 1
… and so on.

Here, we can see that the value of xn repeats after every 5 terms.

∴ x₂₀₂₁ = x₂₀₂₀₊₁ = x₁ = 2

Hence, option (d).

Answer: Option B

Explanation :

Here, 
1^st group has 1 integer
2^nd group has 3 integers
3^rd group has 5 integers
∴ nth groups will have (2n – 1) integers.

Total inters used till
1^st group = 1 = 1²
2^nd group = 1 + 3 = 2²
3^rd group = 1 + 3 + 5 = 9 = 3²

∴ Total integers used till n^th group = n²

∴ Sum of integers in 15^th group = Sum of all integers used till 15^th group - Sum of all integers used till14^th group

= (1 + 2 + 3 + … + 15²) - (1 + 2 + 3 + … + 14²)

= $\frac{225\times 226}{2}$ - $\frac{196\times 197}{2}$

= 25425 – 19306 = 6119

Hence, option (b).

Answer: Option A

Explanation :

Given, x₁ - x₂ + x₃ - … + (-1)⁽ⁿ⁺¹⁾ x_n = n² + 2n

Put n = 1, ⇒ x₁ = 1² + 2 × 1 = 3

Put n = 2, ⇒ x₁ - x₂ = 2² + 2 × 2 = 8 ⇒ x₂ = 3 – 8 = -5

Put n = 3, ⇒ x₁ - x₂ + x₃ = 3² + 2 × 3 = 15 ⇒ x₃ = 8 – 15 = -7
…
⇒ x_n = (-1)ⁿ⁺¹ × (2n+1)

∴ x₄₉ = 99 and x₅₀ = -101

⇒ x₄₉ + x₅₀ = 99 – 101 = -2

Hence, option (a).

Answer: Option C

Explanation :

Let the three integers x, y and z be (a - d), a, (a + d) respectively.

[(a – d), a and (a + d) are all positive integers]

Since y – x > 2, hence d > 2.

Given, xyz = 5(x + y + z) 

⇒ (a – d) × a × (a + d) = 5 × 3a

⇒ (a – d)(a + d) = 15

Here (a – d) and (a + d) are positive integers

∴ We need to write 15 as product of 2 positive integers. This can be done in two ways, 1 × 15 or 3 × 5

Hence, (a, d) is either (8, 7) or (4, 1).

Since d > 0 hence, (4, 1) is rejected.

∴ (a, d) = (8, 7)

∴ z – x = (a + d) – (a - d) = 2d = 14

Hence, option (c).

Answer: Option A

Explanation :

By substituting x = 1 or 2 or 3 and so on, we get

x₂ = x₁ + 1 – 1 = x₁ + 0
x₃ = x₂ + 2 – 1 = x₂ + 1 = x₁ + 0 + 1
x₄ = x₃ + 3 – 1 = x₃ + 2 = x₁ + 0 + 1 + 2
x₅ = x₄ + 4 – 1 = x₄ + 3 = x₁ + 0 + 1 + 2 + 3
…
x_n = x_n-1 +  0 + 1 + 2 + … + (n – 2)

⇒ x₁₀₀ = x₁ + 0 + 1 + 2 + … + 98 = x₁ + (98 × 99)/2

⇒ x₁₀₀ = -1 + 4851 = 4850

Hence, option (a).

Answer: Option D

Explanation :

If the first term is a and common ratio is r (an integer)

T_m = ¾ = ar^m-1 and

T_n = 12 = ar^n-1

Now, $\frac{T_n}{T_m}=\frac{12}{3/4}=\frac{a{r}^{n-1}}{a{r}^{m-1}}=r^{n-m}$

∴ r^(n-m) = 16

Since r is an integer, r can be either ± 2, ± 4 or 16.

If r = ± 2, n – m = 4 ⇒ least value of r + n – m = -2 + 4 = 2

If r = ± 4, n – m = 2 ⇒ least value of r + n – m = -4 + 2 = - 2

If r = 16, n – m = 1 ⇒ least value of r + n – m = 16 + 1 = 17

∴ Least possible value of r + n – m = - 2

Hence, option (d).

Answer: Option C

Explanation :

Given, x_m = x_m+1 + (m + 1)

⇒ x_m+1 = x_m - (m + 1)

Put m = 1 ⇒ x₂ = -1 – 2
Put m = 2 ⇒ x₃ = -1 - 2 – 3
Put m = 3 ⇒ x₄ = - 1 – 2 - 3 – 4

And so on.

∴ x₁₀₀ = - 1 – 2 – 3 -4 … - 100

⇒ x₁₀₀ = - (100 × 101)/2 = - 5050

Hence, option (c).

Answer: 6144

Explanation :

Eleventh term of the series (a₁₁) = S₁₁ − S₁₀  (where S_n represents the sum of n terms of the series)

S₁₁ = 3(2¹² − 2) and S₁₀ = 3(2¹¹ − 2).

∴ a₁₁ = 3(2¹² − 2) - 3(2¹¹ - 2) = 3(2¹² − 2¹¹) = 3 × 2¹¹ (2 − 1) = 6144.

Hence, 6144. 

Answer: Option C

Explanation :

From the given data, populaiton in 

2019 : p

after 1 year i.e. in 2020 : 2p + 3

after 2 years i.e. in 2021 : 2(2p + 3) + 3 = 2²p + 2×3 + 3

after 3 years i.e. in 2022 : 2(2²p + 2×3 + 3) = 2³p + 2²×3 + 2×3 + 3

...

after n years i.e. : = 2ⁿp + 2^n-1×3 + 2^n-2×3 + ... + 3

Hence, population in 2034 i.e. after 15 years = 2¹⁵p + 2¹⁴×3 + 2¹³×3 + ... + 3

= 2¹⁵p + 3(2^n-1 + 2^n-2 + ... + 1)

= 2¹⁵p + 3(2¹⁵ - 1)/(2 - 1)

= 2¹⁵ × 1000 + 3(2¹⁵ - 1)

= 2¹⁵ × 1003 - 3

Hence, option (c).

Answer: Option B

Explanation :

The best approach to solving such questions in exams is to put values and then cross checking the options.

Let n = 2, so we will have three terms in AP (a₁, a₂ and a₃). Let a₁ = a₂ = a₃ = 1.

1/(√a₁ + √a₂) = 1/2.

1/(√a₂ + √a₃) = 1/2.

∴ [1/(√a₁ + √a₂)] + [1/(√a₂ + √a₃)] = (1/2) + (1/2) = 1.

Put n = 2 in;

Option 1: 2/0 = not defined. So this option is incorrect.

Option 2: 2/(1 + 1) = 2/2 = 1. So this option is correct.

Option 3: (2 − 1)/(1 + 1) = 1/2. So this option is incorrect.

Option 4: (2 − 1)/(1 + 1) = 1/2. So this option is incorrect.

Hence, option (b).

Answer: Option B

Explanation :

Given that a₁ - a₂ + a₃ - a₄ + ........ + (-1)^n-1 a_n = n

Put n = 1 ⇒ a₁ = 1

Put n = 2 ⇒ a₁ - a₂ = 2 ⇒ a₂ = 1 - 2 = -1

Put n = 3 ⇒ a₁ - a₂ + a₃ = 3 ⇒ a₃ = 3 – 1 -1 = 1

Hence, the series proceeds as 1, -1, 1, -1, ...

i.e. odd term of the series = +1

& even terms of the series = -1

Then a₅₁ + a₅₂ + ........ + a₁₀₂₃ = 1 + (-1) + .... + 1

⇒ a₅₁ + a₅₂ + ........ + a₁₀₂₃ = 1

Hence, option (b).

Answer: Option A

Explanation :

First series - 15, 19, 23, 27, ......, 415 ⇒ Common difference = 4

Second series - 14, 19, 24, 29, ......, 464 ⇒ Common difference = 5

Common terms in both sequences = 19, 39, 59, ...., (Common difference = LCM (4,5) = 20)

Now :

19, 39, 59, ...., = (20 - 1), (40 -1), (60 -1), ...., (400-1) (There is no room for 419, as the first series ends at 415)

399 = 400 - 1 = 20 × 20 - 1

Hence, the number of common terms in the two sequences = 20

Hence, option (a).

Answer: 4851

Explanation :

Given: (2n + 1) + (2n + 3) + (2n + 5) + ... + (2n + 47) = 5280

⇒ (2n + 2n +.... + 2n) + (1 +3 + 5 +.... + 47) = 5280

Odd numbers from 1 to 47 are added in the above series.

n^th odd natural number = 2n - 1

∴ Number of terms from 1 to 47 = (47 + 1)/2 = 24 terms

∴ Given equation becomes 2n × 24 + (1 + 3 + 5 + … + 47) = 5280

Sum of first n odd natural numbers = n².

⇒ 2n × 24 + 24² = 5280

⇒ 2n × 24 = 5280 - 24²

⇒ 2n = 220 - 24

⇒ n = 98

Hence, 1 + 2 + 3 + ... + 98 = $\frac{98\times 99}{2}$ = 4851.

Hence, 4851.

Answer: 40

Explanation :

Let X and Y are the two real numbers.

X × Y × 73 − X × Y × 37 = 720

X × Y × (73 – 37) = 720

X × Y × 36 = 720

X × Y = 20

AM(X², Y²) ≥ GM(X², Y²)

∴ $\frac{X^2+Y^2}{2}$ ≥ $\sqrt{X^2\times Y^2}$

∴ (X² + Y²) ≥ 40

Alternately, 
We know, X × Y = 40

Also, (X - Y)² ≥ 0 [Square of any number is always greater than or equal to zero]

∴ X² + Y² - 2XY ≥ 0

∴ X² + Y² ≥ 2XY

∴ X² + Y² ≥ 40

Hence, 40.

Answer: Option B

Explanation :

Let the common ratio be ‘k’.
⇒ x = y/k and z = yk

Now considering the AP, 5x = 5y/k, 16y and 12z = 12yk are in arithmetic progression,
∴ 2 × 16y = 5y/k + 12yk
⇒ 32 = 5/k + 12k
⇒ 12k² – 32k + 5 = 0
⇒ (2k – 5)(6k – 1) = 0
⇒ k = 5/2 or 1/6

As x < y < z, k has to be greater than 1.
∴ k = 5/2

Hence, option (b).

Answer: 105

Explanation :

As the AM of x, y and z is 80,
⇒ x + y + z = 80 × 3 = 240   … (1)

As the AM of x, y, z, u and v is 75
⇒ x + y + z + u + v = 75 × 5 = 375   … (2)

Subtracting equation (1) from (2), we get
u + v = 135

$\therefore \frac{x+y}{2}+\frac{y+z}{2}=135$

∴ x + 2y + z = 270 … (3)

Solving equations (1) and (3), we get
⇒ y = 30

∴ x + z = 210.

Now, since x  ≥  z, the minimum value of x = 105.

Hence, 105.

Answer: 24

Explanation :

S_k = 2k² + 9k + 13

S_k-1 = 2(k-1)² + 9(k-1) + 13

⇒ T_k = S_k - S_k-1 = 2k² + 9k + 13 - [2(k-1)² + 9(k-1) + 13]

⇒ T_k = S_k - S_k-1 = 4k + 7

If t_k = 103

⇒ 4k + 7 = 103 or k = 24.

Hence, 24.

Answer: Option D

Explanation :

The first number in the terms are 7, 11, 15, …, 95, which form an AP with a = 7 and d = 4.

The n^th term for this AP is T_n = 3 + 4n and the number of terms is

n = $\frac{95-7}{4}$ + 1 = 23

The second number in the terms are 11,15,19,...,99, which form an AP with a = 11 and d = 4. The nth term for this AP is Tn= 7 + 4n and the number of terms is
n = $\frac{99-11}{4}$ + 1 = 23

Therefore the required sum
= $\sum _{n=1}^{23}(3+4n)(7+4n)$

 = $\sum _{n=1}^{23}(16n^2+40n+21)$

= $16\sum _{n=1}^{23}{n}^2+40\sum _{n=1}^{23}n+21\sum _{n=1}^{23}1$

= $16\times \frac{23\times 24\times 47}{6}+40\times \frac{23\times 24}{2}+21\times 23$

= 69184 + 11040 + 483

= 80707

Hence, option (d).

Answer: Option D

Explanation :

log₃x = a ⇒ x = 3^a.

log₁₂y = a ⇒ y = 12^a.

∴ xy = 36^a and √(xy) = G = 6^a.

∴ log₆G = a.

Hence, option (d).

Answer: Option A

Explanation :

Let the first term be a and the common difference be d.

(a + 6d)² = (a + 2d) (a + 16d)

⇒ a² + 12ad + 36d² = a² + 18ad + 32d²

⇒ 4d² = 6ad

⇒ a/d = 2/3

Hence, option (a).

Answer: Option B

Explanation :

a₁ = 3, a₂ = 7, …..
a_n = 3 + (n - 1) × 4 = 4n – 1,
a_3n = 3 + (3n – 1) × 4 = 12n - 1

a₁ + a₂ + a₃ + … + a_3n = $\frac{3\mathrm{n}}{2}(12\mathrm{n}-1+3)$ = 1830
⇒ n(6n + 1) = 610
⇒ 6n + n – 610 = 0
⇒ (6n + 61)(n - 10) = 0
⇒ n - 10 = 0

Now a₁₀ = 3 + (10 - 1) × 4 = 39

∴ a₁ + a₂ + a₃ + … + a₁₀ = 3 + 7 + … + 39 = $\frac{10}{2}(3+39)$ = 210.

210 × m > 1830

⇒ n > 1830/210 = 8.7.

The minimum integral value of m is 9.

Hence, option (b).