Algebra - Progressions - Previous Year CAT/MBA Questions
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For any natural number n, suppose the sum of the first n terms of an arithmetic progression is (n+ 2n2). If the nth term of the progression is divisible by 9, then the smallest possible value of n is
- A.
9
- B.
8
- C.
4
- D.
7
Answer: Option D
Explanation :
Sn = n + 2n2
Tn = Sn – Sn-1
⇒ Tn = n + 2n2 – [(n-1) + 2(n-1)2]
⇒ Tn = n + 2n2 – [n-1 + 2n2 – 4n + 2]
⇒ Tn = n + 2n2 – n + 1 - 2n2 + 4n – 2
⇒ Tn = 4n – 1
Tn is divisible by 9 when n = 7.
Hence, option (d).
Workspace:
Consider the arithmetic progressions 3, 7, 11, ... and let An dentoe the sum of the first n terms of this progression. Then the value of
- A.
415
- B.
404
- C.
455
- D.
442
Answer: Option C
Explanation :
An = 3 + 7 + 11 + ...
⇒ An = [2 × 3 + (n - 1) × 4] = [4n + 2] = 2n2 + n
Now, [2n2 + n]
= [2 × + ]
= [25 × 26 × 17 + 25 × 13]
= 26 × 17 + 13 = 455
Hence, option (c).
Workspace:
On day one, there are 100 particles in a laboratory experiment. On day n, where n greater than or 2, one out of every n particles produces another particle. If the total number of particles in the laboratory experiment increases to 1000 on day m, then m equals.
- A.
16
- B.
17
- C.
19
- D.
18
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 times the number of particles of previous day.
⇒ Number of particles after day 2 = 100 ×
⇒ Number of particles after day 3 = 100 × ×
⇒ Number of particles after day 4 = 100 × × ×
...
⇒ Number of particles after day m = 100 × × × × ... × = 100 ×
⇒ 100 × = 1000
⇒ = 10
⇒ m = 19
Hence, option (c).
Workspace:
The average of all 3-digit terms in the arithmetic progression 38, 55, 72, ..., is
Answer: 548
Explanation :
38, 55, 72, … forms an AP whose first term is 38 and common difference is 17.
∴ Tn = 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 = = 548
Hence, 548.
Workspace:
If x0 = 1, x1 = 2 and xn+2 = , n = 0, 1, 2, 3, …, then x2021 is equal to
- A.
1
- B.
3
- C.
4
- D.
2
Answer: Option D
Explanation :
Given, xn+2 =
x0 = 1, x1 = 2
Substituting n = 0 we get
x2 = = 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).
Workspace:
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.
4941
- B.
6119
- C.
7471
- D.
6090
Answer: Option B
Explanation :
Here,
1st group has 1 integer
2nd group has 3 integers
3rd group has 5 integers
∴ nth groups will have (2n – 1) integers.
Total inters used till
1st group = 1 = 12
2nd group = 1 + 3 = 22
3rd group = 1 + 3 + 5 = 9 = 32
∴ Total integers used till nth group = n2
∴ Sum of integers in 15th group = Sum of all integers used till 15th group - Sum of all integers used till14th group
= (1 + 2 + 3 + … + 152) - (1 + 2 + 3 + … + 142)
= -
= 25425 – 19306 = 6119
Hence, option (b).
Workspace:
For a sequence of real numbers x1, x2, …, xn, if x1 - x2 + x3 - … + (-1)(n+1) xn = n2 + 2n for all natural numbers n, then the sum x49 + x50 equals.
- A.
-2
- B.
2
- C.
-200
- D.
200
Answer: Option A
Explanation :
Given, x1 - x2 + x3 - … + (-1)(n+1) xn = 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
…
⇒ xn = (-1)n+1 × (2n+1)
∴ x49 = 99 and x50 = -101
⇒ x49 + x50 = 99 – 101 = -2
Hence, option (a).
Workspace:
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.
12
- B.
8
- C.
14
- D.
10
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).
Workspace:
Consider a sequence of real numbers x1, x2, x3, … such that xn+1 = xn + n – 1 for all n ≥ 1. If x1 = -1 then x100
- A.
4850
- B.
4849
- C.
4950
- D.
4949
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
…
xn = xn-1 + 0 + 1 + 2 + … + (n – 2)
⇒ x100 = x1 + 0 + 1 + 2 + … + 98 = x1 + (98 × 99)/2
⇒ x100 = -1 + 4851 = 4850
Hence, option (a).
Workspace:
Let the m-th and n-th terms of a geometric progression be 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
Answer: Option D
Explanation :
If the first term is a and common ratio is r (an integer)
Tm = ¾ = arm-1 and
Tn = 12 = arn-1
Now,
∴ 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).
Workspace:
If x₁ = - 1 and xm = xm+1 + (m + 1) for every positive integer m, then x100 equals
- A.
-5151
- B.
-5150
- C.
-5050
- D.
-5051
Answer: Option C
Explanation :
Given, xm = xm+1 + (m + 1)
⇒ xm+1 = xm - (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).
Workspace:
If a1 + a2 + a3 + … + an = 3(2n+1 – 2), for every n ≥ 1, then a11 equals
Answer: 6144
Explanation :
Eleventh term of the series (a11) = S11 − S10 (where Sn 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.
Workspace:
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.
(1003)215 - 3
- D.
(997)214 + 3
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 + 2n-1×3 + 2n-2×3 + ... + 3
Hence, population in 2034 i.e. after 15 years = 215p + 214×3 + 213×3 + ... + 3
= 215p + 3(2n-1 + 2n-2 + ... + 1)
= 215p + 3(215 - 1)/(2 - 1)
= 215 × 1000 + 3(215 - 1)
= 215 × 1003 - 3
Hence, option (c).
Workspace:
If a1, a2, ... are in A.P., then, + + ... + is equal to
- A.
- B.
- C.
- D.
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).
Workspace:
Let a1 , a2 be integers such that a1 - a2 + a3 - a4 + ........ + (-1)n-1 an = n , for n ≥ 1. Then a51 + a52 + ........ + a1023 equals
- A.
-1
- B.
1
- C.
0
- D.
10
Answer: Option B
Explanation :
Given that a1 - a2 + a3 - a4 + ........ + (-1)n-1 an = 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).
Workspace:
The number of common terms in the two sequences: 15, 19, 23, 27, ...... , 415 and 14, 19, 24, 29, ...... , 464 is
- A.
20
- B.
18
- C.
21
- D.
19
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).
Workspace:
If (2n+1) + (2n+3) + (2n+5) + ... + (2n+47) = 5280 , then what is the value of 1 + 2 + 3 + ... + n?
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.
nth 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 = = 4851.
Hence, 4851.
Workspace:
While multiplying three real numbers, Ashok took one of the numbers as 73 instead of 37. As a result, the product went up by 720. Then the minimum possible value of the sum of squares of the other two numbers is
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)
∴ ≥
∴ (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.
Workspace:
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.
1/6
- B.
5/2
- C.
3/6
- D.
3/2
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).
Workspace:
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
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
∴ 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.
Workspace:
Let t1, t2,… be real numbers such that t1 + t2 + … + tn = 2n2 + 9n + 13, for every positive integer n ≥ 2. If tk = 103, then k equals
Answer: 24
Explanation :
Sk = 2k2 + 9k + 13
Sk-1 = 2(k-1)2 + 9(k-1) + 13
⇒ Tk = Sk - Sk-1 = 2k2 + 9k + 13 - [2(k-1)2 + 9(k-1) + 13]
⇒ Tk = Sk - Sk-1 = 4k + 7
If tk = 103
⇒ 4k + 7 = 103 or k = 24.
Hence, 24.
Workspace:
The value of the sum 7 × 11 + 11 × 15 + 15 × 19 + ...+ 95 × 99 is
- A.
80730
- B.
80773
- C.
80751
- D.
80707
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 nth term for this AP is Tn = 3 + 4n and the number of terms is
n = + 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 = + 1 = 23
Therefore the required sum
=
=
=
=
= 69184 + 11040 + 483
= 80707
Hence, option (d).
Workspace:
Suppose, log3x = log12y = a, where x, y are positive numbers. If G is the geometric mean of x and y, and log6G is equal to:
- A.
√a
- B.
2a
- C.
a/2
- D.
a
Answer: Option D
Explanation :
log3x = a ⇒ x = 3a.
log12y = a ⇒ y = 12a.
∴ xy = 36a and √(xy) = G = 6a.
∴ log6G = a.
Hence, option (d).
Workspace:
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.
2 : 3
- B.
3 : 2
- C.
3 : 4
- D.
4 : 3
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).
Workspace:
Let a1, a2,.......a3n be an arithmetic progression with a1 = 3 and a2 = 7. If a1 + a2 + ......+ a3n = 1830, then what is the smallest positive integer m such that m(a1 + a2 + ..... + an) > 1830?
- A.
8
- B.
9
- C.
10
- D.
11
Answer: Option B
Explanation :
a1 = 3, a2 = 7, …..
an = 3 + (n - 1) × 4 = 4n – 1,
a3n = 3 + (3n – 1) × 4 = 12n - 1
a1 + a2 + a3 + … + a3n = = 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 = = 210.
210 × m > 1830
⇒ n > 1830/210 = 8.7.
The minimum integral value of m is 9.
Hence, option (b).
Workspace:
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