CAT — Sequences and Series
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Shruti travels a distance of 224 km in four parts for a total travel time of 3 hours. Her speeds in these four parts follow an arithmetic progression, and the corresponding time taken to cover these four parts follow another arithmetic progression. If she travels at a speed of 960 meters per minute for 30 minutes to cover the first part, then the distance, in meters, she travels in the fourth part is
76800
112000
96000
86400
86400
Let the speeds (in m/min) be 960, 960+d, 960+2d, 960+3d and times (in minutes) be 30, 30+t, 30+2t, 30+3t.
Total time = 180 minutes: 30 + (30+t) + (30+2t) + (30+3t) = 180 → 120 + 6t = 180 → t = 10. So times are 30, 40, 50, 60 minutes.
Total distance = 224000 m: 28800 + 40(960+d) + 50(960+2d) + 60(960+3d) = 224000 → 172800 + 320d = 224000 → d = 160.
Fourth part: speed = 960 + 3×160 = 1440 m/min, time = 60 min. Distance = 1440 × 60 = 86400 meters.
− In an arithmetic progression, if the sum of fourth, seventh and tenth terms is 99, and the sum of the first fourteen terms is 497, then the sum of first five terms is
Step 1: Let first term = a, common difference = d.
4th term = a+3d, 7th term = a+6d, 10th term = a+9d
Step 2: First condition
(a+3d) + (a+6d) + (a+9d) = 99 → 3a + 18d = 99 → a + 6d = 33
Step 3: Sum of first 14 terms
14/2 × [2a + 13d] = 497 → 7(2a + 13d) = 497 → 2a + 13d = 71
Step 4: Solve
From a + 6d = 33 → 2a + 12d = 66
Subtract: d = 5
a + 30 = 33 → a = 3
Step 5: Sum of first five terms
= 5/2 × [2(3) + 4(5)] = 5/2 × 26 = 65
Suppose x₁, x₂, x₃, ..., x₁₀₀ are in arithmetic progression such that x₅ = −4 and 2x₆ + 2x₉ = x₁₁ + x₁₃. Then, x₁₀₀ equals
-194
-196
204
206
-194
Step 1: Let the first term be a and common difference be d.
Using x₅ = −4: a + 4d = −4
Step 2: Use the given relation
2(a + 5d) + 2(a + 8d) = (a + 10d) + (a + 12d)
4a + 26d = 2a + 22d
2a + 4d = 0 → a + 2d = 0 → a = −2d
Step 3: Find d
Substitute a = −2d into a + 4d = −4:
−2d + 4d = −4 → 2d = −4 → d = −2
Hence, a = 4
Step 4: Find x₁₀₀
x₁₀₀ = a + 99d = 4 + 99(−2) = 4 − 198 = −194
For any natural number k, let aₖ = 3ᵏ. The smallest natural number m for which {(a₁)¹ × (a₂)² × ... × (a₂₀)²⁰} < {a₂₁ × a₂₂ × ... × a₂₀₊ₘ}, is
58
59
56
57
58
LHS = (3¹)¹ × (3²)² × ... × (3²⁰)²⁰ = 3^(1² + 2² + ... + 20²) = 3^2870 (using sum of squares formula: 20×21×41/6 = 2870).
RHS = 3²¹ × 3²² × ... × 3^(20+m) = 3^(21 + 22 + ... + (20+m)) = 3^[m(m+41)/2].
Condition: m(m+41)/2 > 2870 → m(m+41) > 5740.
For m = 57: 57×98 = 5586 < 5740. For m = 58: 58×99 = 5742 > 5740. Hence smallest m = 58.
Consider the sequence t₁ = 1, t₂ = −1 and t = ((n − 3)/(n − 1)) t₋₂ for n ≥ 3. Then, the value of the sum 1/t₂ + 1/t₄ + 1/t₆ + ....... + 1/t₂₀₂₂ + 1/t₂₀₂₄, is
-1024144
-1022121
-1023132
-1026169
-1024144
Given:
t₁ = 1
t₂ = −1
and for n ≥ 3,
t = ((n − 3)/(n − 1)) t₋₂
Find
1/t₂ + 1/t₄ + 1/t₆ + ... + 1/t₂₀₂₂ + 1/t₂₀₂₄
Step 1: Find the even terms
For even indices,
t₄ = (1/3)t₂ = −1/3
t₆ = (3/5)t₄ = −1/5
t₈ = (5/7)t₆ = −1/7
The pattern is
t₂ = −1/1
t₄ = −1/3
t₆ = −1/5
t₈ = −1/7
Hence,
t₂ = −1/(2k − 1)
Step 2: Find the reciprocals
Therefore,
1/t₂ = −(2k − 1)
Step 3: Form the required sum
The last term is
t₂₀₂₄
= t₂×₁₀₁₂
So,
k = 1 to 1012.
Hence,
S = −(1 + 3 + 5 + ... + 2023)
Step 4: Sum the odd numbers
The sum of the first n odd numbers is
n²
Here,
n = 1012
Therefore,
1 + 3 + 5 + ... + 2023
= 1012²
= 1024144
Hence,
S = −1024144
Final Answer
Answer:
A. −1024144
In the set of consecutive odd numbers {1, 3, 5, ..., 57}, there is a number k such that the sum of all the elements less than k is equal to the sum of all the elements greater than k. Then, k equals
41
39
43
37
41
The set has 29 odd numbers with total sum = 29² = 841. Let k be the m-th odd number, so k = 2m−1.
Sum before k = (m−1)². Sum after k = 841 − m². Setting equal: (m−1)² = 841 − m² → 2m² − 2m − 840 = 0 → m² − m − 420 = 0 → (m−21)(m+20) = 0 → m = 21.
k = 2×21 − 1 = 41.
The amount of job that Amal, Sunil and Kamal can individually do in a day, are in harmonic progression. Kamal takes twice as much time as Amal to do the same amount of job. If Amal and Sunil work for 4 days and 9 days, respectively, Kamal needs to work for 16 days to finish the remaining job. Then the number of days Sunil will take to finish the job working alone, is
Let the amount of work done per day by Amal, Sunil and Kamal be A, S and K respectively.
Since the work rates are in Harmonic Progression (HP), their reciprocals are in Arithmetic Progression (AP).
Kamal takes twice as much time as Amal.
Therefore, K = A/2
Taking reciprocals, 1/K = 2/A
Since 1/A, 1/S, 1/K are in AP,
2/S = (1/A + 1/K) = (1/A + 2/
A) = 3/A
Hence, S = 2A/3
Now use the work done.
Amal works for 4 days. Work done = 4A
Sunil works for 9 days. Work done = 9 × (2A/3) = 6A
Kamal works for 16 days. Work done = 16 × (A/2) = 8A
Together they complete the entire job.
So, 4A + 6A + 8A = 1 → 18A = 1 → A = 1/18
Therefore, S = 2A/3 = 2/3 × 1/18 = 1/27
Thus, Sunil alone completes the work in 1 ÷ (1/27) = 27 days
Answer: 27
A container has 40 liters of milk. Then, 4 liters are removed from the container and replaced with 4 liters of water. This process of replacing 4 liters of the liquid in the container with an equal volume of water is continued repeatedly. The smallest number of times of doing this process, after which the volume of milk in the container becomes less than that of water, is
After n replacements, milk remaining = 40 × (9/10)ⁿ We need milk < water → milk < 20 → (9/10)ⁿ < 1/2 n = 6: (0.9)⁶ = 0.531441 > 0.5 → Milk ≈ 21.26 L > Water. Not yet. n = 7: (0.9)⁷ = 0.4782969 < 0.5 → Milk ≈ 19.13 L < Water ≈ 20.87 L ✓ Smallest required number of replacements = 7
In a rectangle ABCD, AB = 9 cm and BC = 6 cm. P and Q are two points on BC such that the areas of the figures ABP, APQ, and AQCD are in geometric progression. If the area of the figure AQCD is four times the area of triangle ABP, then BP : PQ : QC is
1:2:4
1:2:1
2:4:1
1:1:2
2:4:1
Step 1: Total area of rectangle
AB = 9 cm, BC = 6 cm → Area = 9 × 6 = 54 sq cm
Step 2: Determine areas using geometric progression
Let A1 = Area(ABP), A2 = Area(APQ), A3 = Area(AQCD).
A1, A2, A3 are in GP with common ratio r, and A3 = 4 × A1.
A1 × r² = 4 × A1 → r² = 4 → r = 2
So: A1 = a, A2 = 2a, A3 = 4a
Step 3: Solve for a
a + 2a + 4a = 54 → 7a = 54 → a = 54/7 sq cm
Step 4: Find segment lengths
Triangles ABP and APQ have height AB = 9 cm.
BP: Area(ABP) = (1/2) × BP × 9 → 54/7 = 4.5 × BP → BP = 12/7 cm
PQ: Area(APQ) = (1/2) × PQ × 9 → 108/7 = 4.5 × PQ → PQ = 24/7 cm
QC: BC − BP − PQ = 6 − 12/7 − 24/7 = 42/7 − 36/7 = 6/7 cm
Step 5: Ratio
BP : PQ : QC = 12/7 : 24/7 : 6/7 = 12 : 24 : 6 = 2 : 4 : 1
Let aₙ be the nᵗʰ term of a decreasing infinite geometric progression. If a₁ + a₂ + a₃ = 52 and a₁a₂ + a₂a₃ + a₃a₁ = 624, then the sum of this infinite geometric progression is
57
54
60
63
54
Step 1: Let the first term and common ratio be a and r. Then a₁ = a, a₂ = ar, a₃ = ar². Since the GP is decreasing and infinite, 0 < r < 1.
Step 2: Form the given equations. From the first condition: a + ar + ar² = 52 → a(1 + r + r²) = 52. Therefore, a = 52/(1 + r + r²). Also, a₁a₂ + a₂a₃ + a₃a₁ = 624 → a²r + a²r² + a²r³ = 624 → a²(r + r² + r³) = 624.
Step 3: Substitute the value of a. [52²/(1 + r + r²)²] × (r + r² + r³) = 624. Since r + r² + r³ = r(1 + r + r²): 52²r/(1 + r + r²) = 624. Since 52² = 2704: 2704r/(1 + r + r²) = 624. Dividing both sides by 208: 13r/(1 + r + r²) = 3.
Step 4: Solve for r. 13r = 3(1 + r + r²) → 13r = 3 + 3r + 3r² → 3r² − 10r + 3 = 0 → (3r − 1)(r − 3) = 0. So r = 1/3 or r = 3. Since the GP is decreasing, r = 1/3.
Step 5: Find the first term. a = 52/(1 + 1/3 + 1/9) = 52/(13/9) = 36.
Step 6: Find the sum of the infinite GP. Sum = a/(1 − r) = 36/(1 − 1/3) = 36/(2/3) = 54.
For some positive and distinct real numbers x, y and z, if 1/(√y + √z) is the arithmetic mean of 1/(√x + √z) and 1/(√x + √y), then the relationship which will always hold true, is
√x, √z and √y are in arithmetic progression
y, x and z are in arithmetic progression
x, y and z are in arithmetic progression
√x, √y and √z are in arithmetic progression
y, x and z are in arithmetic progression
Let √x = a, √y = b, √z = c, where a, b and c are positive and distinct.
The given condition becomes
1/(b + c) = (1/2) [1/(a + c) + 1/(a + b)]
Multiply both sides by 2, 2/(b + c) = 1/(a + c) + 1/(a + b)
Taking LCM on the RHS, 2/(b + c) = [(a + b) + (a + c)] / [(a + c)(a + b)] = (2a + b + c) / [(a + b)(a + c)]
Cross-multiply, 2(a + b)(a + c) = (b + c)(2a + b + c)
Expand both sides, 2(a² + ab + ac + bc) = 2ab + 2ac + b² + 2bc + c²
Cancel the common terms, 2a² = b² + c²
Since a² = x, b² = y, c² = z, we get 2x = y + z or y + z = 2x
Hence, y, x and z are in arithmetic progression.
Answer:
B. y, x and z are in arithmetic progression
Let both the series a₁, a₂, a₃… and b₁, b₂, b₃… be in arithmetic progression such that the common differences of both the series are prime numbers. If a₅ = b₉, a₁₉ = b₁₉ and b₂ = 0, then a₁₁ equals
86
79
83
84
79
Let aₙ = A + (n−1)d₁ and bₙ = B + (n−1)d₂ where d₁ and d₂ are prime.
b₂ = 0 → B + d₂ = 0 → B = −d₂
So bₙ = (n − 2)d₂
From a₅ = b₉: A + 4d₁ = 7d₂ ...(1)
From a₁₉ = b₁₉: A + 18d₁ = 17d₂ ...(2)
Subtract (1) from (2): 14d₁ = 10d₂ → 7d₁ = 5d₂
Since d₁ and d₂ are prime: d₁ = 5, d₂ = 7
Substitute into (1): A + 20 = 49 → A = 29
a₁₁ = A + 10d₁ = 29 + 50 = 79
Let aₙ and bₙ be two sequences such that aₙ = 13 + 6(n − 1) and bₙ = 15 + 7(n − 1) for all natural numbers n. Then, the largest three digit integer that is common to both these sequences, is
aₙ = 6n + 7 → sequence: 13, 19, 25, ... (numbers ≡ 13 mod 6 ≡ 1 mod 6) bₙ = 7n + 8 → sequence: 15, 22, 29, ... (numbers ≡ 15 mod 7 ≡ 1 mod 7) Common terms satisfy: x ≡ 1 (mod 6) and x ≡ 1 (mod 7) → x ≡ 1 (mod 42) Every common term is of the form x = 42k + 1. Largest three-digit value: 42k + 1 ≤ 999 → 42k ≤ 998 → k ≤ 23 x = 42 × 23 + 1 = 966 + 1 = 967
The sum of the infinite series (1/5)(1/5 − 1/7) + (1/5)²((1/5)² − (1/7)²) + (1/5)³((1/5)³ − (1/7)³) + …… is equal to
7/816
5/408
7/408
5/816
5/408
Given:
Find the sum of the infinite series
(1/5)(1/5 − 1/7) + (1/5)²[(1/5)² − (1/7)²] + (1/5)³[(1/5)³ − (1/7)³] + ...
Step 1: Write the general term
The nth term is
(1/5)ⁿ[(1/5)ⁿ − (1/7)ⁿ]
= (1/25)ⁿ − (1/35)ⁿ
Hence,
S = Σ[(1/25)ⁿ − (1/35)ⁿ], n = 1 to ∞
Step 2: Split into two geometric series
S
= Σ(1/25)ⁿ − Σ(1/35)ⁿ
Step 3: Use the sum of an infinite geometric series
For a geometric series,
Sum = a/(1 − r)
For the first series,
a = r = 1/25
Sum
= (1/25)/(1 − 1/25)
= 1/24
For the second series,
a = r = 1/35
Sum
= (1/35)/(1 − 1/35)
= 1/34
Step 4: Find the required sum
S
= 1/24 − 1/34
= (34 − 24)/816
= 10/816
= 5/408
Answer:
B. 5/408
A lab experiment measures the number of organisms at 8 am every day. Starting with 2 organisms on the first day, the number of organisms on any day is equal to 3 more than twice the number on the previous day. If the number of organisms on the nth day exceeds one million, then the lowest possible value of n is
Let the number of organisms on the nth day be T.
Given, T₁ = 2 and T = 2T₋₁ + 3
Add 3 to both sides. T + 3 = 2(T₋₁ + 3)
Let, U = T + 3
Then, U = 2U₋₁
Also, U₁ = 2 + 3 = 5
Hence, U = 5 × 2ⁿ⁻¹
Therefore, T = 5 × 2ⁿ⁻¹ − 3
We need T > 1,000,000
So, 5 × 2ⁿ⁻¹ − 3 > 1,000,000 → 5 × 2ⁿ⁻¹ > 1,000,003 → 2ⁿ⁻¹ > 200,000.6
Now, 2¹⁷ = 131072, 2¹⁸ = 262144
Since 200,000.6 lies between these two, the smallest possible value is n − 1 = 18
Therefore, n = 19
Answer: 19
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