IIT - JEE - MAINS and ADVANCED : Sequence and Series
50 multiple-choice questions on the topic of , along with their answers and solutions:
1. The sum of the first n terms of an arithmetic sequence is given by Sn = 3n^2 - 5n. Find the common difference.
a) 2
b) 3
c) 4
d) 5
Answer: c) 4
Solution: Since Sn = 3n^2 - 5n, we can find the common difference by subtracting the sum of the first (n - 1) terms from the sum of the first n terms. This gives us Sn - Sn-1 = an, where an represents the nth term. Simplifying the expression, we get 3n^2 - 5n - [3(n-1)^2 - 5(n-1)] = 4n - 1. Equating this to an, we find that the common difference is 4.
2. Find the sum of the series: 2 + 5 + 8 + ... + 23.
a) 155
b) 171
c) 180
d) 192
Answer: b) 171
Solution: This is an arithmetic series with a first term of 2 and a common difference of 3. Using the formula for the sum of an arithmetic series, Sn = (n/2)(2a + (n-1)d), where a is the first term and d is the common difference, we can calculate the sum as Sn = (12/2)(2 + (12-1)3) = 171.
3. Find the sum of the geometric series: 1/3 + 1/9 + 1/27 + ... + 1/6561.
a) 1/2186
b) 1/4372
c) 1/6561
d) 1/8748
Answer: a) 1/2186
Solution: This is a geometric series with a first term of 1/3 and a common ratio of 1/3. The sum of a geometric series can be calculated using the formula Sn = (a(1 - r^n))/(1 - r), where a is the first term, r is the common ratio, and n is the number of terms. Plugging in the values, we get Sn = (1/3)(1 - (1/3)^8)/(1 - 1/3) = 1/2186.
4. If the sum of the first n terms of a series is given by Sn = n^2 + 2n, find the nth term of the series.
a) 2n
b) n + 1
c) 2n + 1
d) 2n + 2
Answer: c) 2n + 1
Solution: To find the nth term of the series, we need to subtract the sum of the first (n - 1) terms from the sum of the first n terms. This gives us Sn - Sn-1 = an, where an represents the nth term. Simplifying the expression, we get n^2 + 2n - [(n-1)^2 + 2(n-1)] = 2n + 1. Thus, the nth term of the series is 2n + 1.
5. Find the sum of the infinite geometric series: 4 + 1 + 1/4 + 1/16 + ...
a) 5/3
b) 4/3
c) 3/4
d) 2/3
Answer: a) 5/3
Solution: This is an infinite geometric series with a first term of 4 and a common ratio of 1/4. The sum of an infinite geometric series can be calculated using the formula S = a/(1 - r), where a is the first term and r is the common ratio. Plugging in the values, we get S = 4/(1 - 1/4) = 5/3.
6. Find the value of x for which the series 4 + x + x^2 + ... converges.
a) 0
b) 1
c) 2
d) 3
Answer: b) 1
Solution: For a geometric series to converge, the absolute value of the common ratio must be less than 1. In this case, the common ratio is x/4. Thus, we have |x/4| < 1, which simplifies to |x| < 4. The series converges for values of x within this range. Therefore, the value of x for which the series converges is 1.
7. The sum of an arithmetic series is 75. If the first term is 5 and the common difference is 2, find the number of terms in the series.
a) 10
b) 12
c) 15
d) 20
Answer: a) 10
Solution: The sum of an arithmetic series can be calculated using the formula Sn = (n/2)(2a + (n-1)d), where a is the first term, d is the common difference, and n is the number of terms. Plugging in the given values, we have 75 = (n/2)(2(5) + (n-1)(2)). Simplifying the equation, we get n^2 + n - 40 = 0, which factors as (n + 8)(n - 5) = 0. The positive solution is n = 5, so the number of terms in the series is 10.
8. Find the value of x for which the series 3 + 3x + 3x^2 + ... diverges.
a) 0
b) 1
c) 2
d) -1
Answer: c) 2
Solution: For a geometric series to diverge, the absolute value of the common ratio must be greater than or equal to 1. In this case, the common ratio is x/3. Thus, we have |x/3| ≥ 1, which simplifies to |x| ≥ 3. The series diverges for values of x outside this range. Therefore, the value of x for which the series diverges is 2.
9. Find the sum of the series: 1 + 1/2 + 1/4 + 1/8 + ... + 1/512.
a) 1/1024
b) 1/512
c) 1/256
d) 1/128
Answer: a) 1/1024
Solution: This is a geometric series with a first term of 1 and a common ratio of 1/2. Using the formula for the sum of an infinite geometric series, S = a/(1 - r), where a is the first term and r is the common ratio, we can calculate the sum as S =
1/(1 - 1/2^10) = 1/1024.
10. Find the value of x for which the series 2 - 2x + 2x^2 - ... converges.
a) 0
b) 1/2
c) 2
d) -1
Answer: b) 1/2
Solution: For a geometric series to converge, the absolute value of the common ratio must be less than 1. In this case, the common ratio is -x/2. Thus, we have |-x/2| < 1, which simplifies to |x| < 2. The series converges for values of x within this range. Therefore, the value of x for which the series converges is 1/2.
11. Find the sum of the series: 7 + 10 + 13 + ... + 46.
a) 316
b) 343
c) 370
d) 397
Answer: c) 370
Solution: This is an arithmetic series with a first term of 7 and a common difference of 3. Using the formula for the sum of an arithmetic series, Sn = (n/2)(2a + (n-1)d), where a is the first term and d is the common difference, we can calculate the sum as Sn = (20/2)(2 + (20-1)3) = 370.
12. Find the sum of the infinite geometric series: 2/3 + 1/9 + 1/27 + ...
a) 3/4
b) 2/3
c) 1/2
d) 1/3
Answer: a) 3/4
Solution: This is an infinite geometric series with a first term of 2/3 and a common ratio of 1/3. The sum of an infinite geometric series can be calculated using the formula S = a/(1 - r), where a is the first term and r is the common ratio. Plugging in the values, we get S = (2/3)/(1 - 1/3) = 3/4.
13. Find the sum of the series: 2 + 4 + 8 + ... + 512.
a) 504
b) 508
c) 512
d) 1024
Answer: d) 1024
Solution: This is a geometric series with a first term of 2 and a common ratio of 2. Using the formula for the sum of a geometric series, S = a(1 - r^n)/(1 - r), where a is the first term, r is the common ratio, and n is the number of terms, we can calculate the sum as S = 2(1 - 2^10)/(1 - 2) = 1024.
14. Find the value of x for which the series 1 - 2x + 4x^2 - ... converges.
a) 0
b) 1/2
c) 2
d) -1
Answer: b) 1/2
Solution: For a geometric series to converge, the absolute value of the common ratio must be less than 1. In this case, the common ratio is -2x. Thus, we have |-2x| < 1, which simplifies to |x| < 1/2. The series converges for values of x within this range. Therefore, the value of x for which the series converges is 1/2.
15. Find the sum of the series: 1 + 3 + 9 + ... + 6561.
a) 8742
b) 8748
c) 8754
d) 8760
Answer: b) 8748
Solution: This is a geometric series with a first term of 1 and a common ratio of 3. Using the formula for the sum of a geometric series, S = a(1 - r^n)/(1 - r), where a is the first term, r is the common ratio, and n is the number of terms, we can calculate the sum as S = 1(1 - 3^9)/(1 - 3) = 8748.
16. Find the sum of the infinite geometric series: 5 - 10 + 20 - ...
a) 15
b) 20
c) -15
d) -20
Answer: a) 15
Solution: This is an infinite geometric series with a first term of 5 and a common ratio of -2. The sum of an infinite geometric series can be calculated using the formula S = a/(1 - r), where a is the first term and r is the common ratio. Plugging in the values, we get S = 5/(1 - (-2)) = 15.
17. The sum of an arithmetic series is 126. If the first term is 4 and the common difference is 6, find the number of terms in the series.
a) 7
b) 8
c) 9
d) 10
Answer: d) 10
Solution: The sum of an arithmetic series can be calculated using the formula Sn = (n/2)(2a + (n-1)d), where a is the first term, d is the common difference, and n is the number of terms. Plugging in the given values, we have 126 = (n/2)(2(4) + (n-1)(6)). Simplifying the equation, we get n^2 + 5n - 72 = 0, which factors as (n - 8)(n + 9) = 0. The positive solution is n = 8, so the number of terms in the series is 10.
18. Find the sum of the series: 1/4 + 1/8 + 1/16 + ... + 1/256.
a) 1/512
b) 1/256
c) 1/128
d) 1/64
Answer: c) 1/128
Solution: This is a geometric series with a first term of 1/4 and a common ratio of 1/2. Using the formula for the sum of a geometric series, S = a(1 - r^n)/(1 - r), where a is the first term, r is the common ratio, and n is the number of terms, we can calculate the sum as S = (1/4)(1 - (1/2)^8)/(1 - 1/2) = 1/128.
19. Find the value of x for which the series 5 + x + x^2 + ... diverges.
a) 0
b) 1
c) 2
d) -1
Answer: b) 1
Solution: For a geometric series to diverge, the absolute value of the common
ratio must be greater than or equal to 1. In this case, the common ratio is x/5. Thus, we have |x/5| ≥ 1, which simplifies to |x| ≥ 5. The series diverges for values of x outside this range. Therefore, the value of x for which the series diverges is 1.
20. Find the sum of the series: 6 + 9 + 12 + ... + 96.
a) 660
b) 720
c) 810
d) 900
Answer: a) 660
Solution: This is an arithmetic series with a first term of 6 and a common difference of 3. Using the formula for the sum of an arithmetic series, Sn = (n/2)(2a + (n-1)d), where a is the first term and d is the common difference, we can calculate the sum as Sn = (31/2)(2 + (31-1)3) = 660.
21. Find the sum of the infinite geometric series: 1/2 + 1/4 + 1/8 + ...
a) 1
b) 2
c) 1/2
d) 1/4
Answer: b) 2
Solution: This is an infinite geometric series with a first term of 1/2 and a common ratio of 1/2. The sum of an infinite geometric series can be calculated using the formula S = a/(1 - r), where a is the first term and r is the common ratio. Plugging in the values, we get S = (1/2)/(1 - 1/2) = 2.
22. Find the value of x for which the series 3 + 4x + 5x^2 + ... converges.
a) 0
b) 1/2
c) 2
d) 3
Answer: a) 0
Solution: For a geometric series to converge, the absolute value of the common ratio must be less than 1. In this case, the common ratio is (x - 3)/x. Thus, we have |(x - 3)/x| < 1, which simplifies to |x - 3| < |x|. Since the absolute value of x is always greater than or equal to 0, the series converges for all values of x. Therefore, the value of x for which the series converges is 0.
23. Find the sum of the series: 4 + 7 + 10 + ... + 37.
a) 183
b) 196
c) 210
d) 224
Answer: b) 196
Solution: This is an arithmetic series with a first term of 4 and a common difference of 3. Using the formula for the sum of an arithmetic series, Sn = (n/2)(2a + (n-1)d), where a is the first term and d is the common difference, we can calculate the sum as Sn = (12/2)(2 + (12-1)3) = 196.
24. Find the sum of the infinite geometric series: 3 - 6 + 12 - ...
a) -3
b) -4
c) 3
d) 4
Answer: a) -3
Solution: This is an infinite geometric series with a first term of 3 and a common ratio of -2. The sum of an infinite geometric series can be calculated using the formula S = a/(1 - r), where a is the first term and r is the common ratio. Plugging in the values, we get S = 3/(1 - (-2)) = -3.
25. The sum of an arithmetic series is 90. If the first term is 6 and the common difference is 4, find the number of terms in the series.
a) 8
b) 10
c) 12
d) 14
Answer: c) 12
Solution: The sum of an arithmetic series can be calculated using the formula Sn = (n/2)(2a + (n-1)d), where a is the first term, d is the common difference, and n is the number of terms. Plugging in the given values, we have 90 = (n/2)(2(6) + (n-1)(4)). Simplifying the equation, we get n^2 + 5n - 36 = 0, which factors as (n - 4)(n + 9) = 0. The positive solution is n = 4, so the number of terms in the series is 12.
26. Find the sum of the series: 1/5 + 1/10 + 1/20 + ... + 1/640.
a) 1/320
b) 1/160
c) 1/80
d) 1/40
Answer: b) 1/160
Solution: This is a geometric series with a first term of 1/5 and a common ratio of 1/2. Using the formula for the sum of a geometric series, S = a(1 - r^n)/(1 - r), where a is the first term, r is the common ratio, and n is the number of terms, we can calculate the sum as S = (1/5)(1 - (1/2)^7)/(1 - 1/2) = 1/160.
27. Find the value of x for which the series 2 - 4x + 8x^2 - ... converges.
a) 0
b) 1/2
c) 2
d) -1
Answer: b) 1/2
Solution: For a geometric series to converge, the absolute value of the common ratio must be less than 1. In this case, the common ratio is -2x/2. Thus, we have |-2x/2| < 1, which simplifies to |x| < 1/2. The series converges for values of x within this range. Therefore, the value of x for which the series converges is 1/2.
28. Find the sum of the series: 9 + 12 + 15 + ... + 99.
a) 1080
b) 1170
c) 1260
d) 1350
Answer: c) 1260
Solution: This is an arithmetic series with a first term of 9 and a common difference of 3. Using the formula for the sum of an arithmetic series, Sn = (n/2)(2a + (n-1)d), where a is the first term and d is the common difference, we can calculate the sum as Sn = (31/2)(2 + (31-1)3) = 1260.
29. Find the sum of the infinite geometric series: 1/3 -
1/6 + 1/12 - ...
a) 1/4
b) 1/3
c) 1/6
d) 1/12
Answer: b) 1/3
Solution: This is an infinite geometric series with a first term of 1/3 and a common ratio of -1/2. The sum of an infinite geometric series can be calculated using the formula S = a/(1 - r), where a is the first term and r is the common ratio. Plugging in the values, we get S = (1/3)/(1 - (-1/2)) = 1/3.
30. Find the value of x for which the series 4 - 5x + 6x^2 - ... diverges.
a) 0
b) 1/2
c) 2
d) 3
Answer: c) 2
Solution: For a geometric series to diverge, the absolute value of the common ratio must be greater than or equal to 1. In this case, the common ratio is (6x - 5)/4. Thus, we have |(6x - 5)/4| ≥ 1, which simplifies to |6x - 5| ≥ 4. Solving this inequality, we find that x ≥ 9/6, or x ≥ 3/2. The series diverges for values of x outside this range. Therefore, the value of x for which the series diverges is 2.
31. Find the sum of the series: 3 + 6 + 12 + ... + 384.
a) 3696
b) 3912
c) 4128
d) 4344
Answer: b) 3912
Solution: This is a geometric series with a first term of 3 and a common ratio of 2. Using the formula for the sum of a geometric series, S = a(1 - r^n)/(1 - r), where a is the first term, r is the common ratio, and n is the number of terms, we can calculate the sum as S = 3(1 - 2^8)/(1 - 2) = 3912.
32. Find the sum of the infinite geometric series: 1 - 1/3 + 1/9 - ...
a) 3/4
b) 2/3
c) 1/2
d) 1/3
Answer: a) 3/4
Solution: This is an infinite geometric series with a first term of 1 and a common ratio of -1/3. The sum of an infinite geometric series can be calculated using the formula S = a/(1 - r), where a is the first term and r is the common ratio. Plugging in the values, we get S = 1/(1 - (-1/3)) = 3/4.
33. Find the value of x for which the series 2 - 6x + 18x^2 - ... converges.
a) 0
b) 1/6
c) 2
d) 3
Answer: a) 0
Solution: For a geometric series to converge, the absolute value of the common ratio must be less than 1. In this case, the common ratio is 18x/2. Thus, we have |18x/2| < 1, which simplifies to |9x| < 1. Since the absolute value of x is always greater than or equal to 0, the series converges for all values of x. Therefore, the value of x for which the series converges is 0.
34. Find the sum of the series: 6 + 10 + 14 + ... + 50.
a) 240
b) 250
c) 260
d) 270
Answer: c) 260
Solution: This is an arithmetic series with a first term of 6 and a common difference of 4. Using the formula for the sum of an arithmetic series, Sn = (n/2)(2a + (n-1)d), where a is the first term and d is the common difference, we can calculate the sum as Sn = (12/2)(2 + (12-1)4) = 260.
35. Find the sum of the infinite geometric series: 3/4 + 1/8 + 1/16 + ...
a) 1
b) 2
c) 1/2
d) 1/4
Answer: a) 1
Solution: This is an infinite geometric series with a first term of 3/4 and a common ratio of 1/2. The sum of an infinite geometric series can be calculated using the formula S = a/(1 - r), where a is the first term and r is the common ratio. Plugging in the values, we get S = (3/4)/(1 - 1/2) = 1.
36. Find the value of x for which the series 1 - 5x + 25x^2 - ... diverges.
a) 0
b) 1/5
c) 2
d) 3
Answer: a) 0
Solution: For a geometric series to diverge, the absolute value of the common ratio must be greater than or equal to 1. In this case, the common ratio is 25x/1. Thus, we have |25x/1| ≥ 1, which simplifies to |25x| ≥ 1. Solving this inequality, we find that x ≥ 1/25. The series diverges for values of x outside this range. Therefore, the value of x for which the series diverges is 0.
37. Find the sum of the series: 5 + 7 + 9 + ... + 101.
a) 1180
b) 1210
c) 1240
d) 1270
Answer: b) 1210
Solution: This is an arithmetic series with a first term of 5 and a common difference of 2. Using the formula for the sum of an arithmetic series, Sn = (n/2)(2a + (n-1)d), where a is the first term and d is the common difference, we can calculate the sum as Sn = (49/2)(2 + (49-1)2) = 1210.
38. Find the sum of the infinite geometric series: 2/5 - 4/25 + 8/125 - ...
a) 2/5
b) 4/5
c) 8/25
d) 16/125
Answer: b) 4/5
Solution: This is an infinite geometric series with a first term of 2/5 and a common ratio of -2/5. The sum of an infinite geometric series can be calculated
using the formula S = a/(1 - r), where a is the first term and r is the common ratio. Plugging in the values, we get S = (2/5)/(1 - (-2/5)) = 4/5.
39. Find the value of x for which the series 1 - 4x + 16x^2 - ... converges.
a) 0
b) 1/4
c) 2
d) 3
Answer: a) 0
Solution: For a geometric series to converge, the absolute value of the common ratio must be less than 1. In this case, the common ratio is 16x/1. Thus, we have |16x/1| < 1, which simplifies to |16x| < 1. Solving this inequality, we find that |x| < 1/16. The series converges for values of x within this range. Therefore, the value of x for which the series converges is 0.
40. Find the sum of the series: 10 + 15 + 20 + ... + 95.
a) 510
b) 560
c) 610
d) 660
Answer: c) 610
Solution: This is an arithmetic series with a first term of 10 and a common difference of 5. Using the formula for the sum of an arithmetic series, Sn = (n/2)(2a + (n-1)d), where a is the first term and d is the common difference, we can calculate the sum as Sn = (18/2)(2 + (18-1)5) = 610.
41. Find the sum of the infinite geometric series: 1/2 + 1/6 + 1/18 + ...
a) 1/4
b) 1/3
c) 2/3
d) 3/4
Answer: c) 2/3
Solution: This is an infinite geometric series with a first term of 1/2 and a common ratio of 1/3. The sum of an infinite geometric series can be calculated using the formula S = a/(1 - r), where a is the first term and r is the common ratio. Plugging in the values, we get S = (1/2)/(1 - 1/3) = 2/3.
42. Find the value of x for which the series 3 - 3x + 3x^2 - ... diverges.
a) 0
b) 1/3
c) 2
d) 3
Answer: a) 0
Solution: For a geometric series to diverge, the absolute value of the common ratio must be greater than or equal to 1. In this case, the common ratio is 3x/3. Thus, we have |3x/3| ≥ 1, which simplifies to |x| ≥ 1. The series diverges for values of x outside this range. Therefore, the value of x for which the series diverges is 0.
43. Find the sum of the series: 7 + 11 + 15 + ... + 47.
a) 310
b) 328
c) 346
d) 364
Answer: b) 328
Solution: This is an arithmetic series with a first term of 7 and a common difference of 4. Using the formula for the sum of an arithmetic series, Sn = (n/2)(2a + (n-1)d), where a is the first term and d is the common difference, we can calculate the sum as Sn = (11/2)(2 + (11-1)4) = 328.
44. Find the sum of the infinite geometric series: 3/2 - 3/4 + 3/8 - ...
a) 1/4
b) 1/2
c) 3/4
d) 1
Answer: c) 3/4
Solution: This is an infinite geometric series with a first term of 3/2 and a common ratio of -1/2. The sum of an infinite geometric series can be calculated using the formula S = a/(1 - r), where a is the first term and r is the common ratio. Plugging in the values, we get S = (3/2)/(1 - (-1/2)) = 3/4.
45. Find the value of x for which the series 1 - 3x + 9x^2 - ... converges.
a) 0
b) 1/3
c) 2
d) 3
Answer: a) 0
Solution: For a geometric series to converge, the absolute value of the common ratio must be less than 1. In this case, the common ratio is 9x/1. Thus, we have |9x/1| < 1, which simplifies to |9x| < 1. Solving this inequality, we find that |x| < 1/9. The series converges for values of x within this range. Therefore, the value of x for which the series converges is 0.
46. Find the sum of the series: 2 + 6 + 18 + ... + 486.
a) 648
b) 729
c) 810
d) 891
Answer: a) 648
Solution: This is a geometric series with a first term of 2 and a common ratio of 3. Using the formula for the sum of a geometric series, S = a(1 - r^n)/(1 - r), where a is the first term, r is the common ratio, and n is the number of terms, we can calculate the sum as S = 2(1 - 3^6)/(1 - 3) = 648.
47. Find the sum of the infinite geometric series: 1/7 + 2/49 + 4/343 + ...
a) 1/3
b) 2/3
c) 1/7
d) 2/7
Answer: b) 2/3
Solution: This is an infinite geometric series with a first term of 1/7 and a common ratio of 2/7. The sum of an infinite geometric series can be calculated using the formula S = a/(1 - r), where a is the first term and r is the common ratio. Plugging in the values, we get S = (1/7)/(1 - 2/7) = 2/3.
48. Find the value of x for which the series 5 - 10x + 20x^2 - ... diverges.
a) 0
b) 1/10
c) 2
d) 3
Answer: a) 0
Solution: For a geometric series to diverge, the absolute value of the common ratio must be greater than or equal to 1. In this case, the common ratio is 20x/5. Thus, we have |20x/5| ≥ 1, which simplifies to |4x| ≥ 1. Solving this inequality, we find that |x| ≥ 1/4. The series diverges for values of x outside this range. Therefore, the value of x for which the series diverges is 0.
49. Find the sum of the series: 3 + 5 + 7 + ... + 99.
a) 2480
b) 2550
c) 2620
d) 2690
Answer: b) 2550
Solution: This is an arithmetic series with a first term of 3 and a common difference of 2. Using the formula for the sum of an arithmetic series, Sn = (n/2)(2a + (n-1)d), where a is the first term and d is the common difference, we can calculate the sum as Sn = (49/2)(2 + (49-1)2) = 2550.
50. Find the sum of the infinite geometric series: 1 - 2 + 4 - ...
a) 2
b) 1
c) -2
d) -1
Answer: a) 2
Solution: This is an infinite geometric series with a first term of 1 and a common ratio of -2. The sum of an infinite geometric series can be calculated using the formula S = a/(1 - r), where a is the first term and r is the common ratio. Plugging in the values, we get S = 1/(1 - (-2)) = 2.
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