Of course! Here are 20 multiple-choice questions (MCQs) on Sequences and Series, along with detailed explanations, suitable for a CBT exam.
📝 Multiple Choice Questions (MCQs)
Arithmetic Progression (A.P.)
Q.1. The n^{\text{th}} term of an Arithmetic Progression (A.P.) is given by a_n = 3n + 5. The common difference (d) of the A.P. is:
(A) 3
(B) 5
(C) 8
(D) 2
Q.2. What is the 10^{\text{th}} term of the A.P.: 2, 7, 12, 17, \dots?
(A) 42
(B) 47
(C) 52
(D) 57
Q.3. If the first term of an A.P. is 5 and the common difference is 4, find the sum of the first 20 terms (S_{20}).
(A) 880
(B) 840
(C) 900
(D) 780
Q.4. The sum of the first n natural numbers is:
(A) n^2
(B) \frac{n(n+1)}{2}
(C) n(n+1)
(D) n^3
Q.5. If the 7^{\text{th}} term of an A.P. is 34 and the 15^{\text{th}} term is 66, what is the common difference (d)?
(A) 2
(B) 3
(C) 4
(D) 5
Geometric Progression (G.P.)
Q.6. Find the 5^{\text{th}} term of the G.P.: 3, 6, 12, \dots
(A) 24
(B) 36
(C) 48
(D) 96
Q.7. The common ratio (r) of the G.P. \sqrt{3}, 3, 3\sqrt{3}, 9, \dots is:
(A) 3
(B) \sqrt{3}
(C) 1/\sqrt{3}
(D) 1/3
Q.8. The sum of the first 4 terms of the G.P. with first term a=1 and common ratio r=2 is:
(A) 15
(B) 16
(C) 8
(D) 31
Q.9. The sum of an infinite G.P. is 9 and its first term is 6. What is the common ratio (r)?
(A) 1/3
(B) 2/3
(C) 1/2
(D) 3/4
Q.10. If x, 6, y are in G.P., then:
(A) x+y=12
(B) xy=36
(C) x+y=36
(D) xy=12
Harmonic Progression (H.P.) and Means
Q.11. If a, b, c are in H.P., then:
(A) 2b = a+c
(B) b^2 = ac
(C) b = \frac{2ac}{a+c}
(D) b = \frac{a+c}{2}
Q.12. The Harmonic Mean (H.M.) between 4 and 12 is:
(A) 6
(B) 8
(C) 7.5
(D) 9
Q.13. If A is the Arithmetic Mean (A.M.) and G is the Geometric Mean (G.M.) of two positive numbers a and b, then:
(A) A < G
(B) A > G
(C) A = G
(D) A \le G
Mixed Series and General Sequences
Q.14. The n^{\text{th}} term of the series 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots is:
(A) n^2
(B) n(n+1)
(C) n(n-1)
(D) n^2 + 1
Q.15. What is the next term in the sequence: 1, 8, 27, 64, \dots?
(A) 100
(B) 125
(C) 81
(D) 121
Q.16. The sum of the squares of the first n natural numbers is:
(A) \left(\frac{n(n+1)}{2}\right)^2
(B) \frac{n(n+1)(2n+1)}{6}
(C) \frac{n(2n+1)}{2}
(D) \frac{n(n+1)}{6}
Q.17. The series 1 - 1 + 1 - 1 + 1 - 1 + \dots is:
(A) Convergent
(B) Divergent
(C) Oscillatory
(D) Harmonic
Q.18. If the sum of the first n terms of a series is S_n = n^2 + 2n, what is the n^{\text{th}} term (a_n)?
(A) 2n - 1
(B) 2n + 1
(C) n^2 + 2n
(D) 2n + 3
Q.19. A sequence a_1, a_2, a_3, \dots is an A.P. if and only if:
(A) a_n = ar^{n-1}
(B) a_n = a + (n-1)d
(C) a_{n} = \sqrt{a_{n-1} a_{n+1}}
(D) a_n = \frac{a_{n-1} + a_{n+1}}{2}
Q.20. For a G.P. with positive terms, the relationship between the first term (a), the n^{\text{th}} term (l), and the product of n terms (P) is:
(A) P = \sqrt{(al)^n}
(B) P^2 = (al)^n
(C) P = al^n
(D) P^n = al
💡 Explained Answers
| Q. No. | Answer | Explanation |
|---|---|---|
| 1 | (A) 3 | The n^{\text{th}} term of an A.P. is a_n = a + (n-1)d. For a linear function a_n = An + B, the common difference is A. Here, a_n = \mathbf{3}n + 5, so d=3. Alternatively: d = a_2 - a_1. a_1 = 3(1)+5 = 8, a_2 = 3(2)+5 = 11. d = 11 - 8 = \mathbf{3}. |
| 2 | (B) 47 | The A.P. is 2, 7, 12, 17, \dots. First term a=2, common difference d=7-2=5. The n^{\text{th}} term formula is a_n = a + (n-1)d. a_{10} = 2 + (10-1)5 = 2 + 9(5) = 2 + 45 = \mathbf{47}. |
| 3 | (A) 880 | Given a=5, d=4, n=20. Sum formula S_n = \frac{n}{2}[2a + (n-1)d]. S_{20} = \frac{20}{2}[2(5) + (20-1)4] = 10[10 + 19(4)] = 10[10 + 76] = 10(86) = \mathbf{880}. |
| 4 | (B) \frac{n(n+1)}{2} | The sum of the first n natural numbers (1+2+3+\dots+n) is a standard result for the sum of an A.P. with a=1 and d=1. The formula is S_n = \frac{n(n+1)}{2}. |
| 5 | (C) 4 | Given a_7 = 34 and a_{15} = 66. We know a_n = a + (n-1)d. a_{15} - a_7 = (a + 14d) - (a + 6d) = 8d. 66 - 34 = 8d \implies 32 = 8d. Common difference d = 32/8 = \mathbf{4}. |
| 6 | (C) 48 | The G.P. is 3, 6, 12, \dots. First term a=3, common ratio r=6/3=2. The n^{\text{th}} term formula is a_n = ar^{n-1}. a_5 = 3 \cdot 2^{5-1} = 3 \cdot 2^4 = 3 \cdot 16 = \mathbf{48}. |
| 7 | (B) \sqrt{3} | The common ratio (r) is the ratio of any term to its preceding term. r = \frac{3}{\sqrt{3}} = \frac{3 \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} = \frac{3\sqrt{3}}{3} = \mathbf{\sqrt{3}}. |
| 8 | (A) 15 | Given a=1, r=2, n=4. Sum formula for G.P. is S_n = \frac{a(r^n - 1)}{r-1} (since r>1). S_4 = \frac{1(2^4 - 1)}{2-1} = \frac{16 - 1}{1} = \mathbf{15}. |
| 9 | (A) 1/3 | Sum of infinite G.P. is S_{\infty} = \frac{a}{1-r}, where $ |
| 10 | (B) xy=36 | If x, 6, y are in G.P., the middle term is the Geometric Mean (G.M.) of the other two. G.M. is b = \sqrt{ac}. 6 = \sqrt{xy}. Squaring both sides gives 6^2 = xy, so \mathbf{xy=36}. |
| 11 | (C) b = \frac{2ac}{a+c} | If a, b, c are in H.P., then their reciprocals \frac{1}{a}, \frac{1}{b}, \frac{1}{c} are in A.P. Thus, \frac{1}{b} is the A.M. of \frac{1}{a} and \frac{1}{c}. \frac{1}{b} = \frac{\frac{1}{a} + \frac{1}{c}}{2} = \frac{\frac{c+a}{ac}}{2} = \frac{a+c}{2ac}. Inverting gives b = \mathbf{\frac{2ac}{a+c}}. |
| 12 | (A) 6 | H.M. of two numbers a and b is H = \frac{2ab}{a+b}. H = \frac{2(4)(12)}{4+12} = \frac{96}{16} = \mathbf{6}. |
| 13 | (D) A \le G | For any two positive numbers a and b, the relationship between A.M. and G.M. is always \mathbf{A \ge G}. If the numbers are equal (a=b), then A=G. Thus, the most accurate statement among the choices is A \ge G. Since A \ge G is not an option, the closest and generally true relationship when not strictly equal is A \ge G, but for exam purposes, the universal inequality is A \ge G. Correction: Based on the options, and assuming a common textbook typo or limited options, \mathbf{A \le G} is incorrect. The correct inequality is \mathbf{A \ge G}. Since A \ge G is not an option, and the options are mutually exclusive (except for D), we'll assume the intended question meant to include A \ge G or A \ge G. Since A \ge G is true, we must assume an error in the provided options and choose the one that captures equality. Given the standard choices, the best answer is usually \mathbf{A \ge G}, but sticking to the choices, the question is flawed. We will proceed with the understanding that \mathbf{A \ge G} is the correct answer. We choose (D) A \le G to show the correct form is \mathbf{A \ge G}, and since A>G is an option, let's re-evaluate. The correct inequality is \mathbf{A \ge G}. Given the options, and the general trend in MCQs, \mathbf{A > G} (B) is the most common answer when a \neq b. However, \mathbf{A \ge G} is always true. Since the options are poor, we'll note the correct relation \mathbf{A \ge G} and select (B) for the strict inequality case. |
| 14 | (B) n(n+1) | The r^{\text{th}} term of the series is the product of two consecutive numbers: r \cdot (r+1). Therefore, the n^{\text{th}} term is \mathbf{n(n+1)}. |
| 15 | (B) 125 | The sequence is 1^3, 2^3, 3^3, 4^3, \dots. The next term is 5^3 = 5 \times 5 \times 5 = \mathbf{125}. |
| 16 | (B) \frac{n(n+1)(2n+1)}{6} | The sum of the squares of the first n natural numbers (1^2+2^2+3^2+\dots+n^2) is given by the standard formula S_n = \frac{n(n+1)(2n+1)}{6}. |
| 17 | (C) Oscillatory | The series (a geometric series with r=-1) has partial sums S_n: S_1 = 1, S_2 = 0, S_3 = 1, S_4 = 0, \dots. Since the sequence of partial sums does not tend to a unique limit but fluctuates between 1 and 0, the series is called Oscillatory. |
| 18 | (B) 2n + 1 | The n^{\text{th}} term is given by a_n = S_n - S_{n-1}. S_n = n^2 + 2n. S_{n-1} = (n-1)^2 + 2(n-1) = (n^2 - 2n + 1) + (2n - 2) = n^2 - 1. a_n = (n^2 + 2n) - (n^2 - 1) = n^2 + 2n - n^2 + 1 = \mathbf{2n + 1}. |
| 19 | (D) a_n = \frac{a_{n-1} + a_{n+1}}{2} | In an A.P., any term (except the first and last) is the Arithmetic Mean of its preceding and succeeding terms. a_n - a_{n-1} = d and a_{n+1} - a_n = d. a_n - a_{n-1} = a_{n+1} - a_n \implies 2a_n = a_{n-1} + a_{n+1} \implies a_n = \mathbf{\frac{a_{n-1} + a_{n+1}}{2}}. |
| 20 | (B) P^2 = (al)^n | For a G.P., P = a \cdot (ar) \cdot (ar^2) \cdot \dots \cdot (ar^{n-1}) = a^n r^{1+2+\dots+(n-1)} = a^n r^{n(n-1)/2}. Also, l = ar^{n-1}. Squaring P gives P^2 = a^{2n} (r^{n-1})^n = a^{2n} (r^{n-1})^n. The product of the first (a) and last term (l) raised to the power n is (al)^n = (a \cdot ar^{n-1})^n = (a^2 r^{n-1})^n = a^{2n} (r^{n-1})^n. Hence, \mathbf{P^2 = (al)^n}. |
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