Class 11 Mathematics Sequences and Series MCQs Set 12

Answer: (b) 3, 11, 35, 107, 323

Explanation:
1. a₁ = 3
2. a₂ = 3(3) + 2 = 11, a₃ = 3(11) + 2 = 35
3. a₄ = 3(35) + 2 = 107, a₅ = 3(107) + 2 = 323

Teacher's Note:
1. In a recursive definition each term is found from the previous one, so work strictly in order.
2. One wrong term spoils all later terms; recheck each multiplication.

Answer: (a) 25, −125, 625, −3125, 15625

Explanation:
1. a₁ = (−1)⁰ 5² = 25, a₂ = (−1)¹ 5³ = −125
2. a₃ = (−1)² 5⁴ = 625, a₄ = (−1)³ 5⁵ = −3125
3. a₅ = (−1)⁴ 5⁶ = 15625

Teacher's Note:
1. The factor (−1)ⁿ⁻¹ makes the signs alternate, starting with +.
2. Odd-numbered terms are positive and even-numbered terms are negative here.

Answer: (a) 5ⁿ

Explanation:
1. a = 5, r = 25/5 = 5
2. aₙ = arⁿ⁻¹ = 5 × 5ⁿ⁻¹ = 5ⁿ

Teacher's Note:
1. The nth term of a GP is arⁿ⁻¹.
2. Check with n = 1: 5¹ = 5, which is the first term.

Answer: (c) 4⁵

Explanation:
1. ar² = 4
2. Product = a · ar · ar² · ar³ · ar⁴ = a⁵r¹⁰ = (ar²)⁵
3. = 4⁵

Teacher's Note:
1. For an odd number of terms of a GP, the product is (middle term) raised to the number of terms.
2. Here the middle of the first 5 terms is the 3rd term.

Answer: (c) 3072

Explanation:
1. ar² = 24 and ar⁵ = 192
2. Dividing: r³ = 8 ⇒ r = 2, so a = 24/4 = 6.
3. a₁₀ = ar⁹ = 6 × 512 = 3072

Teacher's Note:
1. Dividing two terms of a GP eliminates a and gives a power of r.
2. Verify: 6 × 2⁵ = 192.

Answer: (b) −2

Explanation:
1. ar² − a = 9 ⇒ a(r² − 1) = 9
2. ar − ar³ = 18 ⇒ −ar(r² − 1) = 18
3. Dividing: −r = 2 ⇒ r = −2

Teacher's Note:
1. Factorise each condition so that a common factor a(r² − 1) appears, then divide.
2. Check: a(4 − 1) = 9 gives a = 3; terms 3, −6, 12, −24 satisfy both conditions.

Answer: (c) 3072

Explanation:
1. a₁₂ = a₈ × r⁴
2. = 192 × 2⁴ = 192 × 16 = 3072

Teacher's Note:
1. Moving k terms forward in a GP multiplies by rᵏ.
2. There is no need to find the first term.

Answer: (a) 11th

Explanation:
1. a = 5, r = 2; 5 × 2ⁿ⁻¹ = 5120
2. 2ⁿ⁻¹ = 1024 = 2¹⁰ ⇒ n − 1 = 10 ⇒ n = 11

Teacher's Note:
1. Divide by the first term first, then express the result as a power of r.
2. Knowing powers of 2 up to 2¹⁰ = 1024 helps in such questions.

Answer: (a) 1

Explanation:
1. Read from the end: \( \frac{1}{16}, \frac{1}{8}, \frac{1}{4}, \ldots \) is a GP with first term \( \frac{1}{16} \) and ratio 2.
2. 5th term of this reversed GP = \( \frac{1}{16} \times 2^4 = 1 \)

Teacher's Note:
1. Reversing a GP gives another GP with ratio 1/r.
2. Listing the sequence fully (16, 8, 4, 2, 1, 1/2, 1/4, 1/8, 1/16) is a quick check.

Answer: (a) 393216

Explanation:
1. The sequence has 25 terms, so the 8th term from the end is the (25 − 8 + 1) = 18th term from the beginning.
2. a₁₈ = 3 × 2¹⁷ = 3 × 131072 = 393216

Teacher's Note:
1. kth term from the end of n terms = (n − k + 1)th term from the beginning.
2. 2¹⁷ = 131072.

Answer: (d) Common ratio is \( \frac{3}{4} \) or \( \frac{4}{3} \)

Explanation:
1. Take the terms as \( \frac{a}{r}, a, ar \). Product a³ = −1 ⇒ a = −1.
2. Sum: \( -\left(\frac{1}{r} + 1 + r\right) = \frac{13}{12} \Rightarrow 12r^2 + 25r + 12 = 0 \Rightarrow r = -\frac{3}{4} \text{ or } -\frac{4}{3} \).
3. For r = −3/4 the terms are 4/3, −1, 3/4; for r = −4/3 they are 3/4, −1, 4/3. So (a), (b), (c) are correct and (d) is incorrect.

Teacher's Note:
1. Taking three terms of a GP as a/r, a, ar makes the product simply a³.
2. Both values of r give the same three numbers in opposite order.

Answer: (a) 14762

Explanation:
1. a = 1/2, r = 3, n = 10
2. \( S_{10} = \frac{a(r^{10} - 1)}{r - 1} = \frac{\frac{1}{2}(59049 - 1)}{2} = \frac{59048}{4} = 14762 \)

Teacher's Note:
1. Use S = a(rⁿ − 1)/(r − 1) when r > 1 to keep the numbers positive.
2. 3¹⁰ = 59049.

Answer: (b) 2046

Explanation:
1. The numbers 2, 4, 8, … form a GP with a = 2, r = 2, n = 10.
2. \( S_{10} = \frac{2(2^{10} - 1)}{2 - 1} = 2 \times 1023 = 2046 \)

Teacher's Note:
1. Recognise doubling in each step as a GP with ratio 2.
2. 1024 is only the 10th generation, not the total of all ten.

Answer: (c) 4

Explanation:
1. \( S_n = \frac{3(3^n - 1)}{3 - 1} = 120 \Rightarrow 3^n - 1 = 80 \)
2. 3ⁿ = 81 = 3⁴ ⇒ n = 4

Teacher's Note:
1. Check directly: 3 + 9 + 27 + 81 = 120.
2. Write the number as a power of the common ratio to find n.

Answer: (d) 16

Explanation:
1. \( S_\infty = \frac{a}{1 - r} = \frac{a}{1 + \frac{4}{5}} = \frac{5a}{9} \)
2. \( \frac{5a}{9} = \frac{80}{9} \Rightarrow a = 16 \)

Teacher's Note:
1. The formula a/(1 − r) is valid only when |r| < 1, which holds here.
2. With a negative r, 1 − r becomes 1 + |r|.

Answer: (b) 3

Explanation:
1. Product = \( 3^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots} \)
2. The exponent is an infinite GP: \( \frac{1/2}{1 - 1/2} = 1 \)
3. So the product = 3¹ = 3

Teacher's Note:
1. When multiplying powers of the same base, add the exponents.
2. 1/2 + 1/4 + 1/8 + … = 1 is a standard infinite GP sum.

Answer: (a) 4

Explanation:
1. GM of a and b = \( \sqrt{ab} \)
2. \( \sqrt{2 \times 8} = \sqrt{16} = 4 \)

Teacher's Note:
1. Do not confuse GM (√ab = 4) with AM ((a + b)/2 = 5).
2. For positive numbers, GM ≤ AM; here 4 < 5.

Answer: (a) 6, 12, 24, 48

Explanation:
1. With 4 GMs, 3 and 96 are the 1st and 6th terms of a GP: 3r⁵ = 96 ⇒ r⁵ = 32 ⇒ r = 2.
2. The GMs are 6, 12, 24, 48.

Teacher's Note:
1. Inserting n GMs between a and b makes a GP of n + 2 terms, so b = arⁿ⁺¹.
2. Check that each consecutive ratio is the same (here 2).

Answer: (b) 4

Explanation:
1. Both 4ˣ and 4¹⁻ˣ are positive, so AM ≥ GM applies.
2. \( \frac{4^x + 4^{1-x}}{2} \ge \sqrt{4^x \cdot 4^{1-x}} = \sqrt{4} = 2 \)
3. So 4ˣ + 4¹⁻ˣ ≥ 4, with equality when 4ˣ = 4¹⁻ˣ, i.e. x = 1/2. Minimum value = 4.

Teacher's Note:
1. AM ≥ GM is a quick way to find the least value of a sum of two positive terms whose product is constant.
2. Equality in AM ≥ GM holds when the two terms are equal.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. a₁ = 2 − 1 + 1 = 2, a₂ = 8 − 2 + 1 = 7, so A is true.
2. a₃ = 18 − 3 + 1 = 16, a₄ = 32 − 4 + 1 = 29, so R is true.
3. R gives other terms; it does not explain A.

Teacher's Note:
1. To find a term of a sequence from its formula, simply substitute n.
2. Two correct calculations about different terms do not explain each other.

Answer: (c) A is true, but R is false.

Explanation:
1. \( a_7 = \frac{49}{2^7} = \frac{49}{128} \), so A is true.
2. \( a_{20} = \frac{20 \times 18}{23} = \frac{360}{23} \), not \( \frac{323}{22} \), so R is false.

Teacher's Note:
1. Substitute carefully in every factor of the formula.
2. For n = 20, n − 2 = 18 and n + 3 = 23.

Answer: (c) A is true, but R is false.

Explanation:
1. \( a_1 = \frac{6}{4} = \frac{3}{2} \), \( a_2 = \frac{2 \times 9}{4} = \frac{9}{2} \), \( a_3 = \frac{3 \times 14}{4} = \frac{21}{2} \). So x = 9/2 and A is true.
2. a₃ = (−1)² 5⁴ = 625, not 620, so R is false.

Teacher's Note:
1. x is the second term, so put n = 2 in the formula.
2. 5⁴ = 625; watch out for such near-miss values.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. a₁₇ = 68 − 3 = 65 and a₂₄ = 96 − 3 = 93, so A is true.
2. a₉ = (−1)⁸ × 9³ = 729, so R is true.
3. R is about a different sequence, so it does not explain A.

Teacher's Note:
1. (−1) raised to an even power is +1.
2. Substitute n directly into the given formula for each term.