CBSE Class 11 Mathematics Sequences and Series Competitive Exam MCQs Set 01

Answer: (c) \( -\frac{27}{2} \)

Explanation:
1. For a GP, (2x + 2)² = x(3x + 3), i.e. 4(x + 1)² = 3x(x + 1).
2. x = −1 makes the terms 0, which is not allowed, so divide by (x + 1): 4(x + 1) = 3x, giving x = −4.
3. The terms are −4, −6, −9, so the common ratio is \( \frac{3}{2} \).
4. 4^th term \( = -9 \times \frac{3}{2} = -\frac{27}{2} \).

Teacher's Note:
1. Always reject a value that makes a term zero, because a GP cannot contain 0.
2. Use the middle-term test b² = ac for three consecutive terms instead of finding r twice.

Answer: (c) 4

Explanation:
1. \( r = \frac{n^n}{n^{-4}} = n^{n+4} \)
2. 8^th term \( = ar^7 = n^{-4} \cdot n^{7(n+4)} = n^{7n+24} \)
3. \( n^{7n+24} = n^{52} \) ⇒ 7n + 24 = 52 ⇒ n = 4

Teacher's Note:
1. When terms are powers of the same base, compare exponents after writing every term with that base.
2. The exponents of a GP of powers of n themselves form an AP, which is a quick check: −4, n, …, 52.

Answer: (a) 12 or \( \frac{108}{7} \)

Explanation:
1. \( \frac{a_4}{a_6} = \frac{1}{r^2} = \frac{1}{4} \) ⇒ r = ±2
2. a₂ + a₅ = ar + ar⁴ = 216
3. r = 2: 2a + 16a = 18a = 216 ⇒ a = 12
4. r = −2: −2a + 16a = 14a = 216 ⇒ \( a = \frac{108}{7} \)

Teacher's Note:
1. r² = 4 gives two values of r, and both must be followed through.
2. Writing every term as a power of r times a turns the conditions into simple linear equations in a.

Answer: (c) r > 3 or r < 1

Explanation:
1. a₁r² > 4a₁r − 3a₁; divide by a₁ > 0.
2. r² − 4r + 3 > 0 ⇒ (r − 1)(r − 3) > 0
3. Hence r < 1 or r > 3.

Teacher's Note:
1. Dividing an inequality by a₁ keeps the sign only because a₁ is positive.
2. A product (r − 1)(r − 3) is positive outside the roots and negative between them.

Answer: (b) (ab)ⁿ

Explanation:
1. P = a · ar · ar² … arⁿ⁻¹ = \( a^n r^{\frac{n(n-1)}{2}} \)
2. \( P^2 = a^{2n} r^{n(n-1)} = \left(a \cdot ar^{n-1}\right)^n \)
3. Since b = arⁿ⁻¹, P² = (ab)ⁿ.

Teacher's Note:
1. Pair the first and last terms, the second and second-last and so on; every pair has product ab.
2. With n terms there are n such pairings when the product is squared, so P² = (ab)ⁿ.

Answer: (c) 3

Explanation:
1. Let the AP have first term a and common difference d ≠ 0; the terms are a + d, a + 2d, a + 5d.
2. (a + 2d)² = (a + d)(a + 5d) ⇒ 4ad + 4d² = 6ad + 5d² ⇒ d = −2a
3. The terms become −a, −3a, −9a.
4. Common ratio = \( \frac{-3a}{-a} = 3 \)

Teacher's Note:
1. Write the AP terms in a and d, apply b² = ac, and cancel d because it is non-zero.
2. Once d is known in terms of a, the common ratio comes out as a pure number.

Answer: (b) q² = pr

Explanation:
1. p = AR³, q = AR⁶, r = AR⁹
2. pr = A²R¹² = (AR⁶)² = q²

Teacher's Note:
1. Terms taken at equal gaps from a GP (here every third term) again form a GP.
2. So the 4th, 7th and 10th terms satisfy the middle-term rule q² = pr.

Answer: (b) 1

Explanation:
1. From the AP: y − z = (q − r)d, z − x = (r − p)d, x − y = (p − q)d.
2. From the GP: x = AR^p−1, y = AR^q−1, z = AR^r−1.
3. Power of A: (y − z) + (z − x) + (x − y) = 0.
4. Power of R: d[(p − 1)(q − r) + (q − 1)(r − p) + (r − 1)(p − q)] = 0, so the expression = A⁰R⁰ = 1.

Teacher's Note:
1. Expressions of the form (p − 1)(q − r) + (q − 1)(r − p) + (r − 1)(p − q) always simplify to 0.
2. Separate the base A and the ratio R and show each total exponent is zero.

Answer: (a) 1 if n is odd and 0 when n is even

Explanation:
1. This is a GP with a = 1, r = −1, so \( S_n = \frac{1 - (-1)^n}{1 - (-1)} = \frac{1 - (-1)^n}{2} \).
2. n odd: \( S_n = \frac{1 + 1}{2} = 1 \); n even: \( S_n = \frac{1 - 1}{2} = 0 \).

Teacher's Note:
1. For r = −1 the partial sums keep switching between two values, so the series has no sum to infinity.
2. Check small cases: S₁ = 1, S₂ = 0, S₃ = 1 confirm the pattern.

Answer: (a) 1

Explanation:
1. \( S_n = \frac{a(r^n - 1)}{r - 1} = \frac{ar^n - a}{r - 1} = \frac{l \cdot r - a}{r - 1} \), where l = arⁿ⁻¹ = 128.
2. \( 255 = \frac{128 \times 2 - a}{2 - 1} = 256 - a \)
3. ∴ a = 1

Teacher's Note:
1. When the last term l is known, use Sₙ = (lr − a)/(r − 1) directly.
2. Check: 1 + 2 + 4 + … + 128 has 8 terms and sum 255.

Answer: (c) 2

Explanation:
1. \( \frac{S_6}{S_3} = \frac{r^6 - 1}{r^3 - 1} = r^3 + 1 \)
2. r³ + 1 = 9 ⇒ r³ = 8 ⇒ r = 2

Teacher's Note:
1. r⁶ − 1 factorises as (r³ − 1)(r³ + 1), which makes the ratio of sums very simple.
2. r = 1 is not possible here because then S₆ would be only twice S₃.

Answer: (c) 4

Explanation:
1. Let the odd-placed terms be a, ar², ar⁴, … with sum S₁.
2. The even-placed terms are ar, ar³, … with sum rS₁.
3. Total = S₁(1 + r) = 5S₁ ⇒ r = 4

Teacher's Note:
1. With an even number of terms, every odd-placed term has a partner just after it that is r times larger.
2. So the sum of even-placed terms is always r times the sum of odd-placed terms.

Answer: (c) \( \frac{S - 1}{S} \)

Explanation:
1. \( S = \frac{1}{1 - a} \)
2. \( 1 - a = \frac{1}{S} \) ⇒ \( a = 1 - \frac{1}{S} = \frac{S - 1}{S} \)

Teacher's Note:
1. The infinite sum formula a/(1 − r) needs |r| < 1; here the first term is 1 and the ratio is a.
2. Rearranging to make a the subject is just simple algebra once the formula is written.

Answer: (b) 3

Explanation:
1. Product = \( 9^{\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots} \)
2. The exponent is an infinite GP: \( \frac{1/3}{1 - 1/3} = \frac{1}{2} \)
3. Value = \( 9^{1/2} = 3 \)

Teacher's Note:
1. A product of powers of the same base becomes a sum in the exponent.
2. Here the exponents form an infinite GP with ratio 1/3, whose sum is 1/2.

Answer: (d) \( -\frac{1}{2} \)

Explanation:
1. \( \frac{a}{1 - r} = 4 \) and \( \frac{a^3}{1 - r^3} = 192 \)
2. Cubing the first and dividing: \( \frac{(1 - r)^2}{1 + r + r^2} = 3 \)
3. 1 − 2r + r² = 3 + 3r + 3r² ⇒ 2r² + 5r + 2 = 0 ⇒ \( r = -\frac{1}{2} \) or r = −2
4. Since |r| < 1, \( r = -\frac{1}{2} \) (then a = 6, and \( \frac{216}{9/8} = 192 \) checks).

Teacher's Note:
1. The cubes of a GP form a GP with first term a³ and ratio r³.
2. Reject any root with |r| ≥ 1, since the infinite sum would not exist.

Answer: (b) \( \frac{108}{13} \)

Explanation:
1. \( \frac{a}{1 - r} = 3 \), \( \frac{a^2}{1 - r^2} = \frac{9}{2} \)
2. Dividing the square of the first by the second: \( \frac{1 + r}{1 - r} = 2 \) ⇒ \( r = \frac{1}{3} \), a = 2
3. Sum of cubes = \( \frac{a^3}{1 - r^3} = \frac{8}{1 - \frac{1}{27}} = \frac{8 \times 27}{26} = \frac{108}{13} \)

Teacher's Note:
1. The squares form a GP with ratio r² and the cubes a GP with ratio r³.
2. Dividing the two given equations eliminates a and leaves a simple equation in r.

Answer: (d) \( \frac{3}{4} \)

Explanation:
1. For the first term: \( a = 2 \cdot \frac{ar}{1 - r} \) ⇒ 1 − r = 2r ⇒ \( r = \frac{1}{3} \)
2. a + ar = 1 ⇒ \( a \cdot \frac{4}{3} = 1 \) ⇒ \( a = \frac{3}{4} \)

Teacher's Note:
1. The terms after the first form another infinite GP starting at ar.
2. The same condition then holds for every term, so using the first term is enough.

Answer: (b) \( \frac{423}{999} \)

Explanation:
1. 0.423423… = 0.423 + 0.000423 + … , an infinite GP with a = 0.423 and r = 0.001
2. Sum = \( \frac{0.423}{1 - 0.001} = \frac{0.423}{0.999} = \frac{423}{999} \)

Teacher's Note:
1. A recurring block of k digits becomes that block over k nines.
2. The fraction may be simplified further: 423/999 = 47/111.

Answer: (b) G.P.

Explanation:
1. Let b = ar, c = ar², d = ar³.
2. aⁿ + bⁿ = aⁿ(1 + rⁿ), bⁿ + cⁿ = aⁿrⁿ(1 + rⁿ), cⁿ + dⁿ = aⁿr²ⁿ(1 + rⁿ)
3. Each term is rⁿ times the previous one, so they are in GP.

Teacher's Note:
1. Write every term using a and r; a common factor such as (1 + rⁿ) then shows the pattern.
2. Powers of GP terms form a GP, and sums of corresponding terms keep the same ratio.

Answer: (b) G.P.

Explanation:
1. With b = ar, c = ar², d = ar³: a³ + b³ = a³(1 + r³), b³ + c³ = a³r³(1 + r³), c³ + d³ = a³r⁶(1 + r³)
2. These are in GP with ratio r³.
3. Reciprocals of a GP form a GP (ratio \( \frac{1}{r^3} \)), so the given terms are in GP.

Teacher's Note:
1. The reciprocals of the terms of a GP always form a GP.
2. So it is enough to check that a³ + b³, b³ + c³, c³ + d³ are in GP.

Answer: (a) A.P.

Explanation:
1. Let \( a^{1/x} = b^{1/y} = c^{1/z} = k \); then a = k^x, b = k^y, c = k^z.
2. b² = ac ⇒ k^2y = k^{x + z} ⇒ 2y = x + z
3. Hence x, y, z are in AP.

Teacher's Note:
1. Introduce a common value k for equal expressions; it turns the condition into simple powers.
2. When numbers are powers of one base and in GP, their exponents are in AP.

Answer: (b) mⁿ

Explanation:
1. Take the terms as \( \frac{m}{r^k}, \ldots, \frac{m}{r}, m, mr, \ldots, mr^k \) with n = 2k + 1.
2. The powers of r cancel in pairs, so the product = m^2k+1 = mⁿ.

Teacher's Note:
1. For an odd number of terms, write them symmetrically about the middle term.
2. Every pair of terms equidistant from the middle multiplies to m².

Answer: (b) 1

Explanation:
1. \( \frac{a^n + b^n}{a^{n-1} + b^{n-1}} = \frac{a + b}{2} \)
2. 2aⁿ + 2bⁿ = aⁿ + abⁿ⁻¹ + aⁿ⁻¹b + bⁿ ⇒ (a − b)(aⁿ⁻¹ − bⁿ⁻¹) = 0
3. For a ≠ b, aⁿ⁻¹ = bⁿ⁻¹ ⇒ n − 1 = 0 ⇒ n = 1

Teacher's Note:
1. Put n = 1 to check quickly: (a + b)/(1 + 1) is exactly the AM.
2. Similarly n = 1/2 gives the GM, which is a standard companion result.

Answer: (c) ab

Explanation:
1. a, A₁, A₂, b are in AP: \( A_1 = \frac{2a + b}{3} \), \( A_2 = \frac{a + 2b}{3} \)
2. 2A₁ − A₂ = \( \frac{4a + 2b - a - 2b}{3} \) = a; 2A₂ − A₁ = b
3. Product = ab

Teacher's Note:
1. With two AMs, the common difference is (b − a)/3.
2. 2A₁ − A₂ is the term just before A₁ in the AP, which is a itself.

Answer: (b) 2

Explanation:
1. There are 6 terms, so \( d = \frac{3 - \frac{1}{2}}{4 + 1} = \frac{1}{2} \)
2. 3^rd AM = \( \frac{1}{2} + 3 \times \frac{1}{2} = 2 \)

Teacher's Note:
1. With n means inserted, d = (b − a)/(n + 1).
2. The k-th mean is a + kd.

Answer: (c) 3

Explanation:
1. \( d = \frac{\frac{9}{2} - 1}{6 + 1} = \frac{1}{2} \)
2. 4^th AM = \( 1 + 4 \times \frac{1}{2} = 3 \)

Teacher's Note:
1. Count the means plus one to get the number of gaps between a and b.
2. The 4th mean is the 5th term of the AP, i.e. a + 4d.

Answer: (c) 8

Explanation:
1. Third AM = 2 + 3d = 14 ⇒ d = 4
2. b = a + (n + 1)d ⇒ 38 = 2 + 4(n + 1) ⇒ n + 1 = 9 ⇒ n = 8

Teacher's Note:
1. Use the given mean to find d first, then use the last term to find n.
2. Remember b is the (n + 2)th term, so b = a + (n + 1)d.

Answer: (c) 20

Explanation:
1. 160, G₁, G₂, G₃, G₄, 5 are in GP, so 160r⁵ = 5 ⇒ \( r^5 = \frac{1}{32} \) ⇒ \( r = \frac{1}{2} \)
2. G₃ = 160r³ = \( 160 \times \frac{1}{8} = 20 \)

Teacher's Note:
1. With n GMs, r = (b/a)^(1/(n+1)).
2. The means here are 80, 40, 20, 10.