CBSE Class 11 Mathematics Sequences and Series Competitive Exam MCQs Set 02

Answer: (a) a + b

Explanation:
1. a, g₁, g₂, b are in GP, so \( g_1^2 = a g_2 \) and \( g_2^2 = g_1 b \).
2. \( \frac{g_1^2}{g_2} = a \), \( \frac{g_2^2}{g_1} = b \)
3. Sum = a + b

Teacher's Note:
1. Apply the middle-term rule to each set of three consecutive terms a, g₁, g₂ and g₁, g₂, b.
2. This avoids fractional powers like a^(2/3) b^(1/3).

Answer: (c) \( \frac{a + b}{ab} \)

Explanation:
1. Sum of n AMs = n × (single AM): \( A_1 + A_2 = 2 \cdot \frac{a + b}{2} = a + b \)
2. Product of n GMs = (single GM)ⁿ: \( G_1 G_2 = (\sqrt{ab})^2 = ab \)
3. Ratio = \( \frac{a + b}{ab} \)

Teacher's Note:
1. Sum of n AMs is n times the single AM; product of n GMs is the nth power of the single GM.
2. These two results save the work of finding each mean separately.

Answer: (b) 2apq

Explanation:
1. Let the numbers be x and y: x + y = 2a, and x, p, q, y are in GP.
2. p² = xq and q² = py ⇒ p³ = xpq, q³ = pqy
3. p³ + q³ = pq(x + y) = 2apq

Teacher's Note:
1. Multiply each middle-term relation by the remaining mean to get p³ and q³ directly.
2. Then only x + y is needed, which comes from the AM.

Answer: (b) \( 1 - \frac{1}{2^n} \)

Explanation:
1. Terms: \( 1 - \frac{1}{2}, 1 - \frac{1}{4}, 1 - \frac{1}{8}, \ldots \)
2. So the n^th term is \( 1 - \frac{1}{2^n} = \frac{2^n - 1}{2^n} \)

Teacher's Note:
1. Look at how far each term is from 1; the gaps 1/2, 1/4, 1/8 form a GP.
2. Checking n = 1, 2, 3 in the chosen formula confirms the pattern.

Answer: (a) \( \frac{1}{n^2 + n} \)

Explanation:
1. Denominators 2, 6, 12 are 1 × 2, 2 × 3, 3 × 4.
2. So the n^th term is \( \frac{1}{n(n + 1)} = \frac{1}{n^2 + n} \)

Teacher's Note:
1. Factorise the denominators to spot the product of consecutive integers.
2. Test each option at n = 1, 2, 3 to eliminate wrong choices quickly.

Answer: (b) 34

Explanation:
1. Each term from the third onwards is the sum of the two preceding terms (Fibonacci sequence).
2. Terms: 1, 1, 2, 3, 5, 8, 13, 21, 34
3. 9^th term = 34

Teacher's Note:
1. A sequence can be defined by a rule connecting terms (a recurrence) instead of a formula for aₙ.
2. For the Fibonacci sequence, a₁ = a₂ = 1 and aₙ = aₙ₋₁ + aₙ₋₂ for n > 2.

Answer: (c) a = 2, b = 32

Explanation:
1. Let α, β, γ, δ = α, αr, αr², αr³.
2. α + β = α(1 + r) = 3 and γ + δ = αr²(1 + r) = 12 ⇒ r² = 4
3. Increasing GP ⇒ r = 2, α = 1; the terms are 1, 2, 4, 8.
4. a = αβ = 2, b = γδ = 32

Teacher's Note:
1. Dividing the two sums of roots removes α and gives r² directly.
2. Use 'increasing' to choose the sign of r before finding the products of roots.

Answer: (d) 1 : 2 : 3

Explanation:
1. Let b − a = c − b = d (d ≠ 0, as the numbers are unequal).
2. d, d, a in GP ⇒ d² = da ⇒ a = d
3. Then b = 2a, c = 3a, so a : b : c = 1 : 2 : 3

Teacher's Note:
1. For numbers in AP, the consecutive differences are both equal to d.
2. Three terms d, d, a are in GP only if a = d (since d ≠ 0).

Answer: (c) \( \frac{n(1 - x) - x(1 - x^n)}{(1 - x)^2} \)

Explanation:
1. r^th term = \( \frac{1 - x^r}{1 - x} \)
2. \( S_n = \frac{1}{1 - x}\left[n - (x + x^2 + \ldots + x^n)\right] = \frac{1}{1 - x}\left[n - \frac{x(1 - x^n)}{1 - x}\right] \)
3. \( = \frac{n(1 - x) - x(1 - x^n)}{(1 - x)^2} \)

Teacher's Note:
1. Each bracket is itself a GP, so first find a formula for the r-th bracket.
2. Summing then splits into n ones minus a single GP x + x² + … + xⁿ.

Answer: (a) 1 : 2 : 3

Explanation:
1. Let b − a = c − b = d ≠ 0; then d, d, 4a − c in GP ⇒ 4a − c = d.
2. c = a + 2d, so 4a − a − 2d = d ⇒ 3a = 3d ⇒ a = d
3. b = 2a, c = 3a ⇒ a : b : c = 1 : 2 : 3

Teacher's Note:
1. Two equal consecutive terms in a GP force ratio 1, so the third term equals them too.
2. Express c through a and d before solving.

Answer: (b) 9

Explanation:
1. \( S_n = \frac{3^n - 1}{2} > 7000 \) ⇒ 3ⁿ > 14001
2. 3⁸ = 6561 and 3⁹ = 19683
3. Least n = 9

Teacher's Note:
1. Turn the condition on the sum into a condition on 3ⁿ and then test powers.
2. Remember 3⁸ = 6561 and 3⁹ = 19683 as handy reference values.

Answer: (b) G.P.

Explanation:
1. The terms are AR^p−1, AR^q−1, AR^r−1.
2. (AR^q−1)² = A²R^2q−2 and AR^p−1 · AR^r−1 = A²R^p+r−2
3. Since 2q = p + r, these are equal, so the terms are in GP.

Teacher's Note:
1. Terms of a GP whose positions are equally spaced always form a GP.
2. p, q, r in AP means exactly that the positions are equally spaced.

Answer: (a) \( \sqrt{pq} \)

Explanation:
1. The (m − n)^th, m^th and (m + n)^th terms are equally spaced, so they are in GP.
2. (m^th term)² = pq ⇒ m^th term = \( \sqrt{pq} \)

Teacher's Note:
1. The m-th term is the geometric mean of terms equidistant from it.
2. Algebraically: AR^(m+n−1) · AR^(m−n−1) = (AR^(m−1))².

Answer: (a) \( \frac{3}{2}, \frac{3}{4}, \frac{3}{8}, \frac{3}{16}, \ldots \)

Explanation:
1. \( \frac{a}{1 - r} = 3 \) and \( \frac{a^2}{1 - r^2} = 3 \)
2. Squaring the first and dividing by the second: \( \frac{1 + r}{1 - r} = 3 \) ⇒ \( r = \frac{1}{2} \)
3. \( a = 3\left(1 - \frac{1}{2}\right) = \frac{3}{2} \); the series is \( \frac{3}{2}, \frac{3}{4}, \frac{3}{8}, \ldots \)

Teacher's Note:
1. (a/(1 − r))² divided by a²/(1 − r²) simplifies to (1 + r)/(1 − r).
2. You can also verify options directly: 3/2 ÷ (1 − 1/2) = 3 and (9/4) ÷ (3/4) = 3.

Answer: (b) 2ⁿ⁺¹ − n − 2

Explanation:
1. k^th term = 2^k − 1
2. S_n = (2 + 2² + … + 2ⁿ) − n = 2(2ⁿ − 1) − n
3. = 2ⁿ⁺¹ − n − 2

Teacher's Note:
1. Notice each term is one less than a power of 2.
2. Check n = 2: 1 + 3 = 4 and 2³ − 2 − 2 = 4.

Answer: (d) \( \left(\frac{S}{R}\right)^n \)

Explanation:
1. \( S = \frac{a(r^n - 1)}{r - 1} \); the inverses form a GP with first term \( \frac{1}{a} \) and ratio \( \frac{1}{r} \), so \( R = \frac{r^n - 1}{a r^{n-1}(r - 1)} \)
2. \( \frac{S}{R} = a^2 r^{n-1} \)
3. \( P = a^n r^{\frac{n(n-1)}{2}} \) ⇒ \( P^2 = (a^2 r^{n-1})^n = \left(\frac{S}{R}\right)^n \)

Teacher's Note:
1. S/R equals the product of the first and last terms, a · arⁿ⁻¹.
2. Squaring the product of n terms gives the n-th power of that same first × last product.

Answer: (b) 2

Explanation:
1. Let the terms be a, ar, ar² with a(1 + r + r²) = 14.
2. a + 1, ar + 1, ar² − 1 in AP ⇒ 2(ar + 1) = a + ar² ⇒ a(1 − r)² = 2
3. Dividing: \( \frac{1 + r + r^2}{(1 - r)^2} = 7 \) ⇒ 2r² − 5r + 2 = 0 ⇒ r = 2 or \( \frac{1}{2} \)
4. The terms are 2, 4, 8 (or 8, 4, 2); the lowest is 2.

Teacher's Note:
1. Dividing the two equations removes a and leaves a quadratic in r.
2. Both values of r give the same three numbers, just in reverse order.

Answer: (a) 9

Explanation:
1. AM ≥ GM: \( a + b + c \geq 3\sqrt[3]{abc} \) and \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq \frac{3}{\sqrt[3]{abc}} \)
2. Multiplying: \( (a + b + c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \geq 9 \)
3. Equality when a = b = c, so the least value is 9.

Teacher's Note:
1. Applying AM ≥ GM to the numbers and to their reciprocals makes the cube roots cancel.
2. Equality in AM ≥ GM holds only when all the numbers are equal.

Answer: (a) 3

Explanation:
1. AM ≥ GM: \( \frac{\frac{a}{b} + \frac{b}{c} + \frac{c}{a}}{3} \geq \sqrt[3]{\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a}} = 1 \)
2. Hence \( \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq 3 \)

Teacher's Note:
1. Choose the terms so that their product is a constant; then AM ≥ GM gives a fixed lower bound.
2. Here the product of a/b, b/c, c/a is exactly 1.

Answer: (d) 6.64474

Explanation:
1. \( r = \frac{1.212}{5.05} = 0.24 \) (also \( \frac{0.29088}{1.212} = 0.24 \))
2. \( S_\infty = \frac{5.05}{1 - 0.24} = \frac{5.05}{0.76} \approx 6.64474 \)

Teacher's Note:
1. Always confirm the ratio using two different pairs of consecutive terms.
2. Since |r| < 1, the infinite sum exists and equals a/(1 − r).

Answer: (a) A_minimum = 1

Explanation:
1. AM ≥ GM on the 50 positive numbers \( x_i^2 \): \( \frac{50}{50} \geq \left(x_1^2 x_2^2 \ldots x_{50}^2\right)^{1/50} \)
2. So \( x_1^2 x_2^2 \ldots x_{50}^2 \leq 1 \) ⇒ A ≥ 1
3. Equality when all \( x_i^2 = 1 \), so A_minimum = 1.

Teacher's Note:
1. A fixed sum gives an upper bound on the product, hence a lower bound on its reciprocal.
2. Squares are non-negative, so AM ≥ GM can be applied to them (they are non-zero since A is defined).

Answer: (b) −1

Explanation:
1. \( p = \frac{2a + b}{3} \), \( q = \frac{a + 2b}{3} \), g² = ab
2. 2p − q = a and p − 2q = −b
3. \( \frac{(2p - q)(p - 2q)}{g^2} = \frac{-ab}{ab} = -1 \)

Teacher's Note:
1. Write both AMs in terms of a and b; the combinations then collapse to a and −b.
2. The square of the single GM between a and b is ab.

Answer: (a) 1

Explanation:
1. \( x = 1 + 10 + \ldots + 10^{19} = \frac{10^{20} - 1}{9} \), \( y = \frac{3(10^{10} - 1)}{9} \), \( z = \frac{2(10^{10} - 1)}{9} \)
2. \( x - y^2 = \frac{10^{20} - 1 - (10^{10} - 1)^2}{9} = \frac{2 \cdot 10^{10} - 2}{9} = \frac{2(10^{10} - 1)}{9} \)
3. So x − y² = z and the value is 1.

Teacher's Note:
1. A number with k repeated digits d equals d(10ᵏ − 1)/9, a GP sum.
2. Writing every number this way turns the digit puzzle into simple algebra.

Answer: (c) 11 − 2⁻¹⁰

Explanation:
1. k^th term = \( 1 + \frac{1}{2^k} \)
2. \( S_{10} = 10 + \left(\frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{2^{10}}\right) = 10 + \left(1 - \frac{1}{2^{10}}\right) \)
3. = 11 − 2⁻¹⁰

Teacher's Note:
1. Split each term into a constant part and a GP part.
2. The sum 1/2 + 1/4 + … + 1/2ⁿ equals 1 − 1/2ⁿ.

Answer: (d) \( 7 + 4\sqrt{3} \)

Explanation:
1. \( \frac{a + b}{2} = 2\sqrt{ab} \); put \( t = \sqrt{\frac{a}{b}} \): t² + 1 = 4t
2. \( t = 2 \pm \sqrt{3} \); a > b ⇒ \( t = 2 + \sqrt{3} \)
3. \( \frac{a}{b} = t^2 = (2 + \sqrt{3})^2 = 7 + 4\sqrt{3} \)

Teacher's Note:
1. Dividing by b turns the condition into a quadratic in √(a/b).
2. If AM : GM = m : n, then a : b = (m + √(m² − n²))² : (m − √(m² − n²))².

Answer: (b) \( \frac{S_2}{S_1} \)

Explanation:
1. Odd places: a, ar², ar⁴, …; even places: ar, ar³, ar⁵, …
2. Each even-place term is r times the odd-place term before it, so S₂ = rS₁.
3. \( r = \frac{S_2}{S_1} \)

Teacher's Note:
1. With an even number of terms, odd- and even-placed terms pair up exactly.
2. For example, if the total is 5 times the odd-place sum, then 1 + r = 5 and r = 4.

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

Explanation:
1. AP: 2(2x + y) = 5x − y + x + 2y ⇒ y = 2x
2. GP: (xy + 1)² = (x − 1)²(y + 1)² ⇒ xy + 1 = ±(x − 1)(y + 1)
3. Negative sign with y = 2x: 2x² + 1 = −(2x² − x − 1) ⇒ 4x² = x ⇒ \( x = \frac{1}{4} \), \( y = \frac{1}{2} \), \( x + y = \frac{3}{4} \)
4. (The positive sign gives x = −2, y = −4, x + y = −6, which is not among the options.)

Teacher's Note:
1. Use the AP condition first to reduce two unknowns to one.
2. Taking a square root introduces ±; examine both signs before choosing the answer.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Let the sides be a, ar, ar² with a > 0, r > 0.
2. If r ≥ 1 the largest side is ar²: a + ar > ar² ⇒ r² − r − 1 < 0 ⇒ \( r < \frac{\sqrt{5} + 1}{2} \)
3. If r < 1 the largest side is a: ar + ar² > a ⇒ r² + r − 1 > 0 ⇒ \( r > \frac{\sqrt{5} - 1}{2} \)
4. So the Assertion is true, the Reason (triangle inequality) is true, and the Assertion is obtained from the Reason.

Teacher's Note:
1. Only the inequality involving the largest side matters; the other two hold automatically.
2. The two bounds are reciprocals of each other, matching the fact that reversing the GP replaces r by 1/r.