CBSE Class 11 Mathematics Sequences and Series JEE Previous Year Questions Set 01

Answer: (b) \( \frac{3}{5} \)

Explanation:
1. \( \frac{a}{1 - r} = 20 \) and \( \frac{a^2}{1 - r^2} = 100 \)
2. Squaring the first and dividing by the second: \( \frac{1 + r}{1 - r} = 4 \)
3. 1 + r = 4 − 4r ⇒ \( r = \frac{3}{5} \) (then a = 8)

Teacher's Note:
1. The squares of a GP form a GP with first term a² and ratio r².
2. The ratio (sum)²/(sum of squares) always equals (1 + r)/(1 − r).

Answer: (b) 512

Explanation:
1. The 5^th term is the middle term of the first 9 terms.
2. Terms equidistant from the middle multiply to (5^th term)².
3. Product = (ar⁴)⁹ = 2⁹ = 512

Teacher's Note:
1. For an odd number of terms, the product is (middle term)^(number of terms).
2. Directly: a⁹r^(0+1+…+8) = a⁹r³⁶ = (ar⁴)⁹.

Answer: (d) \( \sqrt{2} \)

Explanation:
1. (p + q)² = p² + q² + 2pq = 1 + 2pq
2. AM ≥ GM on p², q²: 2pq ≤ p² + q² = 1
3. (p + q)² ≤ 2 ⇒ p + q ≤ \( \sqrt{2} \), with equality at \( p = q = \frac{1}{\sqrt{2}} \)

Teacher's Note:
1. AM ≥ GM for p² and q² is the same as (p − q)² ≥ 0.
2. The maximum occurs when the two numbers are equal.

Answer: (b) \( \frac{1}{2}(\sqrt{5} - 1) \)

Explanation:
1. a = ar + ar² ⇒ r² + r − 1 = 0
2. \( r = \frac{-1 \pm \sqrt{5}}{2} \)
3. Terms are positive so r > 0: \( r = \frac{\sqrt{5} - 1}{2} \)

Teacher's Note:
1. Write the condition for a general term arⁿ⁻¹ and cancel the common factor.
2. Positive terms force r > 0, which rejects the negative root.

Answer: (c) −12

Explanation:
1. a(1 + r) = 12 and ar²(1 + r) = 48 ⇒ r² = 4
2. Alternating signs ⇒ r = −2
3. a(1 − 2) = 12 ⇒ a = −12

Teacher's Note:
1. Dividing the two given sums removes a(1 + r) and gives r² at once.
2. Alternating signs means the common ratio is negative.

Answer: (c) \( \frac{7}{81}(179 + 10^{-20}) \)

Explanation:
1. \( S = \frac{7}{9}(0.9 + 0.99 + 0.999 + \ldots) = \frac{7}{9}\sum_{k=1}^{20}(1 - 10^{-k}) \)
2. \( = \frac{7}{9}\left[20 - \frac{0.1(1 - 10^{-20})}{0.9}\right] = \frac{7}{9}\left[20 - \frac{1}{9}(1 - 10^{-20})\right] \)
3. \( = \frac{7}{81}\left[180 - 1 + 10^{-20}\right] = \frac{7}{81}(179 + 10^{-20}) \)

Teacher's Note:
1. Turn 0.7, 0.77, … into 7/9 times 0.9, 0.99, … so that each term is 1 minus a power of 1/10.
2. The leftover part is a finite GP with a = 0.1 and r = 0.1.

Answer: (a) \( 2 + \sqrt{3} \)

Explanation:
1. Let the numbers be a, ar, ar² (a > 0, r > 1).
2. a, 2ar, ar² in AP ⇒ 4ar = a + ar² ⇒ r² − 4r + 1 = 0
3. \( r = 2 \pm \sqrt{3} \); increasing ⇒ \( r = 2 + \sqrt{3} \)

Teacher's Note:
1. Positive and increasing together mean r > 1.
2. The other root 2 − √3 is the reciprocal of 2 + √3 and corresponds to a decreasing GP.

Answer: (c) 28

Explanation:
1. Sum of reciprocals of a, ar, …, ar⁴ = \( \frac{S}{a^2 r^4} \), so the ratio is a²r⁴ = 49 ⇒ ar² = 7 (ar² = −7 is impossible with a + ar² = 35, since it would need r² < 0)
2. a + ar² = 35 ⇒ a = 35 − 7 = 28

Teacher's Note:
1. For a GP, (sum of terms) ÷ (sum of reciprocals) = (first term × last term).
2. Here first × last = a · ar⁴ = (ar²)², the square of the middle term.

Answer: (a) 1 : 4

Explanation:
1. \( M = \frac{a + b}{2ab} \), so \( \frac{1}{M} = \frac{2ab}{a + b} \); \( G = \sqrt{ab} \)
2. \( \frac{2ab}{(a + b)\sqrt{ab}} = \frac{4}{5} \) ⇒ \( \frac{2\sqrt{ab}}{a + b} = \frac{4}{5} \)
3. With \( t = \sqrt{\frac{a}{b}} \): 4t² − 10t + 4 = 0 ⇒ t = 2 or \( \frac{1}{2} \)
4. a : b = 4 : 1 or 1 : 4; the option is 1 : 4.

Teacher's Note:
1. Dividing numerator and denominator by b reduces the condition to one variable t = √(a/b).
2. Swapping a and b gives the reciprocal ratio, so 4 : 1 and 1 : 4 both satisfy the condition.

Answer: (c) 6

Explanation:
1. \( \frac{2}{3} + \frac{2}{3^2} + \ldots + \frac{2}{3^{n-1}} = \frac{\frac{2}{3}\left(1 - \frac{1}{3^{n-1}}\right)}{1 - \frac{1}{3}} = 1 - \frac{1}{3^{n-1}} \)
2. LHS = \( \frac{1}{3^{n-1}} < \frac{1}{100} \) ⇒ 3ⁿ⁻¹ > 100
3. 3⁴ = 81, 3⁵ = 243 ⇒ n − 1 = 5 ⇒ n = 6

Teacher's Note:
1. Find the GP sum first; the whole left side then becomes a single power of 1/3.
2. Count the terms carefully: the subtracted GP has n − 1 terms.

Answer: (d) 4lm²n

Explanation:
1. l, G₁, G₂, G₃, n are in GP with \( r^4 = \frac{n}{l} \).
2. \( G_1^4 + 2G_2^4 + G_3^4 = l^4(r^4 + 2r^8 + r^{12}) = l^4 r^4(1 + r^4)^2 \)
3. \( = l^4 \cdot \frac{n}{l} \cdot \left(\frac{l + n}{l}\right)^2 = ln(l + n)^2 = ln(2m)^2 = 4lm^2n \)

Teacher's Note:
1. Express every mean as a power of r and use r⁴ = n/l.
2. The expression is a perfect square in disguise: r⁴(1 + r⁴)².

Answer: (d) 320

Explanation:
1. a · ar · ar² = (ar)³ = 1000 ⇒ ar = 10
2. ar² + ar³ = 10r + 10r² = 60 ⇒ r² + r − 6 = 0 ⇒ r = 2 or −3
3. a > 0 needs r = 2 (r = −3 gives a negative), so a = 5
4. 7^th term = 5 × 2⁶ = 320

Teacher's Note:
1. The product of three consecutive GP terms is the cube of the middle term.
2. Use the sign condition on a to decide between the two values of r.

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

Explanation:
1. (a + 4d)² = (a + d)(a + 8d) ⇒ 8ad + 16d² = 9ad + 8d² ⇒ a = 8d (d ≠ 0)
2. Terms: 9d, 12d, 16d
3. Common ratio = \( \frac{12d}{9d} = \frac{4}{3} \)

Teacher's Note:
1. Non-constant AP means d ≠ 0, so we can divide by d.
2. Shortcut: the ratio also equals (9th − 5th)/(5th − 2nd) = 4d/3d = 4/3, without finding a.

Answer: (d) 216

Explanation:
1. Split x + y + z into 12 parts: \( \frac{x}{3} \) (3 times), \( \frac{y}{4} \) (4 times), \( \frac{z}{5} \) (5 times).
2. AM ≥ GM: \( 1 = \frac{12}{12} \geq \left[\left(\frac{x}{3}\right)^3\left(\frac{y}{4}\right)^4\left(\frac{z}{5}\right)^5\right]^{1/12} \) ⇒ x³y⁴z⁵ ≤ 3³ · 4⁴ · 5⁵ = 21600000
3. (0.1)(600)³ = 21600000, so equality holds: \( \frac{x}{3} = \frac{y}{4} = \frac{z}{5} = 1 \)
4. x = 3, y = 4, z = 5 ⇒ x³ + y³ + z³ = 27 + 64 + 125 = 216

Teacher's Note:
1. To match powers 3, 4, 5, split x, y, z into 3, 4 and 5 equal parts before applying AM ≥ GM.
2. When the given product equals the AM ≥ GM bound, all the parts must be equal.

Answer: (d) \( \frac{5\sqrt{6}}{12} \)

Explanation:
1. a + b = 10\( \sqrt{ab} \)
2. (a − b)² = (a + b)² − 4ab = 100ab − 4ab = 96ab ⇒ a − b = \( 4\sqrt{6}\sqrt{ab} \)
3. \( \frac{a + b}{a - b} = \frac{10}{4\sqrt{6}} = \frac{5}{2\sqrt{6}} = \frac{5\sqrt{6}}{12} \)

Teacher's Note:
1. (a − b)² = (a + b)² − 4ab links the AM and GM directly.
2. Since a > b, take the positive square root for a − b.

Answer: (a) 2

Explanation:
1. b is the AM of a and c: \( b = \frac{a + c}{2} \geq \sqrt{ac} \) ⇒ ac ≤ b²
2. 8 = abc ≤ b³ ⇒ b ≥ 2
3. b = 2 when a = b = c = 2

Teacher's Note:
1. In an AP the middle term is the AM of its neighbours, so AM ≥ GM applies.
2. Equivalently, AM of a, b, c is b and GM is 2, so b ≥ 2.

Answer: (c) (0, 10)

Explanation:
1. \( \frac{b}{1 - r} = 5 \) ⇒ b = 5(1 − r)
2. For the infinite sum to exist, −1 < r < 1 (and r ≠ 0 for a genuine GP), so 0 < 1 − r < 2
3. Hence 0 < b < 10

Teacher's Note:
1. Write the first term in terms of r and then use the range of r.
2. The condition |r| < 1 is what makes the sum to infinity meaningful.

Answer: (c) \( \frac{1}{4} - \frac{1}{2\sqrt{2}} \)

Explanation:
1. AP: 3b = \( \frac{3}{4} \) ⇒ \( b = \frac{1}{4} \), \( a + c = \frac{1}{2} \)
2. GP: b⁴ = a²c² ⇒ ac = ±\( \frac{1}{16} \); ac = \( \frac{1}{16} \) gives a = c, so \( ac = -\frac{1}{16} \)
3. a, c are roots of \( t^2 - \frac{1}{2}t - \frac{1}{16} = 0 \): \( t = \frac{1}{4} \pm \frac{1}{2\sqrt{2}} \)
4. a < c ⇒ \( a = \frac{1}{4} - \frac{1}{2\sqrt{2}} \)

Teacher's Note:
1. From squares in GP we only get ac = ±b², so both signs must be checked.
2. Form the quadratic with sum a + c and product ac to find a and c together.

Answer: (b) 7

Explanation:
1. A_n is a GP with a = \( \frac{3}{4} \), r = \( -\frac{3}{4} \): \( A_n = \frac{3}{7}\left[1 - \left(-\frac{3}{4}\right)^n\right] \)
2. B_n > A_n ⇔ A_n < \( \frac{1}{2} \) ⇔ \( \left(-\frac{3}{4}\right)^n > -\frac{1}{6} \)
3. True for every even n; for odd n we need \( \left(\frac{3}{4}\right)^n < \frac{1}{6} \)
4. \( \left(\frac{3}{4}\right)^5 \approx 0.237 \), \( \left(\frac{3}{4}\right)^7 \approx 0.133 \), so the least odd p is 7

Teacher's Note:
1. B > A with A + B = 1 simply means A < 1/2.
2. For a negative ratio, treat even and odd n separately because the sign of rⁿ changes.

Answer: (a) \( 38 + \frac{1}{2^{19}} \)

Explanation:
1. k^th term = \( 2 - \frac{1}{2^{k-1}} \)
2. \( S_{20} = 40 - \left(1 + \frac{1}{2} + \ldots + \frac{1}{2^{19}}\right) = 40 - 2\left(1 - \frac{1}{2^{20}}\right) \)
3. \( = 38 + \frac{2}{2^{20}} = 38 + \frac{1}{2^{19}} \)

Teacher's Note:
1. Each term is 2 minus a power of 1/2, so the sum splits into a constant part and a GP.
2. Count the GP terms carefully: k = 1 to 20 gives powers 1/2⁰ to 1/2¹⁹.

Answer: (c) \( \frac{3}{5} \)

Explanation:
1. \( \frac{S_6}{S_3} = \frac{r^6 - 1}{r^3 - 1} = 1 + r^3 = \frac{152}{125} \)
2. \( r^3 = \frac{27}{125} \) ⇒ \( r = \frac{3}{5} \)

Teacher's Note:
1. S₆/S₃ = 1 + r³ for any GP with r ≠ 1.
2. The last three terms of the six are exactly r³ times the first three.

Answer: (b) 4⁵

Explanation:
1. The 3^rd term is the middle of the first five terms.
2. Product = \( \frac{a}{r^2} \cdot \frac{a}{r} \cdot a \cdot ar \cdot ar^2 \) with middle term a = 4, so product = 4⁵

Teacher's Note:
1. Write five terms symmetrically around the middle term so the powers of r cancel.
2. Product of an odd number of GP terms = (middle term)^(number of terms).

Answer: (b) are in G.P.

Explanation:
1. The left side equals (ap − b)² + (bp − c)² + (cp − d)².
2. A sum of squares ≤ 0 forces each square to be 0: b = ap, c = bp, d = cp.
3. So \( \frac{b}{a} = \frac{c}{b} = \frac{d}{c} = p \) and a, b, c, d are in GP.

Teacher's Note:
1. Group the expression into perfect squares by matching p²a² with b² and the middle term.
2. A sum of real squares can be ≤ 0 only if every square is zero.

Answer: (c) n + 2⁻ⁿ − 1

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

Teacher's Note:
1. Rewrite each term as 1 minus a power of 1/2.
2. Check with n = 1: 1 + 1/2 − 1 = 1/2, the first term.

Answer: (d) never less than n

Explanation:
1. AM ≥ GM: \( \frac{a_1 + a_2 + \ldots + a_n}{n} \geq (a_1 a_2 \ldots a_n)^{1/n} = 1 \)
2. So a₁ + a₂ + … + a_n ≥ n, i.e. the sum is never less than n.

Teacher's Note:
1. When the product of positive numbers is fixed, AM ≥ GM gives the least possible sum.
2. The minimum n is reached only when every number equals 1.

Answer: (a) \( \frac{\alpha}{\beta} \)

Explanation:
1. a_2n = r · a_2n−1 for every n.
2. So α = a₂ + a₄ + … + a₂₀₀ = r(a₁ + a₃ + … + a₁₉₉) = rβ
3. \( r = \frac{\alpha}{\beta} \)

Teacher's Note:
1. Each even-numbered term is r times the odd-numbered term just before it.
2. So the sum of even-placed terms is r times the sum of odd-placed terms.

Answer: (c) 64

Explanation:
1. \( 1 + n + \ldots + n^{127} = \frac{n^{128} - 1}{n - 1} \)
2. n¹²⁸ − 1 = (n⁶⁴ − 1)(n⁶⁴ + 1), and \( \frac{n^{64} - 1}{n - 1} \) is an integer
3. So (n⁶⁴ + 1) divides the sum; repeating, the full factorisation is (1 + n)(1 + n²)(1 + n⁴) … (1 + n⁶⁴), so the largest m is 64.

Teacher's Note:
1. The GP sum formula turns the long sum into (n¹²⁸ − 1)/(n − 1).
2. Repeated use of x² − 1 = (x − 1)(x + 1) gives factors 1 + n, 1 + n², …, 1 + n⁶⁴.

Answer: (a) lie on a straight line

Explanation:
1. x₂ = rx₁, x₃ = r²x₁ and y₂ = ry₁, y₃ = r²y₁
2. So \( \frac{y_1}{x_1} = \frac{y_2}{x_2} = \frac{y_3}{x_3} \): all three points lie on the line \( y = \frac{y_1}{x_1}x \) through the origin.

Teacher's Note:
1. Each point is r times the previous one, i.e. the points are scalings of one point from the origin.
2. Equal ratios y/x mean equal slopes from the origin, hence collinear points.

Answer: (a) 0 < M ≤ 1

Explanation:
1. AM ≥ GM on (a + b) and (c + d): \( \sqrt{(a + b)(c + d)} \leq \frac{(a + b) + (c + d)}{2} = 1 \)
2. So M ≤ 1, and M > 0 as all numbers are positive.

Teacher's Note:
1. Treat a + b and c + d as two positive numbers with a fixed sum 2.
2. Their product is largest (equal to 1) when the two sums are equal.

Answer: (d) \( a = 3, r = \frac{1}{4} \)

Explanation:
1. \( \frac{a}{1 - r} = 4 \) ⇒ a = 4(1 − r); \( ar = \frac{3}{4} \)
2. 4r(1 − r) = \( \frac{3}{4} \) ⇒ 16r² − 16r + 3 = 0 ⇒ \( r = \frac{1}{4} \) or \( \frac{3}{4} \)
3. \( r = \frac{1}{4} \) ⇒ a = 3 (option d); \( r = \frac{3}{4} \) ⇒ a = 1 (not an option)

Teacher's Note:
1. Substitute a from the sum formula into the second-term condition to get a quadratic in r.
2. Both roots are valid GPs; pick the one listed in the options.

Answer: (a) −2, −32

Explanation:
1. Let the terms be α, αr, αr², αr³: α(1 + r) = 1 and αr²(1 + r) = 4 ⇒ r² = 4
2. r = 2: \( \alpha = \frac{1}{3} \), \( p = \alpha^2 r = \frac{2}{9} \), not an integer
3. r = −2: α = −1, terms −1, 2, −4, 8 ⇒ p = αβ = −2, q = γδ = −32

Teacher's Note:
1. Dividing the sums of roots gives r² without knowing α.
2. The word 'integer' is what rules out r = 2.

Answer: (a) n(2c)^1/n

Explanation:
1. Apply AM ≥ GM to the n numbers a₁, a₂, …, a_n−1, 2a_n.
2. Their product is 2a₁a₂ … a_n = 2c.
3. \( a_1 + \ldots + a_{n-1} + 2a_n \geq n(2c)^{1/n} \), with equality when a₁ = … = a_n−1 = 2a_n.

Teacher's Note:
1. Treat 2aₙ as a single number so that there are exactly n terms.
2. The minimum is attained, since all n numbers can be made equal to (2c)^(1/n).

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

Explanation:
1. 3b = \( \frac{3}{2} \) ⇒ \( b = \frac{1}{2} \), a + c = 1
2. b⁴ = a²c² ⇒ ac = ±\( \frac{1}{4} \); \( ac = \frac{1}{4} \) gives a = c, so \( ac = -\frac{1}{4} \)
3. a, c are roots of \( t^2 - t - \frac{1}{4} = 0 \): \( t = \frac{1}{2} \pm \frac{1}{\sqrt{2}} \)
4. a < c ⇒ \( a = \frac{1}{2} - \frac{1}{\sqrt{2}} \)

Teacher's Note:
1. Squares in GP only give ac = ±b², and the positive sign forces a = b = c here.
2. Distinct numbers with a negative product must have opposite signs, so a is negative.

Answer: (a) 2 tan α

Explanation:
1. Both terms are positive. AM ≥ GM: \( \sqrt{x^2 + x} + \frac{\tan^2 \alpha}{\sqrt{x^2 + x}} \geq 2\sqrt{\tan^2 \alpha} \)
2. tan α > 0 in \( \left(0, \frac{\pi}{2}\right) \), so the expression ≥ 2 tan α.

Teacher's Note:
1. For positive t and k, t + k/t ≥ 2√k by AM ≥ GM.
2. The product of the two terms is tan²α, independent of x, which is why AM ≥ GM works.

Answer: (c) cΔ = 0

Explanation:
1. (α² + β²)² = (α + β)(α³ + β³) ⇒ 2α²β² = αβ(α² + β²)
2. ⇒ αβ(α − β)² = 0
3. \( \alpha\beta = \frac{c}{a} \) and \( (\alpha - \beta)^2 = \frac{\Delta}{a^2} \), so \( \frac{c\Delta}{a^3} = 0 \) ⇒ cΔ = 0

Teacher's Note:
1. Expand both sides; the α⁴ and β⁴ terms cancel immediately.
2. Use (α − β)² = (α + β)² − 4αβ = Δ/a² to bring in the discriminant.

Answer: (b) 8; (c) 9; (d) 10

Explanation:
1. Side of S_n = \( \sqrt{2} \) × side of S_n+1, so sides form a GP with ratio \( \frac{1}{\sqrt{2}} \).
2. Areas form a GP with ratio \( \frac{1}{2} \): area of S_n = \( \frac{100}{2^{n-1}} \)
3. \( \frac{100}{2^{n-1}} < 1 \) ⇒ 2ⁿ⁻¹ > 100 ⇒ n − 1 ≥ 7 ⇒ n ≥ 8
4. So n = 8, 9, 10 all work; n = 7 gives area \( \frac{100}{64} > 1 \).

Teacher's Note:
1. The diagonal of a square of side s is s√2, so each new side is the old side divided by √2.
2. Squaring the side ratio gives the area ratio 1/2.

Answer: 8

Explanation:
1. Write 3a⁻³ as a⁻³ + a⁻³ + a⁻³, giving 8 positive numbers: a⁻⁵, a⁻⁴, a⁻³, a⁻³, a⁻³, 1, a⁸, a¹⁰.
2. Their product = a^{−5−4−9+8+10} = a⁰ = 1
3. AM ≥ GM: sum ≥ 8 × 1 = 8, with equality at a = 1. Minimum value = 8.

Teacher's Note:
1. Split a coefficient like 3 into equal terms so that the product of all terms becomes a constant.
2. Equality in AM ≥ GM needs all terms equal, which happens at a = 1.

Answer: 4

Explanation:
1. Let b = ar, c = ar² with r a positive integer.
2. \( \frac{a + ar + ar^2}{3} = ar + 2 \) ⇒ a(r² − 2r + 1) = 6 ⇒ a(r − 1)² = 6
3. The only perfect square dividing 6 is 1, so (r − 1)² = 1 and a = 6 (r = 2).
4. \( \frac{a^2 + a - 14}{a + 1} = \frac{36 + 6 - 14}{7} = \frac{28}{7} = 4 \)

Teacher's Note:
1. Turning the AM condition into a(r − 1)² = 6 makes it a divisibility puzzle.
2. r = 0 is not allowed because c must be positive.

Answer: 2

Explanation:
1. Let the numbers be a, b: \( x = \frac{a + b}{2} \); a, y, z, b in GP ⇒ y² = az, z² = yb
2. y³ = ayz, z³ = byz ⇒ y³ + z³ = yz(a + b) = 2xyz
3. \( \frac{y^3 + z^3}{xyz} = 2 \)

Teacher's Note:
1. Multiply each middle-term relation by the other GM to get y³ and z³.
2. This avoids working with a^(2/3) b^(1/3) and similar powers.