JEE Mathematics Sequence and Series MCQs Set 31

Answer: (a) \( 272.5 \)

Explanation:
1. First term \( -\frac12 \) and common difference \( -\frac18 + \frac12 = \frac38 \).
2. \( S_{40} = 20\left[2\left(-\tfrac12\right) + 39 \cdot \tfrac38\right] = 20 \times \frac{109}{8} = 272.5 \).

Teacher's Note:
1. d = b - a = 3/8.
2. 20 x 109/8 = 272.5.

Answer: (d) \( 12 \)

Explanation:
1. \( \frac n2[18 + 3(n-1)] = 306 \Rightarrow n(3n + 15) = 612 \Rightarrow n^2 + 5n - 204 = 0 \).
2. \( (n - 12)(n + 17) = 0 \Rightarrow n = 12 \).

Teacher's Note:
1. Reject the negative root.
2. Check: terms 9 to 42, average 25.5, 12 x 25.5 = 306.

Answer: (c) \( 8, -4, 2, 8 \)

Explanation:
1. Let the numbers be \( a, ar, ar^2, a \). AP: \( ar^2 - ar = 6 \) and \( a - ar^2 = 6 \).
2. Equating, \( r^2 - r = 1 - r^2 \Rightarrow 2r^2 - r - 1 = 0 \Rightarrow r = 1 \) or \( -\frac12 \). r = 1 gives zero difference, so \( r = -\frac12 \) and \( a = 8 \).
3. The numbers are 8, -4, 2, 8.

Teacher's Note:
1. The common difference of 6 rules out r = 1.
2. Check: -4, 2, 8 has d = 6.

Answer: (b) \( \frac{\sqrt5 - 1}{2} \)

Explanation:
1. \( \sin^4 2x = \sin x \cos x \sin 4x = \frac12 \sin 2x \cdot 2\sin 2x \cos 2x = \sin^2 2x \cos 2x \).
2. So \( \sin^2 2x = \cos 2x \Rightarrow 1 - \cos^2 2x = \cos 2x \Rightarrow \cos^2 2x + \cos 2x - 1 = 0 \), giving \( \cos 2x = \frac{\sqrt5 - 1}{2} \) (the root with modulus at most 1).

Teacher's Note:
1. Use sin 2x = 2 sin x cos x and sin 4x = 2 sin 2x cos 2x.
2. The common ratio is sqrt2 (increasing).

Answer: (a) \( 364(\sqrt3 + 1) \)

Explanation:
1. \( r = \sqrt3 \), so \( r^{12} = 3^6 = 729 \).
2. \( S_{12} = \frac{729 - 1}{\sqrt3 - 1} = \frac{728(\sqrt3 + 1)}{2} = 364(\sqrt3 + 1) \).

Teacher's Note:
1. Rationalise the denominator.
2. (sqrt3 - 1)(sqrt3 + 1) = 2.

Answer: (d) \( 0 \)

Explanation:
1. \( a = \frac34,\ d = \frac23 - \frac34 = -\frac{1}{12} \).
2. \( S_{19} = \frac{19}{2}\left[\frac32 - \frac{18}{12}\right] = 0 \) (the 10th term is zero).

Teacher's Note:
1. If the middle term is 0, the sum of an AP with odd number of terms is 0.
2. a + 9d = 3/4 - 9/12 = 0.

Answer: (c) \( \frac{n(n+1)(n+2)(n+3)(n+4)}{5} \)

Explanation:
1. Write \( r(r+1)(r+2)(r+3) = \frac15\left[r(r+1)(r+2)(r+3)(r+4) - (r-1)r(r+1)(r+2)(r+3)\right] \).
2. The sum telescopes to \( \frac15 n(n+1)(n+2)(n+3)(n+4) \).

Teacher's Note:
1. For a product of r consecutive factors, affix the next factor and divide by (number of factors + 1) x d.
2. Check n = 1: 24 = 120/5.

Answer: (b) \( \frac{1 - x^2}{x+1} = 1 - x \)

Explanation:
1. \( d = \frac{1 - x^2}{x + 1} = 1 - x \).

Teacher's Note:
1. Simplify the difference of squares.
2. Number of means is x, so divide by x + 1.

Answer: (a) \( \frac{n(n+1)(2n+1)}{6} + 2n(n+1) + n \)

Explanation:
1. The differences 7, 9, 11 form an AP, so \( T_n = n^2 + 4n + 1 \) (check: 6, 13, 22, 33).
2. \( S_n = \sum k^2 + 4\sum k + n = \frac{n(n+1)(2n+1)}{6} + 2n(n+1) + n \).

Teacher's Note:
1. Find the nth term using the method of differences.
2. Check n = 2: 5/... 1 + 4 = 5 + 12 + 2 = 19.

Answer: (d) \( 10p - 8 \)

Explanation:
1. \( a_p = S_p - S_{p-1} = p(5p - 3) - (p - 1)(5p - 8) = 10p - 8 \).

Teacher's Note:
1. Expand carefully: (p - 1)(5p - 8) = 5p^2 - 13p + 8.
2. Check: a1 = 2 = S1.

Answer: (c) \( 8, 12, 18 \)

Explanation:
1. Take \( \frac{a}{r}, a, ar \). Product \( a^3 = 1728 \Rightarrow a = 12 \).
2. Sum: \( 12\left(\frac1r + 1 + r\right) = 38 \Rightarrow 6r^2 - 13r + 6 = 0 \Rightarrow r = \frac32 \) or \( \frac23 \).
3. The numbers are 8, 12, 18.

Teacher's Note:
1. The cube of the middle term is the product.
2. Both ratios give the same numbers in opposite order.

Answer: (b) \( \frac{n(n+1)(n^2 + 3n + 1)}{2} \)

Explanation:
1. \( \sum (2k^3 + 3k^2) = 2\left[\frac{n(n+1)}{2}\right]^2 + \frac{n(n+1)(2n+1)}{2} = \frac{n(n+1)}{2}[n(n+1) + 2n + 1] \).
2. \( = \frac{n(n+1)(n^2 + 3n + 1)}{2} \).

Teacher's Note:
1. Check n = 1: 5 and (1 x 2 x 5)/2 = 5.
2. Combine the two parts over n(n+1)/2.

Answer: (a) \( 5 \)

Explanation:
1. The common difference is \( \frac{10 - 28}{12} = -\frac{3}{2} \).
2. The means are \( 28 - \frac{3k}{2},\ k = 1, 2, \ldots, 11 \). They are integers only for even k: 2, 4, 6, 8, 10.
3. So there are 5 integral means.

Teacher's Note:
1. d = (last - first)/(number of means + 1).
2. Half-integers appear for odd k.

Answer: (a) \( 3 and 6 \)

Explanation:
1. \( G^2 = A \cdot H = 4A \), so \( 2A + 4A = 27 \Rightarrow A = \frac92 \) and \( G^2 = 18 \).
2. \( a + b = 9,\ ab = 18 \Rightarrow \) the numbers are 3 and 6.

Teacher's Note:
1. G squared = A x H for two numbers.
2. Check: HM = 2 x 18/9 = 4.

Answer: (d) \( 16, 24, 36, 54, \ldots \)

Explanation:
1. \( r^3 = \frac{81}{24} = \frac{27}{8} \Rightarrow r = \frac32 \). Then \( a = \frac{24}{3/2} = 16 \).

Teacher's Note:
1. The 5th term is 3 steps after the 2nd.
2. Check: 16 x (3/2)^4 = 81.

Answer: (c) \( (14n - 6) : (8n + 23) \)

Explanation:
1. The ratio of the nth terms is the ratio of the sums with n replaced by \( 2n - 1 \).
2. \( \frac{7(2n - 1) + 1}{4(2n - 1) + 27} = \frac{14n - 6}{8n + 23} \).

Teacher's Note:
1. S_(2n-1) = (2n - 1) x nth term.
2. Check n = 1: ratio of first terms = 8 : 31.

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

Explanation:
1. \( \frac{1}{4r^2 - 1} = \frac12\left[\frac{1}{2r-1} - \frac{1}{2r+1}\right] \).
2. The sum telescopes: \( \frac12\left[1 - \frac{1}{2n+1}\right] = \frac{n}{2n+1} \).

Teacher's Note:
1. Use partial fractions.
2. The sum to infinity is 1/2.

Answer: (a) \( \frac13 \text{ and } 28 \)

Explanation:
1. Sum of roots \( = 1 \), sum of products in pairs \( = \frac{9b}{a} \), product \( = \frac ba \). HM \( = \frac{3 \cdot \text{product}}{\text{pair sum}} = \frac{3b/a}{9b/a} = \frac13 \).
2. The AM of the roots is also \( \frac13 \), so AM = HM, which forces all roots equal to \( \frac13 \). Then the product is \( \frac{1}{27} = \frac ba \), so \( a = 27b \).
3. Minimum natural numbers: \( b = 1,\ a = 27 \), so \( a + b = 28 \).

Teacher's Note:
1. AM = HM only when all numbers are equal.
2. The pair sum 1/3 equals 9b/a with a = 27b.

Answer: (a) \( 5 \)

Explanation:
1. \( 600 = \frac{15}{2}[2a + 14 \times 5] \Rightarrow 80 = 2a + 70 \Rightarrow a = 5 \).

Teacher's Note:
1. Divide both sides by 15/2 first.
2. Check: terms 5, 10, ..., 75: average 40, 15 x 40 = 600.

Answer: (d) \( 12 \)

Explanation:
1. Let there be 2m means. The sum of the means is \( 2m \times \frac{\text{sum of the two numbers}}{2} = m \times \frac{13}{6} \).
2. \( \frac{13m}{6} = 2m + 1 \Rightarrow m = 6 \), so the number of means is 12.

Teacher's Note:
1. Sum of n AMs = n times the single AM.
2. 2m means: single AM = 13/12.

Answer: (c) \( \frac{3^{n+1} - 3}{2} - 2^{n+1} + 2 \)

Explanation:
1. \( \sum 3^k = \frac{3(3^n - 1)}{2} \) and \( \sum 2^k = 2(2^n - 1) \).
2. So \( S_n = \frac{3^{n+1} - 3}{2} - 2^{n+1} + 2 \).

Teacher's Note:
1. Sum the two GPs separately.
2. Check n = 2: 1 + 5 = 6; formula: 12 - 8 + 2 = 6.

Answer: (b) \( 25 \)

Explanation:
1. \( \frac n2[4 + 3(n-1)] = \frac{n(3n+1)}{2} = 950 \Rightarrow 3n^2 + n - 1900 = 0 \).
2. \( (n - 25)(3n + 76) = 0 \Rightarrow n = 25 \).

Teacher's Note:
1. Discard the negative root.
2. Check: 25 x 76/2 = 950.

Answer: (a) \( 820a - 1680b \)

Explanation:
1. First term \( a - 3b \), common difference \( a - 2b \).
2. \( S_{40} = 20[2(a - 3b) + 39(a - 2b)] = 20(41a - 84b) = 820a - 1680b \).

Teacher's Note:
1. Subtract consecutive terms to find d.
2. 20 x 84 = 1680.

Answer: (d) \( \sqrt2 \)

Explanation:
1. Let \( c = \cos 2x = \frac{\sqrt5 - 1}{2} \), so \( c^2 + c = 1 \). The ratio is \( r = \frac{\sin^2 2x}{\sin x} = 4\sin x \cos^2 x \).
2. \( r^2 = 16 \sin^2 x \cos^4 x = 16 \cdot \frac{1 - c}{2} \cdot \frac{(1 + c)^2}{4} = 2(1 - c)(1 + c)^2 = 2c(1 + c) = 2(c + c^2) = 2 \).
3. So \( r = \sqrt2 \) (positive for an increasing sequence).

Teacher's Note:
1. Compute the ratio of the second term to the first.
2. Check numerically: 0.437, 0.618, 0.874: ratios 1.414, 1.414.

Answer: (c) \( 8 \)

Explanation:
1. Let there be 2m terms. Even sum - odd sum = \( m d = 6 \). Last - first = \( (2m - 1)d = 10.5 \).
2. \( 2 \times 6 - d = 10.5 \Rightarrow d = 1.5,\ m = 4 \).
3. Number of terms = 8.

Teacher's Note:
1. Each even-numbered term is d more than the preceding odd one.
2. Check: the terms are 1.5, 3, 4.5, 6, 7.5, 9, 10.5, 12: odd-numbered sum 24, even-numbered sum 30.