JEE Mathematics Sequence and Series MCQs Set 29

Answer: (d) \( n^2 \)

Explanation:
1. \( 7(a + 3d) = 49 \Rightarrow a + 3d = 7 \); \( 17(a + 8d) = 289 \Rightarrow a + 8d = 17 \).
2. \( d = 2,\ a = 1 \): \( S_n = \frac n2[2 + 2(n-1)] = n^2 \).

Teacher's Note:
1. The sum of n terms of 1, 3, 5, ... is n squared.
2. Check 7^2 = 49 and 17^2 = 289.

Answer: (c) \( 5, 10, 15, 20 \)

Explanation:
1. Let the least be \( a \) and the greatest \( 4a \). The sum of four AP terms is \( 2(\text{first} + \text{last}) = 2 \cdot 5a = 10a = 50 \), so \( a = 5 \).
2. The greatest is 20, so \( 3d = 15 \), \( d = 5 \): numbers 5, 10, 15, 20.

Teacher's Note:
1. Sum of an AP = (number of terms)/2 x (first + last).
2. The terms equidistant from the ends have a constant sum.

Answer: (b) \( 1, 4, 7 \)

Explanation:
1. Let the numbers be \( 4 - d, 4, 4 + d \). Sum of cubes: \( 3 \cdot 64 + 6 \cdot 4 d^2 = 408 \Rightarrow 24d^2 = 216 \Rightarrow d = \pm 3 \).
2. The numbers are 1, 4, 7 (1 + 64 + 343 = 408).

Teacher's Note:
1. (a - d)^3 + (a + d)^3 = 2a^3 + 6ad^2.
2. Verify the cubes: 1, 64, 343.

Answer: (a) \( 80 \)

Explanation:
1. Let the roots be \( x_i \). Then \( p = -\sum x_i \), \( r = -\sum x_i x_j x_k = -5 \sum \frac{1}{x_i} \) (since the product of the roots is 5).
2. \( pr = 5\left(\sum x_i\right)\left(\sum \frac{1}{x_i}\right) \ge 5 \times 16 = 80 \) by AM-HM, with equality when all roots are equal.

Teacher's Note:
1. (Sum) x (sum of reciprocals) of n positive numbers is at least n squared.
2. Equality when all four roots are equal to the fourth root of 5.

Answer: (d) \( p + q - m \)

Explanation:
1. \( a + (p-1)d = q \) and \( a + (q-1)d = p \). Subtracting, \( (p - q)d = q - p \), so \( d = -1 \) and \( a = p + q - 1 \).
2. \( a_m = p + q - 1 - (m - 1) = p + q - m \).

Teacher's Note:
1. The common difference is -1.
2. Check: m = p gives q.

Answer: (c) \( 6 and 24 \)

Explanation:
1. \( G^2 = A \cdot H \Rightarrow A = \frac{144}{48/5} = 15 \).
2. \( a + b = 30,\ ab = 144 \Rightarrow t^2 - 30t + 144 = 0 \Rightarrow t = 6, 24 \).

Teacher's Note:
1. Use G squared = A x H to get the arithmetic mean.
2. Check: HM = 2 x 144/30 = 9.6.

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

Explanation:
1. The series is arithmetico-geometric: \( 3\cdot 2 + 5\cdot 2^2 + 7\cdot 2^3 + \cdots \). Multiply by 2 and subtract.
2. \( -S = 6 + 2(4 + 8 + \cdots + 2^n) - (2n+1)2^{n+1} \), giving \( S = (2n - 1)2^{n+1} + 2 \).

Teacher's Note:
1. Test n = 1: 3 x 2 = 6, formula: 1 x 4 + 2 = 6.
2. n = 2: 6 + 20 = 26 and 3 x 8 + 2 = 26.

Answer: (a) \( 5 \)

Explanation:
1. The reciprocals \( \frac12, \frac25, \frac{3}{10}, \ldots \) are in AP with \( d = \frac25 - \frac12 = -\frac{1}{10} \).
2. Fourth reciprocal: \( \frac12 - \frac{3}{10} = \frac15 \), so the fourth term is 5.

Teacher's Note:
1. Take reciprocals, find the AP term, then invert.
2. Check: 3/10 - 1/10 = 1/5.

Answer: (d) \( 2000 \)

Explanation:
1. \( \sqrt{1 + \frac{1}{k^2} + \frac{1}{(k+1)^2}} = 1 + \frac1k - \frac{1}{k+1} \).
2. Summing for k = 1 to 1999: \( 1999 + 1 - \frac{1}{2000} = 2000 - \frac{1}{2000} \), so \( n = 2000 \).

Teacher's Note:
1. The expression under the root is a perfect square.
2. Telescoping leaves 1 - 1/2000.

Answer: (c) \( n^4 \)

Explanation:
1. Expand: \( 4n^3 + 4n - 6n^2 - 1 = n^4 - (n-1)^4 \).
2. The series telescopes: \( S_n = n^4 - 0 = n^4 \).

Teacher's Note:
1. Recognise 4n^3 - 6n^2 + 4n - 1 as the expansion of n^4 - (n - 1)^4.
2. Check n = 1: 8 - 7 = 1.

Answer: (b) \( \frac{1}{(1 - x)^3} \)

Explanation:
1. The coefficients 1, 3, 6, 10 are the triangular numbers \( \binom{k+2}{2} \), the coefficients in the expansion of \( (1 - x)^{-3} \).
2. So the sum is \( (1 - x)^{-3} \).

Teacher's Note:
1. Differences of the coefficients are 2, 3, 4, ..., so (1 - x) must be applied three times.
2. Check x = 1/2: sum 8 = 1/(1/2)^3.

Answer: (a) \( 2 \)

Explanation:
1. \( S_6 = S_3 (1 + r^3) \), so \( 1 + r^3 = 9 \Rightarrow r^3 = 8 \Rightarrow r = 2 \).

Teacher's Note:
1. Split the first 6 terms as the first 3 plus r^3 times the first 3.
2. Check: 1,2,4 sums to 7; 1..32 sums to 63 = 9 x 7.

Answer: (d) \( 12 \)

Explanation:
1. The angles are \( 172 - 4k,\ k = 0..n-1 \), with sum \( \frac n2[172 + 172 - 4(n-1)] = n(174 - 2n) \).
2. Sum of interior angles \( = 180(n - 2) \), so \( 174n - 2n^2 = 180n - 360 \Rightarrow n^2 + 3n - 180 = 0 \Rightarrow n = 12 \).
3. The smallest angle is 128 degrees, so the result is valid.

Teacher's Note:
1. Equate the AP sum to (n - 2) x 180.
2. Reject the negative root.

Answer: (c) \( 4, 1, \tfrac14, \ldots \)

Explanation:
1. First term \( a \) equals \( 3 \times \frac{ar}{1 - r} \Rightarrow 1 - r = 3r \Rightarrow r = \frac14 \).
2. \( a + ar = 5 \Rightarrow \frac{5a}{4} = 5 \Rightarrow a = 4 \).

Teacher's Note:
1. Sum of the terms after the first is ar/(1 - r).
2. Check: 1 = 3 x (1/4)/(3/4)... equals 1.

Answer: (b) \( 4, 9, 14 \)

Explanation:
1. Take \( 9 - d, 9, 9 + d \) (sum is 27). Product: \( 9(81 - d^2) = 504 \Rightarrow 81 - d^2 = 56 \Rightarrow d = \pm 5 \).
2. The numbers are 4, 9, 14.

Teacher's Note:
1. The middle term is one third of the sum.
2. Check: 4 x 9 x 14 = 504.

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

Explanation:
1. \( a = ar + ar^2 \Rightarrow r^2 + r - 1 = 0 \).
2. \( r = \frac{-1 \pm \sqrt5}{2} \); positive terms require \( r > 0 \), so \( r = \frac{\sqrt5 - 1}{2} \).

Teacher's Note:
1. The positive root is the golden ratio conjugate.
2. A negative ratio would give alternating signs.

Answer: (d) \( 4, 9, 14 \)

Explanation:
1. Take the numbers as \( a-d,\ a,\ a+d \). Sum: \( 3a = 27 \), so \( a = 9 \).
2. Squares: \( (9-d)^2 + 81 + (9+d)^2 = 243 + 2d^2 = 293 \), so \( d^2 = 25 \), \( d = \pm 5 \).
3. The numbers are 4, 9, 14 (check: 16 + 81 + 196 = 293).

Teacher's Note:
1. The symmetric form a - d, a, a + d makes the sum immediate.
2. d = -5 gives the same numbers in reverse order.

Answer: (c) \( 128 \)

Explanation:
1. \( 49 + 20\sqrt6 = (5 + 2\sqrt6)^2 = (\sqrt3 + \sqrt2)^4 \), so \( A = \sqrt3 + \sqrt2 \) and \( A(\sqrt3 - \sqrt2) = 1 \).
2. The GP has first term \( 8\sqrt3 \) and ratio \( \sqrt{2/3} \): \( B = \frac{8\sqrt3}{1 - \sqrt{2/3}} = \frac{24}{\sqrt3 - \sqrt2} = 24(\sqrt3 + \sqrt2) \), so the coefficient of x is 24.
3. \( \log_6 18 + \log_6 72 = \log_6 1296 = 4 \), so \( k = \log_6 10 - \log_6 5 + \log_6 2 = \log_6 4 \) and \( |\alpha - \beta| = 6^{3k/2} = 4^{3/2} = 8 \).
4. The equation is \( x^2 + 24x + C = 0 \): \( 576 - 4C = 64 \Rightarrow C = 128 \).

Teacher's Note:
1. Recognise 49 + 20 sqrt6 as a perfect fourth power.
2. 6 sqrt6 = 6 to the power 3/2.

Answer: (b) \( \frac{27}{2} \)

Explanation:
1. The product of the roots is 27, so the middle term is 3. Since \( x = 3 \) is a root, the cubic factors as \( (x - 3)(2x^2 - 13x + 18) \), giving the roots \( 2, 3, \frac92 \).
2. For a convergent infinite GP the terms must decrease: \( \frac92, 3, 2, \ldots \) with \( r = \frac23 \).
3. \( S_\infty = \frac{9/2}{1 - 2/3} = \frac{27}{2} \).

Teacher's Note:
1. The product of the roots of ax^3 + ... + d is -d/a.
2. In increasing order r = 3/2 and the series would diverge.

Answer: (a) \( 30 and 120 \)

Explanation:
1. Let \( G \) be the GM. Then \( A = G + 15,\ H = G - 12 \) and \( G^2 = AH = (G + 15)(G - 12) \Rightarrow G = 60 \).
2. \( A = 75 \), so \( a + b = 150 \), \( ab = 3600 \Rightarrow \) the numbers are 30 and 120.

Teacher's Note:
1. H = A - 27 = G - 12.
2. Check: HM = 2 x 3600/150 = 48 = 75 - 27.

Answer: (d) \( 3 \)

Explanation:
1. \( S = \frac n2(a + l) \Rightarrow 155 = \frac n2 \times 31 \Rightarrow n = 10 \).
2. \( 29 = 2 + 9d \Rightarrow d = 3 \).

Teacher's Note:
1. Find n first from S = n(a + l)/2.
2. Then use l = a + (n - 1)d.

Answer: (c) \( -\frac13 \)

Explanation:
1. First term \( -\frac12 \) and common ratio \( a = -\frac12 \).
2. \( S_\infty = \frac{-1/2}{1 + 1/2} = -\frac13 \).

Teacher's Note:
1. The ratio is equal to a, which is -1/2.
2. Check: -0.5 + 0.25 - 0.125 + 0.0625 - ... tends to -0.333.

Answer: (b) \( \frac{3}{16} \)

Explanation:
1. Write \( S_n = 3 - 3\left(\frac{3}{16}\right)^n \). This has the form \( A(1 - r^n) \) with \( r = \frac{3}{16} \).
2. Check: first term \( S_1 = 3 - \frac{9}{16} = \frac{39}{16} \), second term \( S_2 - S_1 = 3(\frac{3}{16})(1 - \frac{3}{16}) = \frac{117}{256} \); the ratio is \( \frac{117}{256} \cdot \frac{16}{39} = \frac{3}{16} \).

Teacher's Note:
1. For a GP, S_n = A(1 - r^n) so r is the base of the power.
2. Verify with S1 and S2.

Answer: (a) \( 6r - 1 \)

Explanation:
1. \( a_r = S_r - S_{r-1} = (2r + 3r^2) - [2(r-1) + 3(r-1)^2] = 2 + 3(2r - 1) = 6r - 1 \).

Teacher's Note:
1. Take the difference of consecutive sums.
2. Check: S1 = 5 = 6 - 1.

Answer: (d) \( \frac{16}{3},\ 8,\ 12,\ 18,\ 27 \)

Explanation:
1. \( 3\frac59 = \frac{32}{9} \) and \( 40\frac12 = \frac{81}{2} \). \( r^6 = \frac{81/2}{32/9} = \frac{729}{64} \Rightarrow r = \frac32 \).
2. Means: \( \frac{32}{9} \times \frac32 = \frac{16}{3},\ 8,\ 12,\ 18,\ 27 \).

Teacher's Note:
1. Six steps from first to last term, so r^6.
2. Check: 27 x 3/2 = 40.5.