JEE Mathematics Sequence and Series MCQs Set 30

Answer: (c) \( n(n+1)a - n^2 b \)

Explanation:
1. First term \( 2a - b \), difference \( 2a - 2b \).
2. \( S_n = \frac n2[2(2a - b) + (n-1)(2a - 2b)] = \frac n2[2na + 2a - 2nb] = n(n+1)a - n^2 b \).

Teacher's Note:
1. The nth term is 2na - (2n - 1)b.
2. Sum the a-parts and b-parts separately: 2a x n(n+1)/2 and b x n^2.

Answer: (b) \( 24.3 \)

Explanation:
1. \( r = \frac13 \).
2. \( S_\infty = \frac{16.2}{1 - \frac13} = 16.2 \times \frac32 = 24.3 \).

Teacher's Note:
1. First term / (1 - r).
2. Each term is one third of the previous.

Answer: (a) \( \frac{n(n+1)(2n+7)}{6} \)

Explanation:
1. \( \sum (k^2 + 2k) = \frac{n(n+1)(2n+1)}{6} + n(n+1) = \frac{n(n+1)(2n + 1 + 6)}{6} = \frac{n(n+1)(2n+7)}{6} \).

Teacher's Note:
1. Check n = 2: 3 + 8 = 11; formula: 2 x 3 x 11/6 = 11.
2. Take n(n+1)/6 out as a common factor.

Answer: (a) \( 518 \)

Explanation:
1. Sum 1 means \( a = 1 - r \), and the second term is \( r(1 - r) \). Two distinct series with the same second term have ratios \( r_1, r_2 \) with \( r_1 + r_2 = 1 \).
2. The third term \( r^2(1 - r) = \frac18 \Rightarrow 8r^3 - 8r^2 + 1 = 0 \Rightarrow (2r - 1)(4r^2 - 2r - 1) = 0 \). \( r = \frac12 \) would not give distinct series, so \( r = \frac{1 \pm \sqrt5}{4} \).
3. The second term is \( r(1 - r) = \frac{\sqrt5 - 1}{8} \): \( m = 5,\ n = 1,\ p = 8 \), so \( 100m + 10n + p = 518 \).

Teacher's Note:
1. r1 + r2 = 1 because the common second term r(1 - r) is symmetric.
2. Check: r = 0.809 gives r(1 - r) = 0.809 x 0.191 = 0.1545 = (sqrt5 - 1)/8.

Answer: (a) \( 33 \)

Explanation:
1. The first AP ends at 200 and the second at 240. Common terms are the common multiples of 2 and 3, that is multiples of 6.
2. Multiples of 6 up to 200: \( 6, 12, \ldots, 198 \), which gives \( \lfloor 200/6 \rfloor = 33 \) terms (all are present in both lists).

Teacher's Note:
1. Common terms of two APs form an AP with common difference equal to the LCM of the differences.
2. The smaller of the two last terms limits the count.

Answer: (d) \( \cot 1^\circ \)

Explanation:
1. Let \( S = \sum_{k=1}^{90} 2k \sin 2k^\circ \). Pairing \( n \) with \( 180 - n \), where \( \sin(180 - n)^\circ = \sin n^\circ \), gives \( 2S = 180 \sum_{k=1}^{90} \sin 2k^\circ \).
2. Using \( \sum_{k=1}^{90} \sin 2k^\circ = \frac{\sin 90^\circ \sin 91^\circ}{\sin 1^\circ} = \cot 1^\circ \), we get \( S = 90 \cot 1^\circ \).
3. The average is \( S/90 = \cot 1^\circ \) (numerically 57.29).

Teacher's Note:
1. The sine values repeat symmetrically about 90 degrees.
2. The sum of sines of an AP of angles uses the formula sin(nd/2) sin(first + last)/2 over sin(d/2).

Answer: (c) \( -\frac12 \)

Explanation:
1. \( -9\tfrac34 = -\tfrac{39}{4} \). With 19 means there are 21 terms, so \( d = \frac{-\frac{39}{4} - \frac14}{20} = \frac{-10}{20} = -\frac12 \).

Teacher's Note:
1. d = (last - first)/(number of means + 1).
2. The 10th mean is -4.5.

Answer: (b) \( \frac97\left[1 - \left(\frac{16}{9}\right)^n\right] \)

Explanation:
1. \( a = 3,\ r = -\frac43 \).
2. \( S_{2n} = \frac{3[1 - r^{2n}]}{1 - r} = \frac{3[1 - (\frac{16}{9})^n]}{\frac73} = \frac97\left[1 - \left(\frac{16}{9}\right)^n\right] \).

Teacher's Note:
1. r^(2n) = (r^2)^n = (16/9)^n.
2. Check n = 1: S2 = 3 - 4 = -1; formula: (9/7)(-7/9) = -1.

Answer: (a) \( 2 \)

Explanation:
1. \( H = \frac{2PQ}{P+Q} \).
2. \( \frac{H}{P} + \frac{H}{Q} = H \cdot \frac{P + Q}{PQ} = 2 \).

Teacher's Note:
1. Write the reciprocal form 1/H = (1/P + 1/Q)/2.
2. The answer is independent of P and Q.

Answer: (d) \( 1 \)

Explanation:
1. The reciprocals form an AP: \( a + 9D = \frac{1}{21},\ a + 20D = \frac{1}{10} \). Subtracting, \( 11D = \frac{11}{210} \), so \( D = \frac{1}{210} \) and \( a = \frac{1}{210} \).
2. So the nth reciprocal is \( \frac{n}{210} \) and the 210th term of the HP is 1.

Teacher's Note:
1. Work with the reciprocal AP.
2. The nth term of the HP is 210/n.

Answer: (a) \( a = -\tfrac12,\ b = -\tfrac18 \)

Explanation:
1. GP: \( (a^2)^2 = ab \Rightarrow b = a^3 \). AP: \( 2b = a + a^2 \Rightarrow 2a^3 = a + a^2 \Rightarrow 2a^2 - a - 1 = 0 \Rightarrow a = 1 \) or \( -\frac12 \).
2. Since a is negative, \( a = -\frac12 \) and \( b = -\frac18 \).

Teacher's Note:
1. Use b = a cubed from the GP.
2. Check the AP: -1/2, -1/8, 1/4 has d = 3/8.

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

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

Teacher's Note:
1. Factorise 1 + r^2 + r^4 as a difference of squares.
2. Check n = 1: 2/(2 x 3) = 1/3.

Answer: (b) \( 1 : 2 \)

Explanation:
1. If the sums of n terms are in the ratio \( f(n) \), the ratio of the mth terms is \( f(2m-1) \).
2. For m = 24, n = 47: \( (3 \cdot 47 - 13) : (5 \cdot 47 + 21) = 128 : 256 = 1 : 2 \).

Teacher's Note:
1. Put n = 2m - 1 because \( S_{2m-1} = (2m-1) a_m \).
2. Do not substitute n = 24.

Answer: (a) \( 6, -3, \tfrac32, \ldots \)

Explanation:
1. \( \frac{a}{1 - r} = 4 \Rightarrow a = 4(1 - r) \). \( \frac{a^3}{1 - r^3} = 192 \).
2. \( \frac{64(1 - r)^3}{1 - r^3} = 192 \Rightarrow \frac{(1 - r)^2}{1 + r + r^2} = 3 \Rightarrow 2r^2 + 5r + 2 = 0 \Rightarrow r = -\frac12 \) (the other root -2 is rejected).
3. \( a = 6 \): the series is 6, -3, 3/2, ...

Teacher's Note:
1. The cubes of a GP form a GP with ratio r cubed.
2. |r| must be less than 1.

Answer: (d) \( \frac{27}{58} \)

Explanation:
1. \( r = \frac{0.015}{0.45} = \frac{1}{30} \).
2. \( S_\infty = \frac{0.45}{1 - \frac{1}{30}} = \frac{0.45 \times 30}{29} = \frac{13.5}{29} = \frac{27}{58} \).

Teacher's Note:
1. Compute the ratio from the first two terms.
2. 13.5/29 = 27/58.

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

Explanation:
1. The odd numbers 1, 3, 5, ... form an AP with a = 1, d = 2.
2. \( S_n = \frac n2[2 + 2(n-1)] = n^2 \).

Teacher's Note:
1. Check: 1 + 3 + 5 + 7 = 16.
2. The sum of the first n even numbers is n(n + 1).

Answer: (b) \( \frac{9 + 3\sqrt3}{2} \)

Explanation:
1. \( r = \frac{1}{\sqrt3} \).
2. \( S_\infty = \frac{3}{1 - \frac1{\sqrt3}} = \frac{3\sqrt3}{\sqrt3 - 1} = \frac{3\sqrt3(\sqrt3 + 1)}{2} = \frac{9 + 3\sqrt3}{2} \).

Teacher's Note:
1. Rationalise by multiplying by the conjugate.
2. About 7.10.

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

Explanation:
1. \( S - xS = 1 + 2x + 2x^2 + 2x^3 + \cdots = 1 + \frac{2x}{1 - x} = \frac{1 + x}{1 - x} \).
2. \( S = \frac{1 + x}{(1 - x)^2} \).

Teacher's Note:
1. Subtract x S from S to get a GP.
2. Check x = 1/2: 1 + 1.5 + 1.25 + 0.875 + ... = 6 = 1.5/0.25.

Answer: (d) \( - \frac{463}{192} \)

Explanation:
1. \( a = -\frac13,\ r = -\frac32 \).
2. \( S_7 = \frac{-\frac13\left[(-\frac32)^7 - 1\right]}{-\frac32 - 1} = - \frac{463}{192} \).

Teacher's Note:
1. (-3/2)^7 = -2187/128.
2. Keep careful track of the signs.

Answer: (c) \( 5049 \)

Explanation:
1. There are 50 odd numbered terms (1st, 3rd, ..., 99th), forming an AP whose average is the middle term, the 50th term of the original AP.
2. So \( 50 \, a_{50} = 2550 \), giving \( a_{50} = 51 \).
3. For 99 terms the sum is \( 99 \, a_{50} = 99 \times 51 = 5049 \).

Teacher's Note:
1. For an odd number of terms the sum is the number of terms times the middle term.
2. Odd numbered terms also form an AP.

Answer: (b) \( \frac{5^p - 1}{4} \)

Explanation:
1. \( a = 1,\ r = 5 \).
2. \( S_p = \frac{5^p - 1}{5 - 1} = \frac{5^p - 1}{4} \).

Teacher's Note:
1. Check p = 2: (25 - 1)/4 = 6 = 1 + 5.
2. Use the formula for r > 1.

Answer: (a) \( 3, 5, 7, 9 \)

Explanation:
1. Take \( a - 3d, a - d, a + d, a + 3d \): the sum \( 4a = 24 \Rightarrow a = 6 \).
2. Product: \( (36 - 9d^2)(36 - d^2) = 945 \Rightarrow d^2 = 1 \) (with integer d).
3. The numbers are 3, 5, 7, 9 (3 x 5 x 7 x 9 = 945).

Teacher's Note:
1. Use the odd multiples of d for four terms.
2. Check the product by multiplying.

Answer: (d) \( 2184 \)

Explanation:
1. The terms \( 6, 18, 54, \ldots \) form a GP with first term 6 and ratio 3.
2. \( S_6 = \frac{6(3^6 - 1)}{3 - 1} = 3 \times 728 = 2184 \).

Teacher's Note:
1. Identify the first term as the value at i = 1.
2. Use S = a(r^n - 1)/(r - 1).

Answer: (c) \( 4950 \)

Explanation:
1. Let \( \alpha = \frac\pi{12} \). GP condition: \( \tan^2\alpha = \tan(\alpha - x)\tan(\alpha + x) = \frac{\tan^2\alpha - \tan^2 x}{1 - \tan^2\alpha \tan^2 x} \).
2. This simplifies to \( \tan^2 x(\tan^4\alpha - 1) = 0 \Rightarrow \tan x = 0 \Rightarrow x = m\pi \).
3. In [0, 314]: \( m\pi \le 314 \Rightarrow m = 0, 1, \ldots, 99 \) (since \( 100\pi \approx 314.16 \)). Sum \( = \pi \cdot \frac{99 \times 100}{2} = 4950\pi \).

Teacher's Note:
1. tan alpha is not 1 so tan^4 alpha - 1 is not zero.
2. 100 pi is slightly larger than 314.

Answer: (b) \( 64 and 4 \)

Explanation:
1. \( a + b = 68 \) and \( ab = 256 \).
2. The numbers are roots of \( t^2 - 68t + 256 = 0 \), that is \( (t - 64)(t - 4) = 0 \).

Teacher's Note:
1. Form the quadratic with sum and product of roots.
2. Check: GM = sqrt(64 x 4) = 16.