JEE Mathematics Sequence and Series MCQs Set 28

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

Explanation:
1. Take \( a, ar, ar^2 \). After doubling, \( 2(2ar) = a + ar^2 \), so \( r^2 - 4r + 1 = 0 \).
2. \( r = 2 \pm \sqrt{3} \); of the options only \( 2 + \sqrt{3} \) appears.

Teacher's Note:
1. Convert the AP condition 2b = a + c.
2. Both roots are reciprocal to each other.

Answer: (b) \( \frac{20(10^n - 1)}{81} - \frac{2n}{9} \)

Explanation:
1. Write each term as \( \frac29 (10^k - 1) \): 2 = (2/9)(9), 22 = (2/9)(99), and so on.
2. \( S_n = \frac29\left[\frac{10(10^n - 1)}{9} - n\right] = \frac{20(10^n - 1)}{81} - \frac{2n}{9} \).

Teacher's Note:
1. Factor out the repeated digit as 2 x (9, 99, 999, ...)/9.
2. Check with n = 1: the formula gives 20 x 9/81 - 2/9 = 2.

Answer: (a) \( 2 \)

Explanation:
1. \( S = 3 \cdot \frac{1}{1 - \frac14} + d \cdot \frac{\frac14}{(1 - \frac14)^2} = 4 + \frac{4d}{9} \).
2. \( 4 + \frac{4d}{9} = \frac{44}{9} \Rightarrow d = 2 \).

Teacher's Note:
1. Split into a GP part and the part with the multiples of d.
2. \( \sum k r^k = r/(1-r)^2 \).

Answer: (a) \( \frac{11}{15} \)

Explanation:
1. \( a + b = \frac{4}{11},\ ab = -\frac{2}{11} \); both roots have modulus less than 1, so \( X = \frac{1}{(1-a)(1-b)} = \frac{1}{1 - \frac{4}{11} - \frac{2}{11}} = \frac{11}{5} \).
2. With base 2000, \( Y = \sum (\log 2)^k 4^k \log 5 = \frac{\log 5}{1 - 4\log 2} = \frac{\log 5}{\log 2000 - \log 16} = \frac{\log 5}{\log 125} = \frac13 \).
3. \( XY = \frac{11}{5} \times \frac13 = \frac{11}{15} \).

Teacher's Note:
1. The logarithm base is 2000 throughout, so 1 - 4 log_2000(2) = log_2000(125).
2. The geometric series has ratio 4 log_2000 2 which is less than 1.

Answer: (d) \( 2 \)

Explanation:
1. \( b_n = (n+1)^2 + 1 - n^2 - 1 = 2n+1 \).
2. \( b_{n+1} - b_n = 2 \), a constant, so \( \{b_n\} \) is an AP with common difference 2.

Teacher's Note:
1. If consecutive differences of a sequence form an AP, the sequence is quadratic in n.
2. Always subtract two successive terms to test for an AP.

Answer: (c) \( 10 \)

Explanation:
1. \( \frac n2[-4 + 4(n-1)] = 160 \Rightarrow 2n(n-2) = 160 \Rightarrow n^2 - 2n - 80 = 0 \).
2. \( (n - 10)(n + 8) = 0 \), so \( n = 10 \).

Teacher's Note:
1. Reject the negative root.
2. Check S10 = 5(-4 + 36) = 160.

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

Explanation:
1. The kth term is \( \frac{k(k+1)}{2} \).
2. \( S_n = \frac12\left[\sum k^2 + \sum k\right] = \frac12\left[\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}\right] = \frac{n(n+1)(n+2)}{6} \).

Teacher's Note:
1. The answer counts the tetrahedral numbers.
2. Check n = 3: 1 + 3 + 6 = 10 = 3 x 4 x 5/6.

Answer: (a) \( 2 \)

Explanation:
1. \( S = \frac{rl - a}{r - 1} \Rightarrow 728 = \frac{3 \times 486 - a}{2} \Rightarrow 1456 = 1458 - a \Rightarrow a = 2 \).

Teacher's Note:
1. Use S = (rl - a)/(r - 1).
2. Check: 2, 6, 18, ..., 486 has 6 terms and sum 728.

Answer: (d) \( 10, 20, 40 \)

Explanation:
1. Take \( a, ar, ar^2 \). AP: \( 2 \cdot 5ar = 4a + 4ar^2 \Rightarrow 2r^2 - 5r + 2 = 0 \Rightarrow r = 2 \) or \( \frac12 \).
2. \( a(1 + r + r^2) = 70 \): for r = 2, \( a = 10 \): the numbers are 10, 20, 40 (r = 1/2 gives 40, 20, 10).

Teacher's Note:
1. Divide the AP condition by 2a.
2. Check: 40, 100, 160 is an AP.

Answer: (c) \( n^2(n+1) \)

Explanation:
1. \( S_n = 3 \cdot \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} = \frac{n(n+1)}{2}[(2n + 1) - 1] = n^2(n+1) \).

Teacher's Note:
1. Use the standard sums of k and k squared.
2. Check n = 2: 2 + 10 = 12 = 4 x 3.

Answer: (b) \( 495 \)

Explanation:
1. \( a_1 = 5,\ a_{15} = 61 \).
2. \( S_{15} = \frac{15}{2}(5 + 61) = 495 \).

Teacher's Note:
1. A series with a linear nth term is an AP.
2. Or compute 4 x 120 + 15.

Answer: (a) \( \frac{23}{48} \)

Explanation:
1. Pair the terms: \( \frac47 - \frac{5}{49} = \frac{23}{49} \), and the next pair is \( \frac{1}{49} \) times this.
2. \( S = \frac{23/49}{1 - 1/49} = \frac{23/49}{48/49} = \frac{23}{48} \).

Teacher's Note:
1. Pairing removes the alternating signs.
2. The common ratio of the paired GP is 1/49.

Answer: (d) \( 153 \)

Explanation:
1. \( a = 49,\ d = -5 \).
2. \( S_{17} = \frac{17}{2}[98 + 16(-5)] = \frac{17}{2} \times 18 = 153 \).

Teacher's Note:
1. The 17th term is -31; average of first and last is 9.
2. 153 = 17 x 9.

Answer: (c) \( 1071 \)

Explanation:
1. The exponents 1, 6, 11, ..., 101 form an AP with 21 terms (\( 1 + 5k,\ k = 0..20 \)).
2. The degree is the sum of the exponents: \( \frac{21}{2}(1 + 101) = 1071 \).

Teacher's Note:
1. The degree of a product is the sum of the degrees of the factors.
2. Count the terms with \( l = a + (n-1)d \).

Answer: (b) \( \frac12 \)

Explanation:
1. \( a = \frac13,\ r = \frac13 \).
2. \( S_\infty = \frac{1/3}{1 - 1/3} = \frac12 \).

Teacher's Note:
1. First term over (1 - ratio).
2. Check: 0.333 + 0.111 + 0.037 + ... tends to 0.5.

Answer: (a) \( \frac{n(2n+1)(4n+1)}{3} - 1 \)

Explanation:
1. \( S_r = \frac{r}{1 - \frac{1}{r+1}} = r + 1 \).
2. \( \sum_{r=1}^{2n-1} (r+1)^2 = \sum_{k=2}^{2n} k^2 = \frac{2n(2n+1)(4n+1)}{6} - 1 = \frac{n(2n+1)(4n+1)}{3} - 1 \).

Teacher's Note:
1. Simplify S_r first: it is r + 1.
2. Sum from k = 2 so subtract the k = 1 term.

Answer: (d) \( 160 \)

Explanation:
1. \( \sum_{p=1}^{35} \left(\frac p7 + 2\right) = \frac17 \cdot \frac{35 \cdot 36}{2} + 70 = 90 + 70 = 160 \).

Teacher's Note:
1. Split the sum into two parts.
2. Sum of first 35 integers = 630.

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

Explanation:
1. Let \( S = 1 + 2a + \cdots + n a^{n-1} \); then \( aS = a + 2a^2 + \cdots + n a^n \).
2. \( (1-a)S = 1 + a + \cdots + a^{n-1} - n a^n = \frac{1 - a^n}{1 - a} - n a^n \).
3. \( S = \frac{1 - a^n - n a^n(1 - a)}{(1-a)^2} = \frac{1 - (n+1)a^n + n a^{n+1}}{(1-a)^2} \).

Teacher's Note:
1. Multiply by the common ratio and subtract.
2. Check n = 2: 1 + 2a. The formula gives (1 - 3a^2 + 2a^3)/(1-a)^2 = (1 + 2a).

Answer: (b) \( -180 \)

Explanation:
1. Let \( S = \sum_{k=1}^{359} k \cos k^\circ \). Replacing k by \( 360 - k \) leaves the cosine unchanged, so \( S = \sum_{k=1}^{359} (360 - k)\cos k^\circ \).
2. Adding the two forms: \( 2S = 360 \sum_{k=1}^{359} \cos k^\circ = 360(0 - 1) = -360 \), because the cosines over a full circle (k = 0 to 359) add to 0 and the k = 0 term is 1.
3. Hence \( S = -180 \).

Teacher's Note:
1. The cosines of equally spaced angles over a full circle add to zero.
2. The k = 0 term of the original sum is zero, so it does not change S.

Answer: (a) \( 5, 8, 12 \)

Explanation:
1. AP: \( b = 2a - 2 \). GP: \( c^2 = 18b \). Sum: \( c = 25 - a - b = 27 - 3a \).
2. \( (27 - 3a)^2 = 18(2a - 2) \Rightarrow a^2 - 22a + 85 = 0 \Rightarrow a = 5 \) (a = 17 gives b = 32, outside the range).
3. So a = 5, b = 8, c = 12.

Teacher's Note:
1. Check: 2, 5, 8 is an AP and 8, 12, 18 is a GP.
2. Reject the solution that violates 2 < number < 18.

Answer: (a) \( 8 and 59 \)

Explanation:
1. From the 2nd and 31st terms, \( 29d = \frac12 - \frac{31}{4} = -\frac{29}{4} \Rightarrow d = -\frac14 \).
2. First term \( = 7\tfrac34 + \tfrac14 = 8 \).
3. \( -6.5 = 8 + (n-1)(-0.25) \Rightarrow n - 1 = 58 \Rightarrow n = 59 \).

Teacher's Note:
1. The 31st term is 29 steps after the 2nd.
2. Check: a + 58d = 8 - 14.5 = -6.5.

Answer: (d) \( \frac{81}{208} \)

Explanation:
1. Use inclusion and exclusion. Let \( T = \sum 3^{-i} = \frac32,\ U = \sum 9^{-i} = \frac98,\ V = \sum 27^{-i} = \frac{27}{26} \).
2. Distinct triples: \( T^3 - 3TU + 2V = \frac{27}{8} - \frac{81}{16} + \frac{27}{13} = \frac{81}{208} \).

Teacher's Note:
1. All triples minus those with at least two equal indices.
2. Triples with all three equal were subtracted three times, so add 2V.

Answer: (c) \( \frac83 \)

Explanation:
1. The nth term is \( \frac{2^n - 1}{4^{n-1}} = 4\left[\left(\tfrac12\right)^n - \left(\tfrac14\right)^n\right] \).
2. \( S = 4\left[1 - \frac13\right] = \frac83 \).

Teacher's Note:
1. Split each term as a difference of two GPs.
2. Check: 1 + 0.75 + 0.4375 + 0.234 + 0.121 + ... approaches 2.67.

Answer: (b) \( 0.999 \)

Explanation:
1. \( r = \frac{-1.11}{1.665} = -\frac23 \).
2. \( S_\infty = \frac{1.665}{1 + \frac23} = 1.665 \times \frac35 = 0.999 \).

Teacher's Note:
1. 1.11 / 1.665 = 2/3.
2. Check the third term: 1.11 x 2/3 = 0.74.

Answer: (a) \( 23 \)

Explanation:
1. The nth term is \( \frac{(2n-1)(2n+1)}{2^n} = \frac{4n^2 - 1}{2^n} \).
2. \( \sum \frac{n^2}{2^n} = 6 \) and \( \sum \frac{1}{2^n} = 1 \), so the sum is \( 4 \times 6 - 1 = 23 \).

Teacher's Note:
1. Write the numerator as a difference of squares.
2. The sum of n squared over 2 to the n is 6.