JEE Mathematics Sequence and Series MCQs Set 27

Answer: (a) A, B and D

Explanation:
1. Since \( \alpha^2=\alpha+1 \), the numbers aₙ are the Fibonacci numbers 1, 1, 2, 3, 5, … and bₙ are the Lucas numbers 1, 3, 4, 7, ….
2. (A) The sum of the first n Fibonacci numbers is \( a_{n+2}-1 \): true. (B) With x = 1/10: \( \sum a_nx^n=\frac x{1-x-x^2}=\frac{0.1}{0.89}=\frac{10}{89} \): true.
3. (C) \( \sum b_nx^n=\frac{x(1+2x)}{1-x-x^2}=\frac{0.12}{0.89}=\frac{12}{89}\ne\frac8{89} \): false. (D) The Lucas numbers equal \( \alpha^n+\beta^n \): true.

Teacher's Note:
1. Identify the Fibonacci and Lucas sequences.
2. Use the generating function x/(1 − x − x²).

Answer: (d) 63

Explanation:
1. The sum is \( \frac{49^{126}-1}{48}=\frac{(49^{63}-1)(49^{63}+1)}{48} \).
2. Since 48 divides \( 49^{63}-1 \) (as 49 ≡ 1 mod 48), the sum equals \( (49^{63}+1)\cdot\frac{49^{63}-1}{48} \), so \( 49^{63}+1 \) is a factor.
3. The greatest such k is 63.

Teacher's Note:
1. Sum the G.P. and factorise 49¹²⁶ − 1 as a difference of squares.
2. 49 ≡ 1 (mod 48).

Answer: (a) 16

Explanation:
1. Let the numbers be 5 − 2d, 5 − d, 5, 5 + d, 5 + 2d (sum 25).
2. Product: \( 5(25-4d^2)(25-d^2)=2520\Rightarrow4d^4-125d^2+121=0 \), so \( d^2=1\text{ or }\frac{121}4 \).
3. The number −1/2 occurs only for d = 11/2 (5 − d = −1/2). The numbers are −6, −1/2, 5, 21/2, 16, so the greatest is 16.

Teacher's Note:
1. The middle term is 25/5 = 5.
2. Test both values of d² for the term −1/2.

Answer: (c) −171

Explanation:
1. \( a_3+a_4=r^2(a_1+a_2) \) gives \( r^2=4 \), r = ±2.
2. r = 2 gives a₁(1 + 2) = 4 so a₁ = 4/3 > 0, not allowed. r = −2 gives a₁ = −4.
3. \( S_9=-4\cdot\frac{1-(-2)^9}{1+2}=-4\cdot171=-684=4\lambda \), so λ = −171.

Teacher's Note:
1. Find r² from the two given sums.
2. Use a₁ < 0 to choose the sign of r.

Answer: (a) 29

Explanation:
1. The smallest subset containing A and B is \( A\cup B \).
2. |A| = 25, |B| = 7 and |A ∩ B| = number of multiples of 14 = 3.
3. |A ∪ B| = 25 + 7 − 3 = 29.

Teacher's Note:
1. The required subset is the union.
2. Use inclusion–exclusion.

Answer: (b) 3

Explanation:
1. \( f(x)=\frac{(2^{1+x}+2^{1-x})+(3^x+3^{-x})}2 \).
2. By AM–GM, \( 2^{1+x}+2^{1-x}\ge4 \) and \( 3^x+3^{-x}\ge2 \), with equality at x = 0.
3. So the minimum of f(x) is (4 + 2)/2 = 3.

Teacher's Note:
1. The middle term of three numbers in A.P. is their mean.
2. Apply AM–GM to each pair.

Answer: (a) \( 100\frac12 \)

Explanation:
1. 10th and 20th terms give \( 10d=\frac1{10}-\frac1{20}=\frac1{20} \), so d = 1/200 and a = 1/20 − 9/200 = 1/200.
2. So \( a_n=\frac n{200} \) and \( S_{200}=\frac1{200}\cdot\frac{200\cdot201}2=100.5 \).

Teacher's Note:
1. Subtract the two terms to get d.
2. Notice a = d.

Answer: (d) 504

Explanation:
1. Each term equals \( \frac32\sum_{j=1}^{n}j^2 \): the values are 1.5, 7.5, 21, 45, 82.5, 136.5, 210.
2. Their sum is 504.

Teacher's Note:
1. n(n + 1)(2n + 1)/6 is the sum of the first n squares.
2. Add the seven terms.

Answer: (c) y(1 − x) = 1

Explanation:
1. \( x=\frac1{1+\tan^2\theta}=\cos^2\theta \) and \( y=\frac1{1-\cos^2\theta}=\frac1{\sin^2\theta} \).
2. Then \( y(1-x)=\frac{1-\cos^2\theta}{\sin^2\theta}=1 \).

Teacher's Note:
1. Both series are infinite G.P.s with |ratio| < 1.
2. Use 1 + tan²θ = sec²θ.

Answer: (b) 150

Explanation:
1. Each odd-indexed term is r times the preceding even-indexed term: \( a_{2n+1}=r\,a_{2n} \), so \( 200=100r \) and r = 2.
2. Each even-indexed term is r times the preceding odd term, so \( \sum_{n=1}^{100}a_{2n-1}=\frac1r\sum_{n=1}^{100}a_{2n}=50 \).
3. Total of 200 terms = 100 + 50 = 150.

Teacher's Note:
1. Relate the two sums through the common ratio.
2. Split the 200 terms into odd and even positions.

Answer: (a) 14

Explanation:
1. Common terms satisfy x ≡ 3 (mod 4) and x ≡ 2 (mod 7), so x ≡ 23 (mod 28): 23, 51, 79, …
2. The first A.P. ends at 407: the common terms 23 + 28m ≤ 407 give m ≤ 13.7, i.e. m = 0 to 13, so there are 14 terms (all ≤ 709 as well).

Teacher's Note:
1. Solve the two congruences.
2. The smaller upper limit decides the count.

Answer: (d) \( \frac{x+y-xy}{(1-x)(1-y)} \)

Explanation:
1. The nth bracket is \( \frac{x^{n+1}-y^{n+1}}{x-y} \).
2. Sum: \( \frac1{x-y}\left[\frac{x^2}{1-x}-\frac{y^2}{1-y}\right]=\frac{x^2(1-y)-y^2(1-x)}{(x-y)(1-x)(1-y)}=\frac{x+y-xy}{(1-x)(1-y)} \).

Teacher's Note:
1. Each bracket is a G.P. with ratio y/x.
2. Sum the two geometric series.

Answer: (c) \( -\frac{72}5 \)

Explanation:
1. \( S_{11}=0\Rightarrow a_1+5d=0\Rightarrow d=-\frac{a_1}5 \).
2. The 12 terms a₁, a₃, …, a₂₃ have common difference 2d: \( S=\frac{12}2[2a_1+11\cdot2d]=12a_1+132d=12a_1-\frac{132}5a_1=-\frac{72}5a_1 \).

Teacher's Note:
1. Use S₁₁ = 0 to express d in terms of a₁.
2. The new A.P. has 12 terms.

Answer: (b) 4

Explanation:
1. The infinite sum is \( \frac{1/3}{1-1/3}=\frac12 \).
2. 0.16 = (2.5)⁻², so \( (0.16)^{\log_{2.5}(1/2)}=(2.5)^{-2\log_{2.5}(1/2)}=\left(\frac12\right)^{-2}=4 \).

Teacher's Note:
1. Sum the G.P. first.
2. Write 0.16 as a power of 2.5.

Answer: (a) \( \text{the }n\text{th term is }-4 \)

Explanation:
1. The terms are 20, 19.6, 19.2, 18.8, …, an A.P. with a = 20 and d = −2/5.
2. Sₙ = 488: \( \frac n2[40-\frac25(n-1)]=488\Rightarrow n^2-101n+2440=0 \), so n = 40 or n = 61.
3. For n = 40 the nth term is 4.4 (positive), so n = 61 and the 61st term is \( 20-\frac25\cdot60=-4 \).

Teacher's Note:
1. Write the mixed numbers as decimals to see d = −0.4.
2. Reject the root giving a positive nth term.

Answer: (d) 39

Explanation:
1. Three G.M.s between 3 and 243: ratio \( r^4=81 \), r = 3, giving 9, 27, 81. The 2nd G.M. is 27.
2. For m A.M.s, \( d=\frac{240}{m+1} \) and the 4th A.M. is \( 3+4d=27 \), so d = 6 and \( m+1=40 \).

Teacher's Note:
1. Find the geometric means first.
2. Use A.M. term = 3 + k·d.

Answer: (c) 9 : 7

Explanation:
1. With first term α and ratio r: \( \alpha(1+r)=3 \) and \( \alpha r^2(1+r)=6 \), so \( r^2=2 \).
2. p = αβ = α²r and q = γδ = α²r⁵, so \( \frac qp=r^4=4 \).
3. Ratio \( \frac{2q+p}{2q-p}=\frac{2\cdot4+1}{2\cdot4-1}=\frac97 \).

Teacher's Note:
1. Use the sums of the two pairs of roots.
2. Express q/p as a power of r.

Answer: (b) (2490, 248)

Explanation:
1. aₙ = 1 + (n − 1)d = 300, so (n − 1)d = 299 = 13 × 23.
2. With 15 ≤ n ≤ 50 and d an integer: n − 1 = 23 and d = 13 (n = 24).
3. Then \( a_{20}=1+19\cdot13=248 \) and \( S_{20}=\frac{20}2(1+248)=2490 \).

Teacher's Note:
1. Factorise 299.
2. Only one factor pair gives n between 15 and 50.

Answer: (a) 66

Explanation:
1. Let t = 2 sin 2α (so t ≤ 2). The terms are \( 3^{t-1},\ 14,\ 3^{4-t} \) and \( 3^{t-1}+3^{4-t}=28 \).
2. With u = 3ᵗ: \( \frac u3+\frac{81}u=28\Rightarrow u^2-84u+243=0 \), so u = 81 or 3. Since t ≤ 2, u = 3 and t = 1.
3. The terms are 1, 14, 27 with d = 13, so the sixth term is 1 + 5 × 13 = 66.

Teacher's Note:
1. Use 2 × middle = sum of ends.
2. t = 2 sin 2α cannot exceed 2.

Answer: (d) \( \frac1{26}(3^{50}-1) \)

Explanation:
1. Let the first term be a: \( ar(1+r+r^2)=3 \) and \( ar^5(1+r+r^2)=243 \), so \( r^4=81 \) and r = 3 (positive terms).
2. Then \( 3a\cdot13=3\Rightarrow a=\frac1{13} \).
3. \( S_{50}=\frac1{13}\cdot\frac{3^{50}-1}2=\frac1{26}(3^{50}-1) \).

Teacher's Note:
1. Divide the two conditions to find r.
2. Then find a and use the G.P. sum formula.

Answer: (b) a, b, c, d are in G.P.

Explanation:
1. Rewrite the left side as \( (ap-b)^2+(bp-c)^2+(cp-d)^2=0 \).
2. Each square is zero: b = ap, c = bp, d = cp, so b/a = c/b = d/c = p.
3. Hence a, b, c, d are in G.P. with common ratio p.

Teacher's Note:
1. Group the terms as a sum of three squares.
2. A sum of squares is zero only if each is zero.

Answer: (c) \( 7^2 \)

Explanation:
1. With L = log₇ x: the terms are 2L, 3L, 4L, …, 21L (20 terms), since \( \log_{7^{1/k}}x=k\log_7x \).
2. Sum = (2 + 3 + … + 21)L = 230L = 460, so L = 2 and \( x=7^2 \).

Teacher's Note:
1. Convert each logarithm to base 7.
2. The coefficients 2 to 21 form an A.P.

Answer: (b) \( \frac16 \)

Explanation:
1. S₂₅ = S₄₀ − S₂₅ means \( 2S_{25}=S_{40} \).
2. \( 25[6+24d]=20[6+39d]\Rightarrow150+600d=120+780d\Rightarrow d=\frac16 \).

Teacher's Note:
1. The next 15 terms sum to S₄₀ − S₂₅.
2. Use S = (n/2)[2a + (n − 1)d].

Answer: (a) 1

Explanation:
1. Left side: \( 2\left[nc+n(n-1)\right] \); right side: \( c(2^n-1) \). So \( c=\frac{2n(n-1)}{2^n-1-2n} \).
2. For n = 1 and 2 the denominator is ≤ 0 and the numerator is 0 or 4, so no positive c arises. n = 3 gives c = 12/1 = 12. For n ≥ 4 the denominator exceeds the numerator, so c < 1 or is not an integer (n = 4: 24/7, n = 5: 40/21, ...).
3. Only c = 12 works, so there is exactly 1 possible value.

Teacher's Note:
1. Solve the equation for c in terms of n.
2. Test small n and note that the denominator grows much faster.