JEE Mathematics Sequence and Series MCQs Set 26

Answer: (d) \( 4\ell m^2n \)

Explanation:
1. With common ratio r where \( r^4=\frac n\ell \): \( G_1=\ell r,\ G_2=\ell r^2,\ G_3=\ell r^3 \).
2. \( G_1^4+2G_2^4+G_3^4=\ell^4r^4(1+2r^4+r^8)=\ell^4r^4(1+r^4)^2=\ell^3n\cdot\frac{(\ell+n)^2}{\ell^2}=\ell n(\ell+n)^2 \).
3. Since \( \ell+n=2m \), this equals \( 4\ell m^2n \).

Teacher's Note:
1. Use the common ratio r with r⁴ = n/ℓ.
2. ℓ + n = 2m.

Answer: (c) 9

Explanation:
1. \( \frac{S_7}{S_{11}}=\frac{7(a+3d)}{11(a+5d)}=\frac6{11}\Rightarrow7a+21d=6a+30d\Rightarrow a=9d \).
2. Seventh term \( a+6d=15d \) lies between 130 and 140, so 130 < 15d < 140, giving d = 9 (since 15 × 9 = 135).

Teacher's Note:
1. Write S₇ and S₁₁ in terms of a and d.
2. d must be a natural number.

Answer: (c) 101

Explanation:
1. The numbers inside the brackets are 8/5, 12/5, 16/5, 20/5, 24/5, …, i.e. \( \frac{4(k+1)}5 \) for k = 1, 2, ….
2. Sum of squares for k = 1 to 10: \( \frac{16}{25}\sum_{k=1}^{10}(k+1)^2=\frac{16}{25}(2^2+\cdots+11^2)=\frac{16}{25}\cdot505 \).
3. This equals \( \frac{16}5\cdot101 \), so m = 101.

Teacher's Note:
1. Convert each mixed number to an improper fraction.
2. The numerators 8, 12, 16, … form an A.P.

Answer: (b) \( s\gt t\text{ and }a_{101}\lt b_{101} \)

Explanation:
1. The b's form a G.P. with ratio 2, a convex sequence. The a's are linear through the same end points (b₁, b₅₁).
2. Between the two end points a convex sequence lies below the chord, so each \( b_k\le a_k \) and hence s > t.
3. Beyond the end points the convex G.P. lies above the extended chord, so \( b_{101}\gt a_{101} \), i.e. \( a_{101}\lt b_{101} \).

Teacher's Note:
1. A G.P. with ratio > 1 is a convex sequence.
2. Compare it with the line joining the two end points.

Answer: (b) b, c and a are in A.P.

Explanation:
1. Rewrite as \( (15a)^2+(3b)^2+(5c)^2-(15a)(3b)-(3b)(5c)-(5c)(15a)=0 \).
2. This is \( \frac12\left[(15a-3b)^2+(3b-5c)^2+(5c-15a)^2\right]=0 \), so 15a = 3b = 5c, i.e. b = 5a and c = 3a.
3. Then b, c, a = 5a, 3a, a are in A.P.

Teacher's Note:
1. Substitute X = 15a, Y = 3b, Z = 5c.
2. X² + Y² + Z² − XY − YZ − ZX = 0 forces X = Y = Z.

Answer: (b) 248

Explanation:
1. The odd-position terms are \( k^2 \) and the even-position terms are \( 2k^2 \): the series is \( \sum n^2 + \sum_{\text{even}}n^2 \).
2. Sum of first 40 terms: \( \sum_{n=1}^{40}n^2+\sum_{k=1}^{20}(2k)^2=22140+4\cdot2870=33620 \). Sum of first 20: \( \sum_{n=1}^{20}n^2+\sum_{k=1}^{10}(2k)^2=2870+1540=4410 \).
3. B − 2A = 33620 − 8820 = 24800 = 100λ, so λ = 248.

Teacher's Note:
1. Split the terms into odd and even positions.
2. Use Σn² = n(n + 1)(2n + 1)/6.

Answer: (a) 34

Explanation:
1. Let the common difference be d. \( a_9+a_{43}=2a_1+50d=66 \) gives \( a_1+25d=33 \).
2. The 13 terms \( a_1,a_5,\ldots,a_{49} \) have sum \( 13(a_1+24d)=416 \), so \( a_1+24d=32 \).
3. Subtracting gives d = 1 and a₁ = 8. Then \( a_i=7+i \) and \( \sum_{i=1}^{17}(7+i)^2=\sum_{j=8}^{24}j^2=4900-140=4760 \) (since \( \sum_{1}^{24}j^2=4900 \) and \( \sum_1^{7}j^2=140 \)).
4. So 140m = 4760 and m = 34.

Teacher's Note:
1. Use the two conditions to find a₁ and d.
2. The squares run from 8² to 24².

Answer: (d) 660

Explanation:
1. The nth term is \( \frac{(2n+1)[n(n+1)/2]^2}{n(n+1)(2n+1)/6}=\frac{3n(n+1)}2 \).
2. Sum for n = 1 to 10: \( \frac32\left(385+55\right)=660 \).

Teacher's Note:
1. Replace the sums of cubes and squares by their formulas.
2. The (2n + 1) factor cancels.

Answer: (c) 76

Explanation:
1. The six terms \( a_1,a_4,\ldots,a_{16} \) form an A.P., so their sum is \( 3(a_1+a_{16})=114 \) and \( a_1+a_{16}=38 \).
2. In an A.P., \( a_6+a_{11}=a_1+a_{16} \), so the required sum is 2 × 38 = 76.

Teacher's Note:
1. The sum of terms equidistant from the ends is constant.
2. Six terms give three such pairs.

Answer: (b) −25

Explanation:
1. Let the terms be a − d, a, a + d: 3a = 33 gives a = 11.
2. Product: \( 11(121-d^2)=1155\Rightarrow d^2=16 \), so d = ±4.
3. Taking d = −4 the terms are 15, 11, 7 and \( a_{11}=15+10(-4)=-25 \) (the choice d = 4 gives 47, which is not an option).

Teacher's Note:
1. Take the three terms symmetric about the middle one.
2. Two values of d are possible.

Answer: (a) \( 2-\frac{11}{2^{19}} \)

Explanation:
1. For the A.G.P. \( \sum_{k=1}^{n}\frac k{2^k}=2-\frac{n+2}{2^n} \).
2. With n = 20: \( 2-\frac{22}{2^{20}}=2-\frac{11}{2^{19}} \).

Teacher's Note:
1. Use the standard A.G.P. result (multiply by 1/2 and subtract).
2. 22/2²⁰ = 11/2¹⁹.

Answer: (d) 3121

Explanation:
1. 91 = 7 × 13, so H.C.F.(91, n) > 1 means n is a multiple of 7 or of 13.
2. Multiples of 7 from 105 to 196: sum = 14·(105 + 196)/2 = 2107. Multiples of 13 from 104 to 195: sum = 8·(104 + 195)/2 = 1196.
3. Multiples of 91 (counted twice): 182. Total = 2107 + 1196 − 182 = 3121.

Teacher's Note:
1. Factorise 91.
2. Use inclusion–exclusion on the multiples of 7 and 13.

Answer: (d) 28

Explanation:
1. Let the terms be 8/r, 8, 8r (product 512).
2. A.P.: \( 2\times12=\frac8r+4+8r \), so \( 2r^2-5r+2=0 \) and r = 2 or 1/2.
3. For r = 2 the terms are 4, 8, 16; for r = 1/2 they are 16, 8, 4. In both cases the sum is 28.

Teacher's Note:
1. Take the middle term as 8 since the product of three terms is (middle)³.
2. Both values of r give the same set of numbers.

Answer: (c) 303

Explanation:
1. \( S_k=\frac{k+1}2 \), so \( S_k^2=\frac{(k+1)^2}4 \).
2. Sum: \( \frac14(2^2+3^2+\cdots+11^2)=\frac14(506-1)=\frac{505}4 \).
3. Then \( \frac{505}4=\frac5{12}A \) gives A = 303.

Teacher's Note:
1. Simplify Sₖ first.
2. Σj² for j = 1 to 11 is 506.

Answer: (c) 3 : 1

Explanation:
1. a₁₉ = 0 means a + 18d = 0.
2. a₄₉ = a + 48d = 30d and a₂₉ = a + 28d = 10d, so the ratio is 3 : 1.

Teacher's Note:
1. Use a = −18d.
2. Express both terms in d.

Answer: (b) \( 5^4 \)

Explanation:
1. \( \frac{a_3}{a_1}=r^2=25 \).
2. \( \frac{a_9}{a_5}=r^4=25^2=625=5^4 \).

Teacher's Note:
1. The ratio a₃/a₁ is r².
2. a₉/a₅ is r⁴.

Answer: (b) 4

Explanation:
1. Let the first term be A and common difference d: a = A + 6d, b = A + 10d, c = A + 12d.
2. G.P.: \( b^2=ac\Rightarrow(A+10d)^2=(A+6d)(A+12d)\Rightarrow2Ad=-28d^2 \), so A = −14d.
3. Then a = −8d and c = −2d, so a/c = 4.

Teacher's Note:
1. Use b² = ac for the G.P.
2. d ≠ 0, so cancel d.

Answer: (a) 52

Explanation:
1. S is the sum of the odd-numbered terms (T) and the even-numbered terms, so \( S-2T=\sum_{i=1}^{15}(a_{2i}-a_{2i-1})=15d=75 \) and d = 5.
2. a₁₀ = a₅ + 5d = 27 + 25 = 52.

Teacher's Note:
1. Pair each even term with the preceding odd term.
2. Each pair differs by d.

Answer: (d) 7820

Explanation:
1. The kth term is \( \frac{3k\cdot k(k+1)(2k+1)/6}{2k+1}=\frac{k^2(k+1)}2 \) (k = 1 gives 1 and k = 2 gives 6).
2. Sum: \( \frac12\left(\sum k^3+\sum k^2\right)=\frac12(14400+1240)=7820 \).

Teacher's Note:
1. Identify the pattern 3k, 2k + 1 and Σj².
2. The (2k + 1) in the denominator cancels.

Answer: (c) \( \frac14 \)

Explanation:
1. \( \frac{x^m}{1+x^{2m}}\le\frac12 \) since \( 1+x^{2m}\ge2x^m \) by AM–GM, and similarly for y.
2. So the expression is at most 1/4, with equality at x = y = 1.

Teacher's Note:
1. Treat the two factors separately.
2. 1 + t² ≥ 2t.

Answer: (a) (A, 50 + 46A)

Explanation:
1. \( a_n=S_n-S_{n-1}=50+\frac A2\left[n(n-7)-(n-1)(n-8)\right]=50+A(n-4) \).
2. So d = A and \( a_{50}=50+46A \).

Teacher's Note:
1. aₙ = Sₙ − Sₙ₋₁.
2. The coefficient of n gives d.

Answer: (b) a

Explanation:
1. With b = ar, c = ar²: \( 2(7ar)=3a+15ar^2\Rightarrow15r^2-14r+3=0 \), so r = 3/5 or 1/3. Since r ≤ 1/2, r = 1/3.
2. The terms are 3a, 7a/3, 5a/3 with common difference −2a/3, so the 4th term is \( \frac53a-\frac23a=a \).

Teacher's Note:
1. Impose the A.P. condition on 3a, 7b, 15c.
2. Use r ≤ 1/2 to select the root.

Answer: (b) \( \beta\gamma \)

Explanation:
1. Since \( \beta^2=\alpha\gamma \), the first equation has equal roots \( x=-\frac\beta\alpha=-r \).
2. This must satisfy \( x^2+x-1=0 \): \( r^2-r-1=0 \), i.e. \( r^2=r+1 \).
3. Then \( \alpha(\beta+\gamma)=\alpha^2(r+r^2)=\alpha^2(2r+1) \) and \( \beta\gamma=\alpha^2r^3=\alpha^2(2r+1) \), using \( r^3=r\cdot r^2=r^2+r=2r+1 \).

Teacher's Note:
1. β² = αγ makes the discriminant zero.
2. Use r² = r + 1 to reduce r³.

Answer: (a) \( -\sqrt2 \)

Explanation:
1. By AM–GM on the four non-negative numbers \( \sin^4\alpha,\ 4\cos^4\beta,\ 1,\ 1 \): \( \ge4\left(4\sin^4\alpha\cos^4\beta\right)^{1/4}=4\sqrt2\sin\alpha\cos\beta \).
2. Equality needs \( \sin^4\alpha=1=4\cos^4\beta \), so sin α = 1 and \( \cos\beta=\frac1{\sqrt2} \) (positive, since the right side is positive). Hence α = π/2, β = π/4.
3. \( \cos(\alpha+\beta)-\cos(\alpha-\beta)=-2\sin\alpha\sin\beta=-\sqrt2 \).

Teacher's Note:
1. The equation is an equality case of AM–GM.
2. Use the product-to-sum identity.