JEE Mathematics Sequence and Series MCQs Set 24

Answer: (b) 2

Explanation:
1. The nth term is \( \frac{1}{1+2+\cdots+n}=\frac{2}{n(n+1)}=2\left(\frac1n-\frac1{n+1}\right) \).
2. The series telescopes: \( S_n=2\left(1-\frac1{n+1}\right)\to2 \).

Teacher's Note:
1. Write the denominator as n(n + 1)/2.
2. Use partial fractions to telescope.

Answer: (b) \( a=-1;\ \frac x2\in I \)

Explanation:
1. Write the equation as \( (a+1)^2+\left(\operatorname{cosec}^2\theta-1\right)=0 \) with \( \theta=\frac\pi2(a+x) \), i.e. \( (a+1)^2+\cot^2\theta=0 \).
2. Both terms are non-negative, so a = −1 and \( \cot\frac\pi2(x-1)=0 \).
3. \( \frac\pi2(x-1)=\frac\pi2+m\pi\Rightarrow x=2+2m\Rightarrow\frac x2=1+m\in I \).

Teacher's Note:
1. Complete the square in a and use cosec²θ − 1 = cot²θ.
2. A sum of squares is zero only when each is zero.

Answer: (a) S + 5050

Explanation:
1. \( \sum_{k=1}^{50}(2k)^2-\sum_{k=1}^{50}(2k-1)^2=\sum_{k=1}^{50}(4k-1)=2\cdot50\cdot51-50=5050 \).
2. So the sum of the even squares is S + 5050.

Teacher's Note:
1. Subtract the odd square from the even square term by term.
2. (2k)² − (2k − 1)² = 4k − 1.

Answer: (c) \( \frac12,\ \frac13 \)

Explanation:
1. \( \rho=\frac{S_n}{S_{2n}-S_n}=\frac{2a+(n-1)d}{2a+(3n-1)d}=\frac{(2a-d)+nd}{(2a-d)+3nd} \).
2. This is independent of n only if 2a − d = 0, i.e. a/d = 1/2.
3. Then ρ = nd/3nd = 1/3.

Teacher's Note:
1. Write both sums using the formula n/2[2a + (n − 1)d].
2. For the ratio to be constant, the constant parts must vanish.

Answer: (d) 0

Explanation:
1. The sum is \( (10-1)+(10^2-1)+\cdots+(10^9-1)=1111111110-9=1111111101 \).
2. Mean = 1111111101/9 = 123456789, which contains every digit except 0.

Teacher's Note:
1. Write each number as 10^k − 1.
2. Divide the total by 9.

Answer: (c) 37

Explanation:
1. Rationalise: \( \frac{1}{\sqrt{k(k+1)}(\sqrt k+\sqrt{k+1})}=\frac1{\sqrt k}-\frac1{\sqrt{k+1}} \).
2. The sum telescopes to \( 1-\frac1{\sqrt{361}}=\frac{18}{19} \), so m + n = 37.

Teacher's Note:
1. Factor √(k(k + 1)) out of the denominator.
2. Telescoping leaves the first and last terms.

Answer: (b) \( [12,\infty) \)

Explanation:
1. Since a/2 is the mean, \( a=5^{1+x}+5^{1-x}+25^x+25^{-x} \).
2. Let \( u=5^x+5^{-x}\ge2 \): \( a=5u+u^2-2 \).
3. This increases for u ≥ 2, so its minimum is 10 + 4 − 2 = 12.

Teacher's Note:
1. Put u = 5^x + 5^−x and use u² − 2 = 25^x + 25^−x.
2. AM–GM gives u ≥ 2.

Answer: (a) 19/8

Explanation:
1. Let the first term be a and the common differences be p (for s) and q (for t). \( s_2=2t_2 \) gives \( a+p=2a+2q\Rightarrow p=a+2q \).
2. Sum condition: \( 10a+45p=15a+105q \). Substituting p: \( 40a=15q\Rightarrow q=\frac{8a}3 \), \( p=\frac{19a}3 \).
3. Ratio \( \frac pq=\frac{19}{8} \).

Teacher's Note:
1. Take a common first term a.
2. Use the two given conditions to find p and q in terms of a.

Answer: (a) \( \frac1{a_1d} \)

Explanation:
1. \( \frac1{a_ka_{k+1}}=\frac1d\left(\frac1{a_k}-\frac1{a_{k+1}}\right) \).
2. The sum telescopes to \( \frac1d\left(\frac1{a_1}-\frac1{a_{n+1}}\right) \).
3. As n → ∞, \( |a_{n+1}|\to\infty \), so the sum approaches \( \frac1{a_1d} \).

Teacher's Note:
1. Use 1/(a_k a_(k+1)) = (1/d)(1/a_k − 1/a_(k+1)).
2. The last term vanishes in the limit.

Answer: (d) \( 3(2^n-1) \)

Explanation:
1. Let the terms be a, ar, ar²: \( a(1+r+r^2)=21 \) and \( a^2(1+r^2+r^4)=189 \).
2. Using \( 1+r^2+r^4=(1+r+r^2)(1-r+r^2) \): \( a(1-r+r^2)=9 \).
3. Dividing: \( \frac{1+r+r^2}{1-r+r^2}=\frac73\Rightarrow2r^2-5r+2=0 \), so r = 2 (increasing), a = 3.
4. \( S_n=3(2^n-1) \).

Teacher's Note:
1. Factorise 1 + r² + r⁴.
2. The G.P. is increasing, so choose r = 2.

Answer: (d) \( \sqrt[4]e \)

Explanation:
1. Let \( x=\ln a \). The left side is \( \frac{2x}{1-2x} \) and the right side is \( \frac{3x}{1-x} \).
2. Since x ≠ 0: \( \frac{2}{1-2x}=\frac3{1-x}\Rightarrow2-2x=3-6x\Rightarrow x=\frac14 \), which satisfies |2x| < 1.
3. So \( a=e^{1/4} \).

Teacher's Note:
1. Sum both geometric series with common ratios 2x and x.
2. Check convergence for the value found.

Answer: (c) 1/2

Explanation:
1. Let \( u_n=\frac{1\cdot3\cdots(2n-1)}{2\cdot4\cdots2n} \), with \( u_0=1 \). The nth term of the series is \( \frac{u_n}{2n+2} \).
2. \( u_n-u_{n+1}=u_n\left(1-\frac{2n+1}{2n+2}\right)=\frac{u_n}{2n+2} \).
3. Sum from n = 1: \( u_1-u_\infty=\frac12-0=\frac12 \).

Teacher's Note:
1. Spot that each term equals the difference of two consecutive products.
2. uₙ → 0 as n → ∞.

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

Explanation:
1. The first square has side 2r. Joining mid-points gives a square whose side is the original side divided by √2.
2. So each inscribed radius is multiplied by \( \frac1{\sqrt2} \): \( r_n=r\left(\frac1{\sqrt2}\right)^{n-1}=2^{\frac{1-n}2}r \).

Teacher's Note:
1. The new square's side is the old diagonal over 2.
2. The radii form a G.P. with ratio 1/√2.

Answer: (b) 10

Explanation:
1. By AM–GM: \( a^2+b^2+c^2+d^2\ge4(abcd)^{1/2}=4 \).
2. The six products satisfy \( ab+ac+ad+bc+bd+cd\ge6(abcd)^{3/6}=6 \).
3. Equality holds for a = b = c = d = 1, so the minimum is 10.

Teacher's Note:
1. Apply AM–GM separately to the four squares and the six products.
2. Check that equality can occur.

Answer: (a) 12

Explanation:
1. Altitude ratio √3 means area ratio 3, so the areas form a G.P. with ratio 3: \( A_6=3^5A_1=243A_1 \).
2. |A₁ − A₆| = 242 A₁ = \( 968\sqrt3 \), so \( A_1=4\sqrt3 \).
3. \( \frac{\sqrt3}{4}s^2=4\sqrt3\Rightarrow s=4 \), perimeter = 12.

Teacher's Note:
1. Area of similar figures scales with the square of the linear ratio.
2. Use A = (√3/4)s².

Answer: (d) \( n=1+3k,\ \text{integer }k\ge0 \)

Explanation:
1. \( \frac{n(n+1)(2n+1)/6}{n(n+1)/2}=\frac{2n+1}{3} \).
2. This is an integer when 3 divides 2n + 1, i.e. n ≡ 1 (mod 3), so n = 1 + 3k with k ≥ 0 (for example n = 1, 4, 7, ...).

Teacher's Note:
1. Use the standard formulas for the sums.
2. Solve 2n + 1 ≡ 0 (mod 3).

Answer: (d) A is false, but R is true.

Explanation:
1. AM–GM gives \( (a+b+c)^3\ge27abc \) for positive reals, so the given condition forces equality, a = b = c.
2. Then 3a + 4a + 5a = 12 gives a = b = c = 1 and the expression equals 3, not 10. The assertion is false.
3. The reason (AM ≥ GM) is a true statement.

Teacher's Note:
1. Equality in AM–GM requires equal numbers.
2. Substitute the common value back.

Answer: (b) \( \frac5{26} \)

Explanation:
1. The graphs meet where \( (a-5)x^2+5bx+(a+b)=0 \) has a real root for every b.
2. For a ≠ 5 the discriminant must satisfy \( 25b^2-4(a-5)(a+b)\ge0 \) for all b, i.e. \( 25b^2-4(a-5)b-4a(a-5)\ge0 \).
3. This quadratic in b is non-negative for all b when \( 16(a-5)^2+400a(a-5)\le0 \), giving \( (a-5)(26a-5)\le0 \), so \( \frac5{26}\le a\le5 \). The value a = 5 fails (take b = 0), so the minimum is 5/26.

Teacher's Note:
1. Equate f and g and demand a real root for every b.
2. Treat the discriminant as a quadratic in b.

Answer: (d) 10

Explanation:
1. From the previous part, \( 5/26\le a\lt 5 \), so the possible integral values are 1, 2, 3, 4 (p = 4) and \( t_r=(r-1)(r-2)(r-3)(r-4) \).
2. t_r = 0 for r = 1, 2, 3, 4, so S₁ = S₂ = S₃ = S₄ = 0, while \( S_n\gt 0 \) for n ≥ 5.
3. The sum of the values of n is 1 + 2 + 3 + 4 = 10.

Teacher's Note:
1. First find the integral values of a.
2. Terms with r equal to any a_m are zero.

Answer: (c) \( \frac1{18} \)

Explanation:
1. The integral values are a = 1, 2, 3, 4, so \( \frac1{t_r}=\frac1{(r-1)(r-2)(r-3)(r-4)} \).
2. Telescoping: \( \frac1{(r-4)(r-3)(r-2)(r-1)}=\frac13\left[\frac1{(r-4)(r-3)(r-2)}-\frac1{(r-3)(r-2)(r-1)}\right] \).
3. Sum from r = 5: \( \frac13\cdot\frac1{1\cdot2\cdot3}=\frac1{18} \).

Teacher's Note:
1. Write the reciprocal of four consecutive factors as a difference of two reciprocals of three factors.
2. The leftover first term is for r = 5.

Answer: (c) \( \sqrt2\text{ and }-\sqrt2 \)

Explanation:
1. \( \sin x=\frac{2\sin(x-y)\sin(x+y)}{\sin(x-y)+\sin(x+y)}=\frac{\sin^2x-\sin^2y}{\sin x\cos y} \).
2. So \( \sin^2x(1-\cos y)=\sin^2y\Rightarrow\sin^2x=1+\cos y=2\cos^2\frac y2 \).
3. Therefore \( \sin x\sec\frac y2=\pm\sqrt2 \).

Teacher's Note:
1. Use H = 2PQ/(P + Q) for the harmonic mean.
2. Use 1 + cos y = 2cos²(y/2).

Answer: (c) 39 or −10.5

Explanation:
1. Let the terms be a, ar, ar², ar³: \( a(r-1)=6 \) and \( ar^2(r-1)=54 \).
2. Dividing: r² = 9, so r = 3 or r = −3.
3. r = 3: a = 3 and the terms are 3, 9, 27 with sum 39. r = −3: a = −3/2 and the terms are −1.5, 4.5, −13.5 with sum −10.5.

Teacher's Note:
1. Divide the two conditions to get r².
2. Both signs of r give valid progressions.

Answer: (b) B and D only

Explanation:
1. log a, log b, log c in A.P. means \( b^2=ac \) (B is true).
2. The second triplet in A.P. gives \( 2(\log2b-\log3c)=\log3c-\log2b \), so 3 log 2b = 3 log 3c and 2b = 3c.
3. Then a : b : c = 9 : 6 : 4. These satisfy the triangle inequality (6 + 4 > 9), so (D) is true.
4. a, 2b, 3c = 9, 12, 12 are not in H.P., and 18(a + b + c)² ≠ 18(a² + b² + c²) + ab for these values, so A and C are false.

Teacher's Note:
1. Translate both A.P. conditions into relations among a, b, c.
2. Test the options with a = 9, b = 6, c = 4.

Answer: (d) all of A, B, C and D

Explanation:
1. \( x_1(1+r)=3 \) and \( x_1r^2(1+r)=12 \), so r² = 4 and r = 2 for an increasing G.P. of positive terms; x₁ = 1.
2. The terms are 1, 2, 4, 8, so A = 1×2 = 2, B = 4×8 = 32, x₁ + x₃ = 5 and x₂ + x₄ = 10.
3. All four statements are true.

Teacher's Note:
1. The sums of the two pairs give r².
2. List the four terms and check each option.

Answer: (b) 4

Explanation:
1. For roots α, β: \( \text{H.M.}=\frac{2\alpha\beta}{\alpha+\beta} \).
2. Here \( \alpha\beta=\frac{8+2\sqrt5}{5+\sqrt2} \) and \( \alpha+\beta=\frac{4+\sqrt5}{5+\sqrt2} \), so \( \text{H.M.}=\frac{2(8+2\sqrt5)}{4+\sqrt5}=\frac{4(4+\sqrt5)}{4+\sqrt5}=4 \).

Teacher's Note:
1. Use αβ and α + β from the coefficients.
2. 8 + 2√5 = 2(4 + √5).