JEE Mathematics Progressions Related Inequalities and Series MCQs Set 07

Answer: (c) AP

Explanation:
1. \( a + \frac{1}{bc} = \frac{abc + 1}{bc} = \frac{abc+1}{abc}\cdot a \), and similarly the other two terms equal \( \frac{abc+1}{abc}\cdot b \) and \( \frac{abc+1}{abc}\cdot c \).
2. A common factor times \( a, b, c \) keeps the AP.

Teacher's Note:
1. Put each term over a common form with abc.
2. A common multiplier keeps an AP.

Answer: (a) \( \frac{2n}{h} \)

Explanation:
1. Each \( a_k + a_{2n+1-k} = a + b \) and each \( g_kg_{2n+1-k} = ab \).
2. The sum is \( n\cdot\frac{a+b}{ab} = n\cdot\frac2h = \frac{2n}{h} \), since \( h = \frac{2ab}{a+b} \).

Teacher's Note:
1. Terms equidistant from the ends have constant sums and products.
2. Use h = 2ab/(a + b).

Answer: (d) none of these

Explanation:
1. In an AP (or GP/HP) any three terms satisfy rational ratios such as \( \frac{\sqrt y - \sqrt x}{\sqrt z - \sqrt y} \) being rational (or the corresponding GP/HP relation).
2. For distinct primes, \( \sqrt x, \sqrt y, \sqrt z \) are linearly independent over the rationals, so none of these relations can hold.

Teacher's Note:
1. Square roots of distinct primes are not rationally related.
2. A rational ratio is needed for three terms of a progression.

Answer: (a) \( \frac{21!-1}{21!} \)

Explanation:
1. \( \frac{n}{(n+1)!} = \frac{(n+1) - 1}{(n+1)!} = \frac{1}{n!} - \frac{1}{(n+1)!} \).
2. The sum telescopes to \( 1 - \frac{1}{21!} = \frac{21! - 1}{21!} \).

Teacher's Note:
1. Write n as (n + 1) - 1.
2. Then the terms telescope.

Answer: (c) \( \frac{r-q}{q-p} \)

Explanation:
1. Let the GP ratio be \( R \). \( a_p, a_q, a_r \) in GP gives \( a_q = Ra_p \) and \( a_r = Ra_q \).
2. Then \( a_r - a_q = a_q(R-1) \) and \( a_q - a_p = a_p(R-1) \), so \( \frac{r-q}{q-p} = \frac{a_q}{a_p} = R \).

Teacher's Note:
1. The same factor (R - 1) appears in both differences.
2. Divide the two AP differences.

Answer: (b) 521

Explanation:
1. The first sequence has terms \( \equiv 1 \pmod{10} \) up to \( 991 \). The second has terms \( \equiv 1 \pmod 5 \) from 31 up to \( 31 + 99 \times 5 = 526 \).
2. Common terms: numbers \( \equiv 1 \pmod{10} \) from 31 to 526, i.e. 31, 41, ..., 521. The largest is 521.

Teacher's Note:
1. Find the common difference of the common terms (the LCM of 10 and 5).
2. The shorter sequence limits the largest value.

Answer: (a) \( \dfrac{n(n+1)}{2}\cdot\dfrac{a_2 - a_1}{a_{n+1}} \)

Explanation:
1. Each denominator \( a_{2n+2-k} + a_k = 2a_{n+1} \), and each numerator \( a_{2n+2-k} - a_k = (2n+2-2k)d \).
2. Sum \( = \frac{d}{a_{n+1}}\sum_{k=1}^{n}(n+1-k) = \frac{d}{a_{n+1}}\cdot\frac{n(n+1)}{2} \), with \( d = a_2 - a_1 \).

Teacher's Note:
1. Terms equidistant from the middle add to twice the middle term.
2. Sum the resulting arithmetic series of numerators.

Answer: (b) \( 4b \)

Explanation:
1. \( 4\sqrt{ab} = 5\cdot\frac{2ab}{a+b} \) gives \( 2(a+b) = 5\sqrt{ab} \).
2. With \( t = \sqrt{a/b} \): \( 2t^2 - 5t + 2 = 0 \), so \( t = 2 \) (as \( a > b \)) and \( a = 4b \).

Teacher's Note:
1. Divide through by b to get a quadratic in √(a/b).
2. Choose the root that gives a > b.

Answer: (a) \( \frac{2000}{9801} \)

Explanation:
1. The terms are \( \frac{2}{10}, \frac{4}{10^3}, \frac{6}{10^5}, \dots \), i.e. \( \frac{2k}{10^{2k-1}} = 20\cdot\frac{k}{100^k} \).
2. \( \sum_{k \geq 1}\frac{k}{100^k} = \frac{1/100}{(1 - 1/100)^2} = \frac{100}{9801} \), so the sum is \( 20\cdot\frac{100}{9801} = \frac{2000}{9801} \).

Teacher's Note:
1. Recognise an arithmetico-geometric series.
2. Use the sum of k x^k.

Answer: (a) 6

Explanation:
1. \( S_{2n} = 3S_n \) gives \( a = \frac{(n+1)d}{2} \).
2. Then \( S_n = n^2d \) and \( S_{3n} = 6n^2d \), so the ratio is 6.

Teacher's Note:
1. Express a in terms of d and n.
2. Substitute into S_n and S_3n.

Answer: (b) 1

Explanation:
1. \( t_m - t_n = (m-n)d \) gives \( \frac1n - \frac1m = (m-n)d \), so \( d = \frac{1}{mn} \).
2. \( t_m = a + (m-1)d = \frac1n \) gives \( a = \frac{1}{mn} \).
3. \( t_{mn} = a + (mn-1)d = \frac{1}{mn} + \frac{mn-1}{mn} = 1 \).

Teacher's Note:
1. Find d first, then a.
2. Substitute into the nth-term formula.

Answer: (b) AP

Explanation:
1. For an AP, \( a_n = a + (n-1)d \) is a linear function of \( n \).
2. Equal steps in the index give equal steps in the term, so \( a_p, a_q, a_r \) are in AP exactly when \( p, q, r \) are in AP.

Teacher's Note:
1. The nth term of an AP is linear in n.
2. Equally spaced indices give equally spaced terms.

Answer: (b) 5

Explanation:
1. The second series has terms \( 3k + 1 \) up to \( 298 \). A term \( 2^j \) equals \( 3k + 1 \) when \( 2^j \equiv 1 \pmod 3 \), i.e. \( j \) is even.
2. \( 2^0, 2^2, 2^4, 2^6, 2^8 = 1, 4, 16, 64, 256 \), all at most 298; \( 2^{10} = 1024 \) is too large. So there are 5 common terms.

Teacher's Note:
1. Powers of 2 are 1 or 2 mod 3.
2. Limit by the largest term of the shorter series.

Answer: (c) \( x \geq \frac{3}{\sqrt[3]2} \)

Explanation:
1. The product of the three logs is \( \log_5 3 \cdot \log_7 5 \cdot \log_9 7 = \log_9 3 = \frac12 \) by change of base.
2. By AM-GM, \( x \geq 3\left(\frac12\right)^{1/3} = \frac{3}{\sqrt[3]2} \).

Teacher's Note:
1. The product of the logs telescopes.
2. Apply AM-GM to three positive terms.

Answer: (b) 2

Explanation:
1. For these \( \theta \), \( \cos\theta > 0 \), so by AM-GM \( \cos^3\theta + \sec^3\theta \geq 2\sqrt{\cos^3\theta\sec^3\theta} = 2 \).
2. Equality holds at \( \theta = 0 \).

Teacher's Note:
1. A positive number plus its reciprocal is at least 2.
2. Check the sign of cos θ first.

Answer: (a) 77

Explanation:
1. \( 360 = 2^3 \cdot 3^2 \cdot 5 \). The odd divisors come from \( 3^2 \cdot 5 \): their sum is \( (1 + 3 + 9)(1 + 5) = 78 \).
2. Excluding 1 (360 is even and not odd): \( 78 - 1 = 77 \).

Teacher's Note:
1. Odd divisors ignore the power of 2.
2. Remove 1 as it is not a proper divisor here.

Answer: (a) 2

Explanation:
1. Both logs are positive and reciprocals of each other.
2. So the sum is at least 2 by AM-GM, with equality at \( a = b \).

Teacher's Note:
1. Log base-change makes them reciprocals.
2. AM-GM applies to positive reciprocals.

Answer: (a) 2

Explanation:
1. For numbers \( a, b \): \( x = \frac{a+b}{2} \), \( y = a^{2/3}b^{1/3} \), \( z = a^{1/3}b^{2/3} \), so \( yz = ab \) and \( y^3 + z^3 = a^2b + ab^2 = ab(a+b) \).
2. The ratio \( = \frac{ab(a+b)}{\frac{a+b}{2}\cdot ab} = 2 \).

Teacher's Note:
1. Two GMs between a and b are a^(2/3)b^(1/3) and a^(1/3)b^(2/3).
2. yz = ab.

Answer: (d) none of these

Explanation:
1. AP: \( 2\log\frac{3b}{5c} = \log\frac{5c}{3b} \), so \( \left(\frac{3b}{5c}\right)^3 = 1 \) and \( 3b = 5c \).
2. GP: \( b^2 = ac \). With \( b = \frac{5c}{3} \): \( a = \frac{b^2}{c} = \frac{25c}{9} \). So \( a : b : c = 25 : 15 : 9 \).
3. \( 15 + 9 = 24 \lt 25 \), so the triangle inequality fails: they cannot be the sides of any triangle.

Teacher's Note:
1. Test the triangle inequality on the ratio.
2. The longest side must be less than the sum of the other two.

Answer: (a) 4 : 1

Explanation:
1. \( \frac{HM}{GM} = \frac{2\sqrt{ab}}{a+b} = \frac45 \). With \( t = \sqrt{a/b} \): \( \frac{2}{t + 1/t} = \frac45 \), so \( t + \frac1t = \frac52 \) and \( t = 2 \).
2. \( a : b = t^2 : 1 = 4 : 1 \).

Teacher's Note:
1. HM/GM depends only on the ratio.
2. Let t = √(a/b).

Answer: (d) \( 5^{\text{th}} \) term

Explanation:
1. The reciprocals form an AP: \( \frac52, \frac{23}{12} \) with common difference \( -\frac{7}{12} \), so \( \frac{1}{t_n} = \frac{37 - 7n}{12} \).
2. The reciprocal is positive for \( n \leq 5 \) and smallest (the term largest) at \( n = 5 \), where it equals \( \frac{2}{12} \), so \( t_5 = 6 \) is the largest positive term.

Teacher's Note:
1. Move to the reciprocal AP.
2. A smaller positive reciprocal gives a larger term.

Answer: (b) \( an^2 + \dfrac{b^n - 1}{b^n(b-1)} \)

Explanation:
1. \( \sum(2r-1)a = an^2 \).
2. \( \sum_{r=1}^n b^{-r} \) is a GP with first term \( \frac1b \) and ratio \( \frac1b \): sum \( = \frac{1 - b^{-n}}{b - 1} = \frac{b^n - 1}{b^n(b-1)} \).

Teacher's Note:
1. Split the sum into an AP part and a GP part.
2. Write the GP sum with negative powers.

Answer: (c) 8

Explanation:
1. Each side is \( \frac{1}{\sqrt2} \) times the previous, so each area is half the previous: area of \( S_n = \frac{100}{2^{n-1}} \).
2. \( \frac{100}{2^{n-1}} \lt 1 \) needs \( 2^{n-1} > 100 \), i.e. \( n - 1 \geq 7 \), so \( n = 8 \).

Teacher's Note:
1. The area halves each time.
2. Solve 2^(n-1) > 100.

Answer: (c) 2

Explanation:
1. The \( n \)th factor is \( (2^n)^{1/2^{n+1}} = 2^{n/2^{n+1}} \).
2. The exponent is \( \sum_{n \geq 1}\frac{n}{2^{n+1}} = \frac12\cdot2 = 1 \), so the product is \( 2^1 = 2 \).

Teacher's Note:
1. Collect the powers of 2.
2. Use Σ n/2ⁿ = 2.

Answer: (b) \( \frac{3}{2} \)

Explanation:
1. Let the numbers be \( a > b \) with \( a + b = 4 \). Then \( (a+1)b = 4 \), so \( (a+1)(4-a) = 4 \), i.e. \( a(3-a) = 0 \) and \( a = 3 \), \( b = 1 \).
2. HM \( = \frac{2ab}{a+b} = \frac{6}{4} = \frac32 \).

Teacher's Note:
1. Set up two equations in a and b.
2. HM = 2ab/(a + b).

Answer: (c) 24

Explanation:
1. Zeros come from factors of 5: \( \lfloor 100/5 \rfloor + \lfloor 100/25 \rfloor = 20 + 4 = 24 \).

Teacher's Note:
1. Count the factors of 5.
2. Add the multiples of 25 for the extra factor.