JEE Mathematics Progressions Related Inequalities and Series MCQs Set 06

Answer: (b) 4

Explanation:
1. \( 2\log_{100}a = \log_{10}a \) and \( \log_a 0.0001 = -\frac{4}{\log_{10}a} \).
2. The expression is \( t + \frac4t \) with \( t = \log_{10}a > 0 \), which is at least \( 2\sqrt4 = 4 \).

Teacher's Note:
1. Change everything to base 10.
2. Apply AM-GM to t + 4/t.

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

Explanation:
1. \( 1 + 2 + \dots + 2^{k-1} = 2^k - 1 \), so the \( k \)th term is \( 1 - \frac{1}{2^k} \).
2. Summing: \( n - \sum_{k=1}^n\frac{1}{2^k} = n - \left(1 - \frac{1}{2^n}\right) = n - 1 + \frac1{2^n} \).

Teacher's Note:
1. Sum the inner GP first.
2. Then sum the resulting terms.

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

Explanation:
1. \( S_r = \left(\frac{r(r+1)}{2}\right)^2 \) and \( s_r = \frac{r(r+1)}{2} \), so \( \frac{S_r}{s_r} = \frac{r(r+1)}{2} \).
2. \( \sum\frac{r(r+1)}{2} = \frac{n(n+1)(n+2)}{6} \).

Teacher's Note:
1. The sum of cubes is the square of the sum.
2. The ratio is a triangular number.

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

Explanation:
1. The \( n \)th term is \( \frac{n+2}{n(n+1)}\left(\frac12\right)^n = \left[\frac{2}{n} - \frac{1}{n+1}\right]\left(\frac12\right)^n = \frac{1}{n\,2^{n-1}} - \frac{1}{(n+1)2^n} \).
2. The sum telescopes to \( 1 - \frac{1}{(n+1)2^n} \).

Teacher's Note:
1. Use partial fractions on (n+2)/(n(n+1)).
2. The terms telescope.

Answer: (a) 255

Explanation:
1. The product of \( (1 + x^{2^k}) \) for \( k = 0, \dots, 7 \) equals \( \frac{1 - x^{256}}{1 - x} = 1 + x + \dots + x^{255} \).
2. So \( n = 255 \).

Teacher's Note:
1. Multiply and divide by (1 - x) repeatedly.
2. The result is a full GP sum.

Answer: (a) \( \frac{\sqrt{a+rx}-\sqrt a}{x} \)

Explanation:
1. Rationalise each term: \( \frac{1}{\sqrt{a+nx} + \sqrt{a+(n-1)x}} = \frac{\sqrt{a+nx} - \sqrt{a+(n-1)x}}{x} \).
2. The sum telescopes to \( \frac{\sqrt{a+rx} - \sqrt a}{x} \).

Teacher's Note:
1. Rationalise the denominator.
2. Intermediate terms cancel.

Answer: (a) 3

Explanation:
1. The product of the three terms is \( 10^{(\log x)(\log y - \log z) + (\log y)(\log z - \log x) + (\log z)(\log x - \log y)} = 10^0 = 1 \).
2. By AM-GM the sum is at least \( 3\cdot1^{1/3} = 3 \).

Teacher's Note:
1. Take logs to find the product.
2. The exponents cancel to give product 1.

Answer: (a) 6

Explanation:
1. By AM-GM, \( ab^2 + ac^2 + bc^2 + ba^2 + ca^2 + cb^2 \geq 6\sqrt[6]{a^6b^6c^6} = 6abc \).
2. So \( \lambda = 6 \).

Teacher's Note:
1. Six terms, so use the six-term AM-GM.
2. The product of all six terms is (abc)^6.

Answer: (c) -2

Explanation:
1. \( a_n = \frac{a_{n+1} + a_{n+2}}{2} \) gives \( 2 = r + r^2 \), so \( r = 1 \) or \( -2 \).
2. The terms alternate in sign, so \( r \lt 0 \) and \( r = -2 \).

Teacher's Note:
1. Divide the AM condition by the first term.
2. Alternating signs mean a negative ratio.

Answer: (c) -2500

Explanation:
1. There are 50 factors, so the coefficient of \( x^{49} \) is the negative of the sum of the roots.
2. Sum \( = 1 + 3 + \dots + 99 = 50^2 = 2500 \), so the coefficient is \( -2500 \).

Teacher's Note:
1. The x^(n-1) coefficient is minus the sum of roots.
2. The sum of the first 50 odd numbers is 2500.

Answer: (c) \( p^3 \)

Explanation:
1. The two given relations are the same formula \( S_r = p\,r^2 \) evaluated at \( r = n \) and \( r = m \), so it holds for every \( r \).
2. Put \( r = p \): \( S_p = p \cdot p^2 = p^3 \).

Teacher's Note:
1. The formula S_r = p r² holds for all r.
2. Replace r by p.

Answer: (d) 37

Explanation:
1. The sum of the first \( n \) odd numbers is \( n^2 \).
2. \( 36^2 = 1296 \lt 1357 \) and \( 37^2 = 1369 > 1357 \), so \( n = 37 \).

Teacher's Note:
1. Sum of first n odd numbers is n².
2. Test neighbouring squares.

Answer: (c) 0

Explanation:
1. For \( n = 0 \) the expression is \( \frac{2}{a^{-1} + b^{-1}} = \frac{2ab}{a+b} \), which is the HM.
2. For other values (such as \( n = 1 \), the AM; or \( n = \frac12 \)) it is not the HM.

Teacher's Note:
1. Test n = 0 first.
2. HM = 2/(1/a + 1/b).

Answer: (c) a rational number

Explanation:
1. \( \int_0^n x\,dx = \frac{n^2}{2} \), so the sum is \( \frac12\sum_{n=1}^{10} n^2 = \frac{385}{2} \).
2. This is a rational number but not an integer.

Teacher's Note:
1. Evaluate the integral first.
2. 385/2 is not an integer.

Answer: (c) \( p \)

Explanation:
1. The common difference is \( \frac{0 - q}{q} = -1 \).
2. \( t_q = t_p + (q - p)(-1) = q - q + p = p \).

Teacher's Note:
1. Find the common difference from the two given terms.
2. Move from the p-th term to the q-th term.

Answer: (b) GP

Explanation:
1. By Cauchy-Schwarz the reverse inequality \( (a^2+b^2+c^2)(b^2+c^2+d^2) \geq (ab+bc+cd)^2 \) always holds.
2. So equality holds, which requires \( \frac ab = \frac bc = \frac cd \): the numbers are in GP.

Teacher's Note:
1. Cauchy-Schwarz gives the opposite inequality.
2. Equality means proportional vectors.

Answer: (c) 9

Explanation:
1. Smallest angle \( = 120^\circ \). Sum of interior angles \( = (n-2)\cdot 180^\circ = \frac n2[240 + 5(n-1)] \).
2. This gives \( n^2 - 25n + 144 = 0 \), so \( n = 9 \) or \( 16 \). For \( n = 16 \) the largest angle is \( 120 + 75 = 195^\circ > 180^\circ \), which is not convex.
3. So \( n = 9 \).

Teacher's Note:
1. A convex polygon needs every angle below 180°.
2. Reject the root that gives a reflex angle.

Answer: (b) GP

Explanation:
1. For a GP, \( T_n = AR^{n-1} \), so \( T_q/T_p = R^{q-p} \), \( T_r/T_q = R^{r-q} \), \( T_s/T_r = R^{s-r} \).
2. \( q - p = r - q = s - r \) (consecutive AP terms), so the ratios are equal: the terms are in GP.

Teacher's Note:
1. Index differences control the ratios of a GP.
2. Equal index gaps give equal ratios.

Answer: (b) \( \dfrac{2^{21} - 1}{2^{10}} + 19 \)

Explanation:
1. \( t_r^2 = 2^r + 2^{-r} + 2 \).
2. \( \sum 2^r = 2046 \), \( \sum 2^{-r} = 1 - 2^{-10} \), \( \sum 2 = 20 \). Total \( = 2067 - 2^{-10} = \frac{2^{21} - 1}{2^{10}} + 19 \).

Teacher's Note:
1. Square the term and sum three pieces.
2. Two GPs and one constant.

Answer: (a) AP

Explanation:
1. GP gives \( (3\,n!)^2 = n!\,(n+1)! \), so \( 9\,n! = (n+1)! \) and \( n + 1 = 9 \).
2. Then the three numbers are \( n!, 5\,n!, 9\,n! \), i.e. \( 1, 5, 9 \) times \( n! \), which are in AP.

Teacher's Note:
1. Solve for n from the GP condition.
2. Then compare the multiples 1, 5, 9.

Answer: (c) \( p^2 - q^2 = p'^2 - q'^2 \)

Explanation:
1. \( p^2 - p'^2 = \left(\frac{a+b}{2}\right)^2 - ab = \frac{(a-b)^2}{4} \) and \( q^2 - q'^2 = \frac{(b-c)^2}{4} \).
2. Since \( a - b = b - c \) in an AP, these are equal, so \( p^2 - p'^2 = q^2 - q'^2 \), i.e. \( p^2 - q^2 = p'^2 - q'^2 \).

Teacher's Note:
1. AM² minus GM² is a perfect square.
2. Equal common differences make the two squares equal.

Answer: (b) \( [3, +\infty) \)

Explanation:
1. The ratios are positive and their product is 1.
2. By AM-GM the sum is at least 3, and 3 is attained when \( x = y = z \). So the interval is \( [3, \infty) \).

Teacher's Note:
1. Same sign means every ratio is positive.
2. Equality at x = y = z closes the interval.

Answer: (c) 6

Explanation:
1. AP: \( d = \frac{3 - 2}{9} = \frac19 \), so \( a_4 = 2 + \frac39 = \frac73 \).
2. HP: the reciprocals \( \frac12, \dots, \frac13 \) form an AP with difference \( -\frac{1}{54} \), so \( \frac{1}{h_7} = \frac12 - \frac{6}{54} = \frac{7}{18} \) and \( h_7 = \frac{18}{7} \).
3. \( a_4h_7 = \frac73\cdot\frac{18}{7} = 6 \).

Teacher's Note:
1. Pair terms symmetric about the ends.
2. The product of corresponding AP and HP terms is constant.

Answer: (c) HP

Explanation:
1. Take \( a_i = \frac{1}{u + (i-1)d} \) (an HP). Then \( \frac{a_2a_3}{a_1a_4} = \frac{u(u+3d)}{(u+d)(u+2d)} \).
2. Also \( \frac{a_2 + a_3}{a_1 + a_4} = \frac{u(u+3d)}{(u+d)(u+2d)} \) and \( 3\cdot\frac{a_2 - a_3}{a_1 - a_4} = \frac{u(u+3d)}{(u+d)(u+2d)} \), so all three agree.
3. Hence the numbers are in HP.

Teacher's Note:
1. Test the conditions on a general HP.
2. Reciprocals in AP simplify all three ratios identically.

Answer: (d) \( A \geq -\frac12 \)

Explanation:
1. Squaring: \( a_i^2 = a_{i-1}^2 + 2a_{i-1} + 1 \). Summing from \( i = 2 \) to \( n \): \( a_n^2 = 2\sum_{i=1}^{n-1}a_i + (n-1) \) (as \( a_1 = 0 \)).
2. Since \( a_n^2 \geq 0 \), \( \sum_{i=1}^{n-1}a_i \geq -\frac{n-1}{2} \), so \( A \geq -\frac12 \).

Teacher's Note:
1. Square the recurrence and add.
2. A square is non-negative.

Answer: (d) \( \dfrac m2\log\dfrac{a^{2m}}{b^{m-1}} \)

Explanation:
1. The sum is \( \log a\sum(2n-1) - \log b\sum(n-1) = m^2\log a - \frac{m(m-1)}{2}\log b \).
2. This equals \( \frac m2\left[2m\log a - (m-1)\log b\right] = \frac m2\log\frac{a^{2m}}{b^{m-1}} \).

Teacher's Note:
1. Separate the log of a quotient.
2. Use the sums of 2n - 1 and n - 1.