JEE Mathematics Progressions Related Inequalities and Series MCQs Set 05

Answer: (b) HP

Explanation:
1. \( \frac{1}{a+b} + \frac{1}{b+c} = \frac{a + 2b + c}{(a+b)(b+c)} \), and \( (a+b)(b+c) = b(a + 2b + c) \) using \( ac = b^2 \).
2. So the sum is \( \frac1b = \frac{2}{2b} \): the reciprocals of the three numbers are in AP, hence HP.

Teacher's Note:
1. Check reciprocals for HP.
2. Use ac = b² to simplify.

Answer: (c) GP

Explanation:
1. Let the HP numbers be \( a, b, c \) with \( b = \frac{2ac}{a+c} \). Then \( \left(a - \frac b2\right)\left(c - \frac b2\right) = ac - \frac{b(a+c)}{2} + \frac{b^2}{4} \).
2. Since \( \frac{b(a+c)}{2} = ac \), this equals \( \frac{b^2}{4} = \left(\frac b2\right)^2 \), so the new numbers are in GP.

Teacher's Note:
1. Use b(a + c) = 2ac for an HP.
2. Check the GP condition on the three new numbers.

Answer: (b) 1 : 1

Explanation:
1. With first term \( a \) and ratio \( r \): \( P = a^n r^{n(n-1)/2} \), \( S = \frac{a(r^n - 1)}{r - 1} \) and \( R = \frac{r^n - 1}{a\,r^{n-1}(r-1)} \).
2. \( P^2R^n = a^{2n}r^{n(n-1)}\cdot\frac{(r^n-1)^n}{a^n r^{n(n-1)}(r-1)^n} = \frac{a^n(r^n-1)^n}{(r-1)^n} = S^n \).

Teacher's Note:
1. Write P, S, R in terms of a and r.
2. The powers of r cancel exactly.

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

Explanation:
1. \( f \) is increasing, so the greatest value is \( f(3) = 27 \), and \( f'(0) = 3 \).
2. \( \frac{a}{1-r} = 27 \) and \( a(1-r) = 3 \). Multiplying: \( a^2 = 81 \), so \( a = 9 \) and \( 1 - r = \frac13 \), giving \( r = \frac23 \).

Teacher's Note:
1. Find the maximum using monotonicity.
2. Multiply the two equations to eliminate r.

Answer: (d) \( 49^2 + 2 \)

Explanation:
1. The differences 1, 3, 5, 7, ... are odd numbers, so \( t_n = 2 + (n-1)^2 \).
2. \( t_{50} = 2 + 49^2 \).

Teacher's Note:
1. Take first differences, then second differences.
2. The second differences are constant, so the term is quadratic.

Answer: (a) 12

Explanation:
1. \( \frac{t_6}{t_4} = r^2 = 4 \), so \( r = \pm2 \).
2. \( t_2 + t_5 = a(r + r^4) = 216 \). For \( r = 2 \): \( 18a = 216 \), so \( a = 12 \). For \( r = -2 \): \( 14a = 216 \) is not an integer.
3. So \( t_1 = 12 \).

Teacher's Note:
1. Get r² from the ratio of terms.
2. Reject the root that gives non-integer terms.

Answer: (c) GP

Explanation:
1. GP: \( a_3^2 = a_2a_4 \). HP: \( a_4 = \frac{2a_3a_5}{a_3 + a_5} \). AP: \( a_2 = \frac{a_1 + a_3}{2} \).
2. Then \( a_3^2 = \frac{a_1 + a_3}{2}\cdot\frac{2a_3a_5}{a_3 + a_5} \) gives \( a_3(a_3 + a_5) = a_5(a_1 + a_3) \), so \( a_3^2 = a_1a_5 \): GP.

Teacher's Note:
1. Substitute the AM and HM expressions into the GP condition.
2. The cross terms cancel.

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

Explanation:
1. Let the GP ratio be \( R = \frac{a_q}{a_p} = \frac{a_r}{a_q} \). Then \( a_q - a_p = a_p(R-1) \) and \( a_r - a_q = a_q(R-1) \).
2. In the AP these differences are \( (q-p)d \) and \( (r-q)d \), so \( \frac{r-q}{q-p} = \frac{a_q}{a_p} \).

Teacher's Note:
1. Compare AP differences with GP differences.
2. Both differences carry the factor (R - 1).

Answer: (b) common ratio 3, first number 4

Explanation:
1. Let the numbers be \( a, ar, ar^2 \). AP: \( 2(ar + 8) = a + ar^2 \), so \( a(r-1)^2 = 16 \). Second GP: \( (ar+8)^2 = a(ar^2 + 64) \), so \( ar + 4 = 4a \), i.e. \( a(4 - r) = 4 \).
2. Dividing: \( (r-1)^2 = 4(4 - r) \), so \( r^2 + 2r - 15 = 0 \), \( r = 3 \) or \( -5 \). Positive numbers need \( r = 3 \) and \( a = 4 \).

Teacher's Note:
1. Set up the AP and GP conditions.
2. Reject r = -5 because the numbers must be positive.

Answer: (b) \( f(2n) - 16f(n) \)

Explanation:
1. \( \sum_{r=1}^{n}(2r-1)^4 \) is the sum of the fourth powers of the first \( n \) odd numbers, i.e. all fourth powers up to \( 2n \) minus the even ones.
2. Even fourth powers: \( \sum(2r)^4 = 16f(n) \). So the sum is \( f(2n) - 16f(n) \).

Teacher's Note:
1. Odd = all minus even.
2. The even terms factor out 2⁴ = 16.

Answer: (c) 17

Explanation:
1. The values up to \( n \) occupy \( \frac{n(n+1)}{2} \) places. \( \frac{16\cdot17}{2} = 136 \lt 150 \) and \( \frac{17\cdot18}{2} = 153 \geq 150 \).
2. So the 150th term is 17.

Teacher's Note:
1. Use triangular numbers for the cumulative count.
2. Find where 150 falls.

Answer: (c) 12 cm

Explanation:
1. Let the edges be \( \frac ar, a, ar \). Volume \( a^3 = 216 \), so \( a = 6 \).
2. Surface area \( 2\cdot36\left(\frac1r + r + 1\right) = 252 \) gives \( r + \frac1r = \frac52 \), so \( r = 2 \) (or \( \frac12 \)). The edges are 3, 6, 12 cm.

Teacher's Note:
1. Take the middle edge as the geometric mean.
2. The surface area gives r + 1/r.

Answer: (a) 2

Explanation:
1. \( H = \frac{2PQ}{P + Q} \) and \( \frac1P + \frac1Q = \frac{P+Q}{PQ} \).
2. Their product is \( 2 \).

Teacher's Note:
1. Write HM in its standard form.
2. The factors cancel.

Answer: (a) \( bc = ad \)

Explanation:
1. AP: \( 2b = a + c \). HP: \( c = \frac{2bd}{b+d} \), i.e. \( c(b + d) = 2bd \).
2. Substitute \( 2b = a + c \): \( cb + cd = (a + c)d \), so \( bc = ad \).

Teacher's Note:
1. Write each progression as an equation.
2. Substitute 2b = a + c into the HP relation.

Answer: (c) \( \dfrac{n(a_1 + a_{2n})}{\sqrt{a_1} + \sqrt{a_{n+1}}} \)

Explanation:
1. In an AP, \( a_k + a_{2n+1-k} = a_1 + a_{2n} \) is constant. Also \( \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} = \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d} \).
2. The sum telescopes to \( \frac{a_1 + a_{2n}}{d}\left(\sqrt{a_{n+1}} - \sqrt{a_1}\right) \), and \( d = \frac{a_{n+1} - a_1}{n} \).
3. This equals \( \frac{n(a_1 + a_{2n})}{\sqrt{a_1} + \sqrt{a_{n+1}}} \).

Teacher's Note:
1. Rationalise each denominator to telescope.
2. Pair terms equidistant from the ends.

Answer: (c) greater than or equal to n

Explanation:
1. By AM-GM, \( \frac{\text{sum}}{n} \geq (\text{product})^{1/n} = 1 \).
2. So the sum is at least \( n \).

Teacher's Note:
1. AM-GM links sum and product.
2. The sum need not be an integer.

Answer: (c) \( x \lt -2 \) or \( x > 2 \)

Explanation:
1. A GP with common ratio \( \frac2x \) converges when \( \left|\frac2x\right| \lt 1 \).
2. That means \( |x| > 2 \), i.e. \( x \lt -2 \) or \( x > 2 \).

Teacher's Note:
1. An infinite GP converges only for
2. r
3. < 1.
4. Solve the modulus inequality.

Answer: (d) \( \frac{1}{3} \)

Explanation:
1. Let \( y = xr \), \( z = xr^2 \). AP: \( x + 3xr^2 = 4xr \), so \( 3r^2 - 4r + 1 = 0 \).
2. \( r = 1 \) or \( \frac13 \). The numbers are distinct, so \( r = \frac13 \).

Teacher's Note:
1. Write y and z in terms of x and r.
2. Reject r = 1 because the numbers are distinct.

Answer: (a) \( \frac{n^2(n+1)}{2} \)

Explanation:
1. For odd \( n \), the sum is the even-case sum for \( n - 1 \) plus the last (odd-indexed) term \( n^2 \): \( \frac{(n-1)n^2}{2} + n^2 = \frac{n^2(n+1)}{2} \).
2. Check \( n = 3 \): \( 1 + 8 + 9 = 18 = \frac{9\cdot4}{2} \).

Teacher's Note:
1. Add the last term to the previous even sum.
2. Verify with a small n.

Answer: (b) 4

Explanation:
1. Sum of all terms \( = \) sum of odd-placed terms \( + \) sum of even-placed terms \( = S_{odd}(1 + r) \).
2. \( 1 + r = 5 \), so \( r = 4 \).

Teacher's Note:
1. Even-placed terms are r times the odd-placed ones.
2. So S = S_odd (1 + r).

Answer: (a) \( \frac{1}{a} \) + \( \frac{1}{c} \)

Explanation:
1. For an HP, \( b = \frac{2ac}{a+c} \), so \( b - a = \frac{a(c-a)}{a+c} \) and \( b - c = \frac{c(a-c)}{a+c} \).
2. The sum \( = \frac{a+c}{c-a}\left(\frac1a - \frac1c\right) = \frac{a+c}{ac} = \frac1a + \frac1c \).

Teacher's Note:
1. Write b as the HM of a and c.
2. Combine the fractions carefully.

Answer: (a) 4

Explanation:
1. By AM-GM, \( 4^x + 4^{1-x} \geq 2\sqrt{4^x\cdot 4^{1-x}} = 2\sqrt4 = 4 \).
2. Equality holds when \( 4^x = 4^{1-x} \), i.e. \( x = \frac12 \).

Teacher's Note:
1. Multiply the two terms to remove x.
2. The product 4^x · 4^(1-x) is constant.

Answer: (c) GP

Explanation:
1. \( t_{n+1} \) is the middle term, so \( 2\log_{10} b = \log_{10} a + \log_{10} c \).
2. This gives \( b^2 = ac \), so \( a, b, c \) are in GP.

Teacher's Note:
1. The middle term of an AP is the average of the ends.
2. Logs of a GP form an AP.

Answer: (a) HP

Explanation:
1. Let \( x = 2a \), \( y = 3b \), \( z = 4c \). The equation becomes \( x^2 + y^2 + z^2 = xy + yz + zx \), which gives \( (x-y)^2 + (y-z)^2 + (z-x)^2 = 0 \).
2. So \( 2a = 3b = 4c \), i.e. \( a : b : c = \frac12 : \frac13 : \frac14 \). The reciprocals \( 2, 3, 4 \) are in AP, so \( a, b, c \) are in HP.

Teacher's Note:
1. Reduce to the symmetric equation x² + y² + z² = xy + yz + zx.
2. Equal values of 2a, 3b, 4c fix the ratios.

Answer: (a) -(p + q)

Explanation:
1. \( S_p - S_q = q - p \) gives \( (p - q)\left[a + \frac{(p+q-1)d}{2}\right] = q - p \), so \( a + \frac{(p+q-1)d}{2} = -1 \).
2. \( S_{p+q} = \frac{p+q}{2}[2a + (p+q-1)d] = \frac{p+q}{2}(-2) = -(p+q) \).

Teacher's Note:
1. Subtract the two sum equations.
2. S_{p+q} contains the same combination 2a + (p+q-1)d.

Answer: (a) -55

Explanation:
1. The sum of all ten numbers is 0, so the sum of pairwise products is \( \frac{0^2 - \sum x^2}{2} \).
2. \( \sum x^2 = 2(1 + 4 + 9 + 16 + 25) = 110 \), so the answer is \( -55 \).

Teacher's Note:
1. Use (Σx)² = Σx² + 2Σxy.
2. The numbers cancel in pairs in Σx.