JEE Mathematics Progressions Related Inequalities and Series MCQs Set 08

Answer: (c) 1

Explanation:
1. Let the AP have first term A and common difference d: \( A+(m-1)d=\tfrac1n \) and \( A+(n-1)d=\tfrac1m \).
2. Subtracting, \( (m-n)d=\tfrac1n-\tfrac1m=\tfrac{m-n}{mn} \), so \( d=\tfrac1{mn} \) and then \( A=\tfrac1n-\tfrac{m-1}{mn}=\tfrac1{mn} \).
3. Hence \( t_{mn}=A+(mn-1)d=\tfrac1{mn}+\tfrac{mn-1}{mn}=1 \).

Teacher's Note:
1. Subtract the two equations to get the common difference first.
2. Then find the first term and evaluate the required term.

Answer: (c) 0

Explanation:
1. GP condition: \( f(2x)^2=f(x)f(4x) \), i.e. \( (4x+1)^2=(2x+1)(8x+1) \).
2. \( 16x^2+8x+1=16x^2+10x+1 \), so \( x=0 \).
3. At \( x=0 \) all three numbers equal 1, so they are not unequal. Hence there is no valid real value.

Teacher's Note:
1. The condition gives only x = 0, so check that the numbers are unequal there.

Answer: (c) HP

Explanation:
1. Let \( a=1,\ b=r,\ c=r^2 \) (general case by scaling): the numbers are \( 1+r,\ 2r,\ r(1+r) \).
2. Their reciprocals are \( \dfrac1{1+r},\ \dfrac1{2r},\ \dfrac1{r(1+r)} \), and \( \dfrac1{1+r}+\dfrac1{r(1+r)}=\dfrac{r+1}{r(1+r)}=\dfrac1r=2\cdot\dfrac1{2r} \).
3. The reciprocals are in AP, so the numbers are in HP.

Teacher's Note:
1. Test the reciprocals of the three numbers for an AP.

Answer: (b) GP

Explanation:
1. By the Cauchy-Schwarz inequality, \( (a^2+b^2+c^2)(b^2+c^2+d^2)\ge(ab+bc+cd)^2 \).
2. The given condition therefore forces equality, which holds when \( \dfrac ab=\dfrac bc=\dfrac cd \).
3. This means a, b, c, d are in GP.

Teacher's Note:
1. The given inequality is the reverse of Cauchy-Schwarz, so equality must hold.

Answer: (c) HP

Explanation:
1. Expanding, \( (2a-3b)^2+(3b-4c)^2+(2a-4c)^2=2\left[(4a^2+9b^2+16c^2)-(6ab+12bc+8ca)\right] \), which is twice the difference of the two sides of the given equation, so it equals 0.
2. A sum of squares is zero only if each is zero: \( 2a=3b=4c \).
3. So \( a:b:c=6:4:3 \) and \( \tfrac1a:\tfrac1b:\tfrac1c=2:3:4 \) (an AP), so a, b, c are in HP.

Teacher's Note:
1. Group the terms into three perfect squares.
2. Compare the reciprocals of a, b, c.

Answer: (b) GP

Explanation:
1. \( t_1,t_{n+1},t_{2n+1} \) are equally spaced terms of an AP, so they are in AP: \( 2\log b=\log a+\log c \).
2. This gives \( b^2=ac \).
3. So a, b, c are in GP.

Teacher's Note:
1. Terms that are equally spaced in an AP form an AP themselves.

Answer: (a) AP

Explanation:
1. GP condition: \( (3\,n!)^2=n!\,(n+1)! \), so \( 9\,n!=(n+1)! \), i.e. \( n+1=9 \) and \( n=8 \).
2. The second set is \( n!,\ 5\,n!,\ 9\,n! \), proportional to 1, 5, 9.
3. These have common difference 4, so they are in AP.

Teacher's Note:
1. Reduce (n + 1)! to (n + 1) times n! to find n.

Answer: (b) p

Explanation:
1. \( A+(p-1)d=q \) and \( A+(p+q-1)d=0 \). Subtracting, \( qd=-q \), so \( d=-1 \).
2. Then \( A=q+p-1 \).
3. The qth term is \( A+(q-1)d=(q+p-1)-(q-1)=p \).

Teacher's Note:
1. Subtract the two conditions to find d and then A.

Answer: (c) HP

Explanation:
1. The right side is \( 15yz+5xz+3xy \). Put \( X=x,\ Y=3y,\ Z=5z \): the equation becomes \( X^2+Y^2+Z^2=XY+YZ+ZX \).
2. This gives \( (X-Y)^2+(Y-Z)^2+(Z-X)^2=0 \), so \( X=Y=Z \), i.e. \( x=3y=5z \).
3. Then \( \tfrac1x:\tfrac1y:\tfrac1z=1:3:5 \) (an AP), so x, y, z are in HP.

Teacher's Note:
1. Substitute X = x, Y = 3y, Z = 5z to get a symmetric equation.

Answer: (d) none of these

Explanation:
1. The first sequence ends at \( 1+10\cdot99=991 \); the second ends at \( 31+5\cdot99=526 \).
2. Common terms are \( \equiv1\pmod{10} \) and lie in the second sequence (which has terms \( \equiv1\pmod5 \)), so they are \( 31,41,51,\dots \).
3. The largest one not exceeding 526 is 521, which is not among the options.

Teacher's Note:
1. Find the range of each sequence and then the common difference of the common terms.

Answer: (a) 9

Explanation:
1. The smallest angle is \( 120^\circ \). The sum of the angles is \( (n-2)180=\dfrac n2\left[240+5(n-1)\right] \).
2. This gives \( 5n^2-125n+720=0 \), i.e. \( n^2-25n+144=0 \), so \( n=9 \) or \( n=16 \).
3. For \( n=16 \) the largest angle is \( 120+75=195^\circ>180^\circ \), which is not convex. So \( n=9 \).

Teacher's Note:
1. Reject a root that makes an interior angle greater than 180 degrees.

Answer: (b) 23952

Explanation:
1. \( 9900=2^2\cdot3^2\cdot5^2\cdot11 \).
2. The sum of all divisors is \( (1+2+4)(1+3+9)(1+5+25)(1+11)=7\cdot13\cdot31\cdot12=33852 \).
3. Subtracting 9900 gives \( 33852-9900=23952 \).

Teacher's Note:
1. Use the multiplicative formula for the sum of divisors.

Answer: (b) 78

Explanation:
1. \( 360=2^3\cdot3^2\cdot5 \), so the odd divisors are the divisors of \( 3^2\cdot5=45 \).
2. Their sum is \( (1+3+9)(1+5)=13\cdot6=78 \).
3. Every odd divisor is smaller than 360, so all are proper divisors and the sum is 78.

Teacher's Note:
1. Ignore the factors of 2 to list the odd divisors.

Answer: (b) \( 2^{10} \)

Explanation:
1. The number of terms up to and including the value \( 2^k \) is \( 1+2+4+\dots+2^k=2^{k+1}-1 \).
2. Through the value \( 2^9 \) there are \( 2^{10}-1=1023 \) terms.
3. So the 1024th to 2047th terms equal \( 2^{10} \), and the 1025th term is \( 2^{10} \).

Teacher's Note:
1. Count how many terms are used up by each value.

Answer: (a) 12

Explanation:
1. \( \dfrac{t_4}{t_6}=\dfrac1{r^2}=\dfrac14 \), so \( r=\pm2 \).
2. \( t_2+t_5=t_1(r+r^4)=216 \). For \( r=2 \): \( 18t_1=216 \), \( t_1=12 \). For \( r=-2 \): \( 14t_1=216 \) is not an integer.
3. Since the terms are integers, \( r=2 \) and \( t_1=12 \).

Teacher's Note:
1. Use the integer condition to reject the other ratio.

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

Explanation:
1. For the AP, \( a_q-a_p=(q-p)d \) and \( a_r-a_q=(r-q)d \).
2. For the GP \( a_p,a_q,a_r \) with ratio \( \rho \): \( \rho=\dfrac{a_r-a_q}{a_q-a_p} \).
3. So \( \rho=\dfrac{r-q}{q-p} \).

Teacher's Note:
1. The common ratio of three GP terms equals the ratio of successive differences.

Answer: (b) \( 17\tfrac17 \)

Explanation:
1. The reciprocals \( \tfrac1{30},\tfrac1{24},\tfrac1{20} \) have common difference \( \tfrac1{120} \), so the progression is an HP.
2. The next reciprocal is \( \tfrac1{20}+\tfrac1{120}=\tfrac7{120} \).
3. So the next term is \( \tfrac{120}7=17\tfrac17 \).

Teacher's Note:
1. Check differences first, then ratios, then reciprocals.

Answer: (c) HP

Explanation:
1. Let the numbers be \( a,ar,ar^2 \). Adding the middle number gives \( a(1+r),\ 2ar,\ ar(1+r) \).
2. The reciprocals are proportional to \( \dfrac1{1+r},\ \dfrac1{2r},\ \dfrac1{r(1+r)} \), i.e. to \( r,\ \dfrac{1+r}2,\ 1 \) (multiplying by \( r(1+r) \)); these are in AP since \( r+1=2\cdot\dfrac{1+r}2 \).
3. The reciprocals are in AP, so the numbers are in HP.

Teacher's Note:
1. Look at the reciprocals when the numbers might be in HP.

Answer: (b) \( \tfrac13 \)

Explanation:
1. \( y=xr,\ z=xr^2 \) and \( 4y=x+3z \) gives \( 4r=1+3r^2 \).
2. \( 3r^2-4r+1=0 \), so \( r=1 \) or \( r=\tfrac13 \).
3. The numbers are distinct, so \( r\ne1 \); hence \( r=\tfrac13 \).

Teacher's Note:
1. Express y and z in terms of x and r.

Answer: (a) \( |y|=3z \)

Explanation:
1. For positive a, b: \( x=\dfrac{a+b}2 \) (AM), \( y^2=ab \) (GM squared) and \( z=\dfrac{2ab}{a+b} \) (HM).
2. \( xz=\dfrac{a+b}2\cdot\dfrac{2ab}{a+b}=ab=y^2 \).
3. With \( x=9z \): \( y^2=9z^2 \), so \( |y|=3z \).

Teacher's Note:
1. Use the relation AM times HM equals GM squared.

Answer: (b) GP

Explanation:
1. AP: \( 2b=a+c \). HP: \( c=\dfrac{2de}{d+e} \) i.e. \( \dfrac2d=\dfrac1c+\dfrac1e \). GP: \( c^2=bd \).
2. Then \( d=\dfrac{c^2}b \), so \( \dfrac{2b}{c^2}=\dfrac1c+\dfrac1e \), giving \( \dfrac1e=\dfrac{2b-c}{c^2}=\dfrac{a}{c^2} \).
3. So \( c^2=ae \), and a, c, e are in GP.

Teacher's Note:
1. Express d and e in terms of a, b, c.

Answer: (d) 6

Explanation:
1. AP: \( a_4=2+\dfrac{3}{9}=\dfrac73 \) (common difference \( \tfrac19 \)).
2. HP: the reciprocals go from \( \tfrac12 \) to \( \tfrac13 \) in 9 steps, so \( \tfrac1{h_7}=\tfrac12+6\cdot\left(-\tfrac1{54}\right)=\tfrac7{18} \) and \( h_7=\tfrac{18}7 \).
3. \( a_4h_7=\dfrac73\cdot\dfrac{18}7=6 \).

Teacher's Note:
1. Handle the HP through its reciprocals, which are in AP.

Answer: (b) 3

Explanation:
1. The bracket is a GP: \( 1+3x+\dots+(3x)^5=\dfrac{1-(3x)^6}{1-3x} \).
2. So \( (1-p)\dfrac{1-(3x)^6}{1-3x}=1-p^6 \), which is satisfied by \( p=3x \) (both sides become \( 1-p^6 \)).
3. Hence \( \dfrac px=3 \).

Teacher's Note:
1. Recognise the bracket as the sum of a GP with ratio 3x.

Answer: (d) x < -2 or x > 2

Explanation:
1. This is a GP with common ratio \( \dfrac2x \).
2. The sum to infinity is finite when \( \left|\dfrac2x\right|<1 \), i.e. \( |x|>2 \).
3. So \( x<-2 \) or \( x>2 \).

Teacher's Note:
1. An infinite GP converges only when the absolute value of the ratio is below 1.

Answer: (c) 5th term

Explanation:
1. The reciprocals are in AP: \( \tfrac52,\ \tfrac{23}{12} \), with common difference \( -\tfrac7{12} \).
2. The nth reciprocal is \( \tfrac52-\tfrac{7(n-1)}{12}>0 \) for \( n-1<\tfrac{30}7 \), i.e. \( n\le5 \). The 5th reciprocal is \( \tfrac52-\tfrac{28}{12}=\tfrac16 \), the smallest positive one.
3. The largest positive term of the HP is the reciprocal of the smallest positive AP term: the 5th term, equal to 6.

Teacher's Note:
1. The largest positive HP term corresponds to the smallest positive reciprocal.