JEE Mathematics Sequence and Series MCQs Set 22

Answer: (a) \( 8 \)

Explanation:
1. Let the numbers be \( a,ar,ar^2 \). The A.P. condition gives \( a(r-1)^2=2 \), and \( a(1+r+r^2)=14 \).
2. So \( 2r^2-5r+2=0 \), \( r=2 \) or \( \frac12 \), giving 2, 4, 8 (or 8, 4, 2).
3. The greatest is 8.

Teacher's Note:
1. Form two equations in a and r.

Answer: (c) \( 2 \)

Explanation:
1. \( \frac ax+\frac cy=\frac{2a}{a+b}+\frac{2c}{b+c} \).
2. With \( b^2=ac \) this simplifies to \( \frac{2(ab+2ac+bc)}{ab+2ac+bc}=2 \).

Teacher's Note:
1. Use b squared = ac in the numerator and denominator.

Answer: (b) G.P.

Explanation:
1. \( c=2b-a \) and \( d=\frac{b^2}a \), so \( d-c=\frac{(b-a)^2}a \).
2. With \( u=a-b \), the three numbers are \( a,u,\frac{u^2}a \), a G.P.

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

Answer: (a) \( 1 \)

Explanation:
1. A.P. case: \( 3y=15 \), \( y=5 \), \( d=-2 \), so \( a=9+4(-2)=1 \).
2. H.P. case: \( \frac3y=\frac53 \) gives \( y=\frac95 \); the reciprocals go from \( \frac19 \) to \( \frac59 \) in 2 steps, \( d'=\frac29 \), so \( \frac1a=\frac19+\frac89=1 \).
3. Both give \( a=1 \).

Teacher's Note:
1. Use the middle term of the three consecutive terms.

Answer: (c) \( 5 \)

Explanation:
1. The \( i^{th} \) A.M. is \( 2+\frac i{10} \). The reciprocals of the H.M.'s are in A.P. from \( \frac12 \) to \( \frac13 \): \( \frac1{H_i}=\frac12-\frac i{60} \).
2. \( A+\frac6H=2+\frac i{10}+3-\frac i{10}=5 \).

Teacher's Note:
1. The i-dependent parts cancel.

Answer: (b) 4, 8, 16

Explanation:
1. The middle term is \( \sqrt[3]{512}=8 \). Terms \( \frac8r,8,8r \): \( \frac8r+8,\ 14,\ 8r \) in A.P. gives \( 8r^2-20r+8=0 \), so \( r=2 \) or \( \frac12 \).
2. The numbers are 4, 8, 16 (check: 12, 14, 16).

Teacher's Note:
1. Take the terms as a/r, a, ar.

Answer: (c) \( x^2-16x+25=0 \)

Explanation:
1. The sum of the roots is 16 and the product is 25.
2. The equation is \( x^2-16x+25=0 \).

Teacher's Note:
1. Sum = 2 A.M., product = G.M. squared.

Answer: (d) \( 6 \)

Explanation:
1. A.P.: \( d=\frac19 \), \( a_4=2+\frac13=\frac73 \).
2. H.P.: \( \frac1{h_7}=\frac12+6\cdot\left(-\frac1{54}\right)=\frac7{18} \), so \( h_7=\frac{18}7 \).
3. \( a_4h_7=\frac73\cdot\frac{18}7=6 \).

Teacher's Note:
1. Use the A.P. of reciprocals.

Answer: (d) a, b, c are in H.P.

Explanation:
1. \( (c-a)^2=(a+c)(a-2b+c) \) gives \( c^2-2ac+a^2=a^2+2ac+c^2-2b(a+c) \).
2. So \( 2b(a+c)=4ac \), that is \( b=\frac{2ac}{a+c} \): H.P.

Teacher's Note:
1. Square the middle term condition.

Answer: (b) H.P.

Explanation:
1. \( \ln x,\ln y,\ln z \) are in A.P., hence so are \( 1+\ln x,1+\ln y,1+\ln z \).
2. Their reciprocals are therefore in H.P.

Teacher's Note:
1. Take logarithms of the G.P. condition.

Answer: (c) G.P.

Explanation:
1. Let \( p,q,r=P,Pk,Pk^2 \) and \( a,b,c=A,Am,Am^2 \).
2. Then \( cp=APm^2,\ bq=APmk,\ ar=APk^2 \), a G.P. with ratio \( \frac km \).

Teacher's Note:
1. Write both progressions with their ratios.

Answer: (c) (a) and (b) both are true

Explanation:
1. G.P.: \( a^2=\frac b{16} \). H.P.: \( b=\frac{2a}{6a+1} \).
2. So \( 16a^2=\frac{2a}{6a+1} \) gives \( 48a^2+8a-1=0 \), so \( a=\frac1{12} \) or \( -\frac14 \).
3. These give \( b=\frac19 \) and \( b=1 \) respectively, and both pairs satisfy the conditions.

Teacher's Note:
1. Form two equations in a and b.

Answer: (c) \( 2A \)

Explanation:
1. With the numbers a and b, \( G_1=ar,\ G_2=ar^2,\ r^3=\frac ba \).
2. \( \frac{G_1^2}{G_2}=a \) and \( \frac{G_2^2}{G_1}=ar^3=b \), so the sum is \( a+b=2A \).

Teacher's Note:
1. Use the common ratio of the G.P.

Answer: (c) H.P.

Explanation:
1. The reciprocals \( \frac{b+c}a,\frac{c+a}b,\frac{a+b}c \) are in A.P. Adding 1 to each gives \( \frac{a+b+c}a,\frac{a+b+c}b,\frac{a+b+c}c \) in A.P.
2. So \( \frac1a,\frac1b,\frac1c \) are in A.P. and a, b, c are in H.P.

Teacher's Note:
1. Add 1 to each reciprocal.

Answer: (d) 17, 9, 1

Explanation:
1. Let the numbers be \( 9+d,9,9-d \) with \( d>0 \). After adding: \( 8+d,8,12-d \), so \( 64=(8+d)(12-d) \), giving \( d^2-4d-32=0 \) and \( d=8 \) or \( -4 \).
2. Decreasing numbers need \( d=8 \): 17, 9, 1 (then 16, 8, 4).

Teacher's Note:
1. The middle number is 9.

Answer: (a) A.P.

Explanation:
1. \( -\frac ba=\frac{b^2-2ac}{c^2} \) gives \( ab^2+bc^2=2a^2c \).
2. Dividing by abc: \( \frac bc+\frac ca=\frac{2a}b \), so \( \frac ca,\frac ab,\frac bc \) are in A.P.

Teacher's Note:
1. Use the sum and product of the roots.

Answer: (d) G.P. for all \( x\ne0 \)

Explanation:
1. The exponents \( ax+1,bx+1,cx+1 \) are in A.P. for every x.
2. So the terms are in G.P. for all \( x\ne0 \).

Teacher's Note:
1. Exponents in A.P. give a G.P.

Answer: (b) \( \frac13 \)

Explanation:
1. \( (1+9d)^2=(1+d)(1+33d) \) gives \( 48d^2=16d \).
2. The non-zero value is \( d=\frac13 \).

Teacher's Note:
1. Square the middle term.

Answer: (b) \( 2 \)

Explanation:
1. With \( a,ar,ar^2 \): \( a(r-1)^2=2 \) and \( a(1+r+r^2)=14 \), so \( 2r^2-5r+2=0 \) and \( r=2 \) or \( \frac12 \).
2. The terms are 2, 4, 8 (or 8, 4, 2), so the lowest is 2.

Teacher's Note:
1. Form two equations in a and r.

Answer: (c) g is the geometric mean between a and h

Explanation:
1. \( g^2=ah \), so g is the geometric mean of a and h.

Teacher's Note:
1. G squared equals A times H.

Answer: (b) G.P.

Explanation:
1. Let \( X=x-a,Y=y-a,Z=z-a \). The H.P. condition gives \( 2Y=\frac{2(Y-X)(Y-Z)}{(Y-X)+(Y-Z)} \), so \( Y(2Y-X-Z)=(Y-X)(Y-Z) \).
2. This simplifies to \( Y^2=XZ \): G.P.

Teacher's Note:
1. Substitute differences of x - a and so on.

Answer: (c) \( A=G \)

Explanation:
1. The A.M. of the roots of the first equation is half the sum: \( A=a \).
2. The product of the roots of the second equation is \( a^2 \), so \( G=\sqrt{a^2}=a \).
3. So \( A=G \).

Teacher's Note:
1. Use sum and product of roots.

Answer: (a) \( 1:2:3 \)

Explanation:
1. Let \( d=b-a=c-b\ne0 \). The G.P. condition gives \( d^2=d\cdot a \), so \( a=d \).
2. Then \( b=2a,\ c=3a \), so the ratio is 1 : 2 : 3.

Teacher's Note:
1. Both of the first two terms equal d.

Answer: (d) \( (a_1+a_2+a_3)\left(\frac1{a_1}+\frac1{a_2}+\frac1{a_3}\right)^3\le27 \)

Explanation:
1. The first three follow from the A.M.-G.M. inequality.
2. For (d), take \( a_1=a_2=a_3=1 \): the value is \( 3\cdot27=81>27 \), so (d) is false.

Teacher's Note:
1. Test equal values.

Answer: (d) \( \frac12-\frac1{\sqrt2} \)

Explanation:
1. The A.P. gives \( b=\frac12 \) and \( a+c=1 \). The G.P. gives \( b^4=a^2c^2 \), so \( ac=\pm\frac14 \).
2. \( ac=\frac14 \) forces \( a=c \), not allowed. So \( ac=-\frac14 \) and a, c are roots of \( t^2-t-\frac14=0 \): \( t=\frac{1\pm\sqrt2}2 \).
3. Hence \( a=\frac12-\frac1{\sqrt2} \).

Teacher's Note:
1. Use b = 1/2 first.