Question 1: If a, b, c are in G.P., prove that \log a, \log b , \log c are in A.P.

Answer:

Given a, b, c are in G.P.

\therefore b^2 = ac

Taking \log on both sides we get

\log b^2 = \log ac

\Rightarrow 2 \log b = \log a + \log b

Hence \log a, \log b , \log c are in A.P.

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Question 2: If a,b, c are in G.P., prove that \frac{1}{\log_a m} , \frac{1}{\log_b m}, \frac{1}{\log_c m} are in A.P.

Answer:

Given a, b, c are in G.P.

\therefore b^2 = ac

Taking \log on both sides we get

\log b^2 = \log ac

\Rightarrow 2 \log b = \log a + \log b

\Rightarrow \frac{2}{ \log_b m} = \frac{1}{ \log_a m} + \frac{1}{ \log_c m}

Hence \frac{1}{\log_a m} , \frac{1}{\log_b m}, \frac{1}{\log_c m} are in A.P.

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Question 3: Find k such that k + 9, k - 6 and 4 form three consecutive terms of a G.P.

Answer:

Given k + 9, k - 6 and 4 form three consecutive terms of a G.P.

\Rightarrow (k-6)^2 = 4 ( k+9)

\Rightarrow k^2 + 36 - 12 k = 4 k + 36

\Rightarrow k^2 - 16k = 0

\Rightarrow k(k-16) = 0

\Rightarrow k = 0  \ or \ k = 16

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Question 4: Three numbers are in A.P. and their sum is 15 . If 1,3,9 be added to them respectively, they form a G.P. Find the numbers.

Answer:

Let the first term of the A.P. be a and the common difference be d

\therefore a_1 + a_2 + a_3 = 15

\Rightarrow a + ( a+d) ( a+2d) = 15

\Rightarrow 3a + 3 d = 15

\Rightarrow a + d = 5      … … … … … i)

Also given a_1 + 1, a_2+3, a_3+9 are in G.P

\Rightarrow a+1, a+d + 3, a+ 2d + 9 are in G.P.

\therefore  (a+d+3)^2 = (a+1) ( a + 2d + 9)

Substituting from i) we get

(5 + 3) ^2 = (5-d+1) ( 5 + d + 9)

\Rightarrow 8^2 = ( 6-d)(14+d)

\Rightarrow 64 = 84 - 14d +6d -d^2

\Rightarrow d^2 + 8d - 20 = 0

\Rightarrow (d-2)(d+10) = 0

\Rightarrow d = 2 \ or \ d = -10

When d = 2, a = 5-2 = 3 . Then A.P. is 3, 5, 7, \ldots

When d = -10, a = 5-(-10) = 15 . Then A.P. is 15, 5, -5, \ldots

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Question 5: The sum of three numbers which are consecutive terms of an A.P. is 21 . If the second number is reduced by 1 and the third is increased by 1 we obtain three consecutive terms of a G.P. Find the numbers.

Answer:

Let the three terms be a, ( a+d), (a+2d) in A.P.

Given a+ ( a+d)+ (a+2d) = 21

\Rightarrow 3a + 3d = 21

\Rightarrow a + d = 7      … … … … … i)

Also given a, ( a+d-1), ( a+2d+1) are in G.P.

\therefore ( a+d-1)^2 = a ( a + 2d + 1)

Substituting d = 7 - a we get

( a + 7 - a -1)^2 = a [ a + 2 ( 7-a) + 1 ]

\Rightarrow 36 = a ( a + 14 - 2a + 1)

\Rightarrow 36 = a ( -a + 15)

\Rightarrow a^2 - 15a + 36 = 0

\Rightarrow (a-3)(a-12) = 0

\therefore a = 3 \ or \ a = 12

When a = 3, d = 7-3 = 4 and the numbers are 3, 7, 11

When a = 12, d = 7 - 12 = -5 and the numbers are 12, 7, 2

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Question 6: The sum of three numbers a, b, c in A.P. is 18 . If a and b are each increased by 4 and c is  increased by 36 , the new numbers form a G.P. Find a, b, c .

Answer:

Let the three terms be a, ( a+d), (a+2d) in A.P.

Given a+ ( a+d)+ (a+2d) = 18

\Rightarrow 3a + 3d = 18

\Rightarrow a + d = 6      … … … … … i)

Also given a+4, ( a+d+4), ( a+2d+36) are in G.P.

\therefore ( a+d+4)^2 = (a+4) ( a + 2d + 36)

Substituting a = 6 - d we get

( 6-d+d+4)^2 = (6-d+4) ( 6-d + 2 d + 36 )

\Rightarrow 100 = (10-d)(42+d)

\Rightarrow 100 = 420 - 42 d + 10 d - d^2

\Rightarrow d^2 + 32 d - 320 = 0

\Rightarrow (d+40)(d-8) = 0

\therefore d = -40 \ or \ d = 8

When d = -40, a = 6-(-40) = 46 and the numbers are 46, 6, -34

When d = 8, a = 6 - 8 = -2 and the numbers are -2, 6, 14

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Question 7: The sum of three numbers in G.P. is 56 .  If we subtract 1, 7, 21 from these numbers in that order, we obtain an A.P. Find the numbers.

Answer:

Let the three number be a, ar, ar^2

Given a + ar + ar^2 = 56

\Rightarrow a ( 1 + r + r^2) = 56      … … … … … i)

Also a-1, ar-7, ar^2 - 21 are in A.P.

\therefore 2 ( ar - 7) = ( a-1) + ( ar^2 - 21)

\Rightarrow 2ar - 14 = a - 22 + ar^2

\Rightarrow ar^2 + a ( 1 - 2r) - 8 = 0

\Rightarrow a ( r^2 - 2r + 1 ) = 8

\Rightarrow a(1-r)^2 = 8

\Rightarrow a = \frac{8}{(1-r)^2}      … … … … … ii)

From i) and ii) we get

\frac{8}{(1-r)^2} = \frac{56}{ 1 + r + r^2}

\Rightarrow (1+r+r^2) = 7 ( 1-r)^2

\Rightarrow 1+r+r^2 = 7 + 7r^2 - 14r

\Rightarrow 6r^2 -15r + 6 = 0

\Rightarrow 2r^2 - 5r + 2 = 0

\Rightarrow (r-2)(2r-1) = 0

\Rightarrow r = 2 \ or \ r = \frac{1}{2}

When r = 2, a = \frac{8}{(1-2)^2} = 8 and the numbers are 8, 16, 32

When r = \frac{1}{2} , a = \frac{8}{(1-\frac{1}{2})^2} and the numbers are 32, 16, 8

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Question 8: If a, b, c are in G.P., prove that:

i) a(b^2 +c^2)= c(a^2 +b^2)

ii) a^2 b^2 c^2 ( \frac{1}{a^3} + \frac{1}{a^3} +\frac{1}{a^3} ) = a^3 + b^3 +c^3

iii) \frac{(a+b+c)^2}{a^2 + b^2 + c^2} = \frac{a+b+c}{a-b+c}

iv) \frac{1}{a^2 - b^2} + \frac{1}{b^2} = \frac{1}{b^2 - c^2}

v) (a + 2b + 2c) (a - 2b + 2c) = a^2 + 4c^2

Answer:

i)       Given a, b, c are in G.P.

Therefore b^2 = ac

To prove: a(b^2 +c^2)= c(a^2 +b^2)

LHS = a(b^2+c^2) = ab^2 + ac^2 = a(ac) + b^2(c) = c( a^2 + b^2) = RHS. Hence proved.

ii)      Given a, b, c are in G.P.

Therefore b^2 = ac

To prove: a^2 b^2 c^2 \Big( \frac{1}{a^3} + \frac{1}{a^3} +\frac{1}{a^3} \Big) = a^3 + b^3 +c^3

LHS = a^2 b^2 c^2 \Big( \frac{1}{a^3} + \frac{1}{a^3} +\frac{1}{a^3} \Big)

 = \frac{b^2c^2}{a} + \frac{a^2c^2}{b} + \frac{a^2b^2}{c}

 = \frac{(ac)c^2}{a} + \frac{(b^2)^2}{b} + \frac{a^2(ac)}{c}

 = a^3 + b^3 + c^3 = RHS. Hence proved.

iii)     Given a, b, c are in G.P.

Therefore b^2 = ac

To prove: \frac{(a+b+c)^2}{a^2 + b^2 + c^2}

= \frac{(a+b+c)^2}{a^2 - b^2 + c^2 + 2b^2}

= \frac{(a+b+c)^2}{a^2 - b^2 + c^2 + 2ac}

= \frac{(a+b+c)^2}{(a+b+c)(a-b+c)}

= \frac{(a+b+c)}{(a+b+c)(a-b+c)} =  RHS. Hence proved.

iv)     Given a, b, c are in G.P.

Therefore b^2 = ac

To prove: \frac{1}{a^2 - b^2} + \frac{1}{b^2} = \frac{1}{b^2 - c^2}

LHS  = \frac{1}{a^2 - b^2} + \frac{1}{b^2}

= \frac{b^2+a^2-b^2}{b^2(a^2-b^2)}

= \frac{a^2}{a^2b^2-b^4}

= \frac{a^2}{a^2(ac) - (ac)^2}

= \frac{1}{ac-c^2}

= \frac{1}{b^2-c^2} =  RHS. Hence proved.

v)       Given a, b, c are in G.P.

Therefore b^2 = ac

To prove: (a + 2b + 2c) (a - 2b + 2c) = a^2 + 4c^2

LHS = (a + 2b + 2c) (a - 2b + 2c)

= a^2 - 4b^2 + 4c^2 + 4ac

= a^2 - 4 ( ac) + 4c^2 + 4ac

= a^2 + 4c^2 = RHS. Hence proved.

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Question 9: If a, b, c, d are in G.P., prove that:

i) \frac{ab-cd}{b^2 - c^2} = \frac{a+c}{b}

ii) ( a + b+c+d)^2 = (a+b)^2 + ( b+c)^2 + (c+d)^2

iii) (b + c) (b + d) = (c + a) (c + d)

Answer:

i)       Given a, b, c, d are in G.P. Therefore

b^2 = ac      … … … … … i)          ad = bc            c^2 = bd      … … … … … ii)

To prove: \frac{ab-cd}{b^2 - c^2} = \frac{a+c}{b}

LHS = \frac{ab-cd}{b^2 - c^2}   = \frac{ab-cd}{ac-bd}   = \frac{(ab-cd)b}{(ac-bd)b}   = \frac{a(ac) - c(bd)}{(ac-bd)b}   = \frac{ab^2 - c(c^2)}{(ac-bd)b}

= \frac{a(ac) - c^3}{(ac-bd)b}   = \frac{c( a^2 - c^2)}{(ac-bd)b}   = \frac{c(a-c)(a+c)}{(ac-bd)b}   = \frac{(ac-c^2)(a+c)}{(ac-bd)b}

= \frac{(ac-bd)(a+c)}{(ac-bd)b}   = \frac{a+c}{b} = RHS. Hence proved.

ii)      Given a, b, c, d are in G.P. Therefore

b^2 = ac      … … … … … i)          ad = bc            c^2 = bd      … … … … … ii)

To prove: ( a + b+c+d)^2 = (a+b)^2 + ( b+c)^2 + (c+d)^2

LHS = ( a + b+c+d)^2

= (a+b)^2 + 2 (a+b)(c+d) + ( c+d)^2

= (a+b)^2 + 2 (ac+bc+ad+bd) + ( c+d)^2

= (a+b)^2 + 2 (b^2+bc+bc+c^2) + ( c+d)^2

= (a+b)^2 + 2 (b+c)^2 + ( c+d)^2 = RHS. Hence proved.

iii)     Given a, b, c, d are in G.P. Therefore

b^2 = ac      … … … … … i)          ad = bc            c^2 = bd      … … … … … ii)

To prove: (b + c) (b + d) = (c + a) (c + d)

LHS   = (b + c) (b + d)

= b^2 + bc + bd + cd

= ac + c^2 + ad + cd

= c ( a+c) + d ( a+c)

= ( a+c)(c+d) = RHS. Hence proved.

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Question 10: If a, b, c are in G.P., prove that the following are also in G.P.:

i) a^2 , b^2, c^2

ii) a^3, b^3, c^3

iii) a^2 + b^2 , ab + bc, b^2 +c^2

Answer:

i)       a^2 , b^2, c^2

Given a, b, c are in G.P.

Therefore b^2 = ac

\Rightarrow (b^2)^2 = (ac)^2

Therefore a^2 , b^2, c^2 are in G.P.

ii)      a^3, b^3, c^3

Given a, b, c are in G.P.

Therefore b^2 = ac

\Rightarrow (b^2)^3 = (ac)^3

To prove: a^3, b^3, c^3 are in G.P.

iii)     a^2 + b^2 , ab + bc, b^2 +c^2

Given a, b, c are in G.P.

Therefore b^2 = ac

(ab+bc)^2 = ( ab)^2 + 2 ab^2 c + ( bc)^2

= a^2 b^2 + ab^2c + ab^2 c + b^2c^2

= a^2b^2 + ac( ac) + b^2 (b^2) + b2c^2

= a^2( b^2 +c^2) + b^2 ( b^2 +c^2)

= ( a^2+b^2)(b^2+c^2)

Therefore a^2 + b^2 , ab + bc, b^2 +c^2 are in G.P.

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Question 11: If a, b, c, d are in G.P., prove that:

i) (a^2 + b^2), (b^2+c^2), (c^2+d^2) are in G.P.

ii) (a^2 - b^2), (b^2-c^2), (c^2-d^2) are in G.P.

iii) \frac{1}{a^2 + b^2}, \frac{1}{b^2+c^2}, \frac{1}{c^2+d^2} are in G.P.

iv) (a^2+b^2+c^2), (ab+bc+cd), (b^2+c^2+d^2) are in G.P.

Answer:

i)       Given a, b, c, d are in G.P. Therefore

b^2 = ac      … … … … … i)          ad = bc            c^2 = bd      … … … … … ii)

To prove: (a^2 + b^2), (b^2+c^2), (c^2+d^2) are in G.P.

(b^2+c^2)^2 = (b^2)^2 + 2 b^2 c^2 + ( c^2)^2

= (ac)^2 + b^2 c^2 + b^2c^2 + (bd)^2

= a^2c^2 + b^2 c^2 + b^2 c^2 + b^2 d^2

= a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2

= a^2 ( c^2 +d^2) + b^2 (c^2 + d^2)

= ( a^2 + b^2)(c^2+d^2)

Therefore (a^2 + b^2), (b^2+c^2), (c^2+d^2) are in G.P.

ii)     Given a, b, c, d are in G.P. Therefore

b^2 = ac      … … … … … i)          ad = bc            c^2 = bd      … … … … … ii)

To prove: (a^2 - b^2), (b^2-c^2), (c^2-d^2) are in G.P.

(b^2-c^2) = (b^2)^2 - 2b^2c^2 + ( c^2)^2

= a^2c^2 - b^2c^2 - b^2 c^2 + b^2 d^2

= c^( a^2 - b^2) - a^2d^2 + b^2 d^2

= c^2 9 a^2 - b^2) - d^2 ( a^2 - b^2)

= ( a^2 - b^2)(c^2 - d^2)

Hence (a^2 - b^2), (b^2-c^2), (c^2-d^2) are in G.P.

iii)     Given a, b, c, d are in G.P.  Let the common ratio be r

\therefore a = a, b = ar, c = ar^2, d = ar^3

 For \frac{1}{a^2 + b^2}, \frac{1}{b^2+c^2}, \frac{1}{c^2+d^2} are in G.P.

\Big[ \frac{1}{b^2+c^2} \Big]^2 = \Big[ \frac{1}{a^2 + b^2} \Big] \Big[ \frac{1}{c^2+d^2} \Big] 

\Rightarrow \Big[ \frac{1}{a^2 r^2+a^2 r^4} \Big]^2 = \Big[ \frac{1}{a^2 + a^2r^2} \Big] \Big[ \frac{1}{a^2r^4+a^2r^6} \Big] 

\Rightarrow \Big[  \frac{1}{a^4r^4 (1+r^2)^2}  \Big] = \Big[  \frac{1}{a^4r^4 (1+r^2)^2} \Big]  

\Rightarrow LHS = RHS . Hence proved.

iv)     Given a, b, c, d are in G.P. Therefore

b^2 = ac      … … … … … i)          ad = bc            c^2 = bd      … … … … … ii)

To prove: (a^2+b^2+c^2), (ab+bc+cd), (b^2+c^2+d^2) are in G.P.

(ab+bc+cd)^2= (ab)^2 + ( bc)^2 + (cd)^2 + 2ab^2c + 2 bc^2d + 2abcd

= a^2b^2 + b^2c^2+ c^2 d^2 + ab^2c + ab^2 c+ bc^2d +bc^2d ++ abcd + abcd

= a^2b^2 + b^2c^2+ c^2 d^2 + b^(b^2) + ac(ac)+c^2(c^2)+bd(bd) + bc(bc)+ad(ad)

= a^2b^2 + b^2c^2+ c^2 d^2 + b^4 +a^2c^2 +c^4 + b^2d^2 + b^2c^2+a^2d^2

= a^2( b^2+c^2+d^2) + b^2( b^2+c^2+d^2) + c^2( b^2+c^2+d^2)

= ( b^2+c^2+d^2) (a^2+b^2+c^2)

Hence (a^2+b^2+c^2), (ab+bc+cd), (b^2+c^2+d^2) are in G.P.

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Question 12: If (a -b),(b - c),(c - a) are in G.P., then prove that (a+b+c)^2= 3 ( ab + bc+cd)

Answer:

Given (a -b),(b - c),(c - a) are in G.P.

\therefore (b-c)^2 = ( a-b)(c-a)

\Rightarrow b^2 + c^2 - 2 bc = bc + ac + ab

\Rightarrow a^2 + b^2 + c^2 = bc + ac + ab    … … … … … i)

LHS = (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac

= bc + ac + ab + 2 ab + 2bc + 2 ac

= 3 ( ab + bc + ac) = RHS . Hence proved.

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Question 13: If a, b, c are in G.P. then prove that \frac{a^2 + ab + b^2}{bc+ca + ab} = \frac{b+a}{c+b}

Answer:

Given a, b, c are in G.P. \therefore b^2 = ac      … … … … … i)

LHS = \frac{a^2 + ab + b^2}{bc+ca + ab} = \frac{a^2 + ab + b^2}{bc+b^2 + ab} = \frac{a(a+b+c)}{b(a+b+c)} = \frac{a}{b}

RHS = \frac{b+a}{c+b} = \frac{b+a}{\frac{b^2}{a}+b} = \frac{a(b+a)}{b^2 + ab} = \frac{a(b+a)}{b(a+b)} = \frac{a}{b}

Hence proved.

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Question 14: If the 4^{th} , 10^{th} and 16^{th} terms of a G.P. are x, y and z respectively. Prove that x, y , z are in G.P.

Answer:

Given 4^{th} , 10^{th} and 16^{th} terms of a G.P. are x, y and z respectively

\therefore ar^3 = x            ar^9 y            ar^{15} = z

\therefore \frac{y}{x} = r^6            \frac{z}{y} = r^6

\Rightarrow  \frac{y}{x} = \frac{z}{y}

\Rightarrow y^2 = xz

\therefore x, y, z are in G .P.

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Question 15: If a, b, c are in A.P. and a, b, d are in G.P., then prove that a, a-b, d-c are in G.P.

Answer:

Given a, b, c are in A.P. \Rightarrow 2b = a + c      … … … … … i)

Also a, b, d are in G.P. \Rightarrow b^2 = ad      … … … … … ii)

To prove: ( a-b)^2 = a ( d-c)

LHS = ( a - b)^2 = a^2 - 2ab + b^2

= a^2 - a( a+c) + b^2

=a^2 - a^2 - ac + ad

=  ac + ad = a ( d-c)= RHS. Hence proved.

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Question 16: If p^{th}, q^{th}, r^{th} and s^{th} terms of an A.P. be in G.P., then prove that p-q, q-r, r-s are in G.P.

Answer:

Given p^{th}, q^{th}, r^{th} and s^{th} terms of an A.P. be in G.P. Therefore

a_p = a + ( p-1) d = A    … … … … … i)

a_q = a ( q-1) d = AR    … … … … … ii)

a_r = a + ( r-1) d = AR^2      … … … … … iii)

a_s = a + ( s-1)d = AR^3    … … … … … iv)

Subtracting ii) from i) we get

[ a +(p-1)d ] - [ a + (q-1)d ] = A - AR

\Rightarrow (p-q) d = A (1-R)      … … … … … v)

Subtracting iii) from ii) we get

[ a +(q-1)d ] - [ a + (r-1)d ] = AR - AR^2

\Rightarrow (q-r) d = AR (1-R)      … … … … … vi)

Subtracting iv from iii) we get

[ a +(r-1)d ] - [ a + (s-1)d ] = AR^2 - AR^3

\Rightarrow (r-s) d = AR^2 (1-R)    … … … … … vii)

From vi) we get

(q-r)^2 d^2 = A^2R^2 (1-R)^2

\Rightarrow (q-r)^2 d^2 = [ A(1-R)][AR^2(1-R)]

\Rightarrow (q-r)^2 d^2 = (p-q)d. (r-s)d

\Rightarrow (q-r)^2 = ( p-q)(r-s)

\therefore p-q, q-r, r-s are in G.P.

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Question 17: If \frac{1}{a+b}, \frac{1}{2b}, \frac{1}{b+c} are three consecutive terms of an A.P., prove that a, b, c are the three consecutive terms of a G.P.

Answer:

Given \frac{1}{a+b}, \frac{1}{2b}, \frac{1}{b+c} are three consecutive terms of an A.P. Therefore

\frac{2}{2b} = \frac{1}{a+b} + \frac{1}{b+c}

\Rightarrow \frac{1}{b} = \frac{b+c+a+b}{(a+b)(b+c)}

\Rightarrow (a+b0(b+c) = 2b^2 + bc + ab

\Rightarrow ab+b^2 +c + bc = b^2 + bc + ab

\Rightarrow ac = b^2

\Rightarrow a, b, c are the three consecutive terms of a G.P.

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Question 18: If x^a = x^{b/2} z^{b/2} = z^c , then prove that \frac{1}{a} , \frac{1}{b} , \frac{1}{c}   are in A.P.

Answer:

Given x^a = x^{b/2} z^{b/2} = z^c

Take \log of both sides

a \log x = \frac{b}{2} ( \log x + \log z) = c \log z

\Rightarrow a \log x = \frac{b}{2} \log x + \frac{b}{2} \log z

\Rightarrow \Big(a - \frac{b}{2} \Big) \log x = \frac{b}{2} \log z 

\Rightarrow \frac{\log x}{\log z} = \frac{\frac{b}{2}}{a - \frac{b}{2} } … … … … … i)

and \frac{b}{2} \log x + \frac{b}{2} \log z = c \log z

\Rightarrow \frac{b}{2} \log x = ( c - \frac{b}{2} ) \log z

\Rightarrow \frac{\log x}{\log z} = \frac{c- \frac{b}{2}}{\frac{b}{2} } … … … … … ii)

From i) and ii) we get

\frac{\frac{b}{2}}{a - \frac{b}{2} } = \frac{c- \frac{b}{2}}{\frac{b}{2} }

\Rightarrow \Big( \frac{b}{2} \Big)^2 = \Big( a - \frac{b}{2} \Big) \Big( c - \frac{b}{2} \Big)

\Rightarrow \frac{b^2}{4} = ac - \frac{bc}{2} - \frac{ab}{2} + \frac{b^2}{4}

\Rightarrow 2ac = bc + ab

\Rightarrow \frac{2}{b} = \frac{1}{a} +\frac{1}{b}

Hence \frac{1}{a} , \frac{1}{b} , \frac{1}{c}   are in A.P.

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Question 19: If a, b, c are in A.P. b, c, d are in G.P. and \frac{1}{c} , \frac{1}{d} , \frac{1}{e} are in A.P., prove that a, c, e are in G.P.

Answer:

Given a, b, c are in A.P. \Rightarrow 2b = a + c       … … … … … i)

Also b, c, d are in G.P. \Rightarrow c^2 = bd       … … … … … ii)

Also \frac{1}{c} , \frac{1}{d} , \frac{1}{e} are in A.P. \Rightarrow \frac{2}{d} = \frac{1}{c} + \frac{1}{e} \Rightarrow d = \frac{2ce}{c+e}

Since c^2 = bd \Rightarrow d = \frac{c^2}{b}

\frac{c^2}{b} = \frac{2ce}{c+e}

\Rightarrow c^2 = \Big( \frac{a+c}{2} \Big ) \Big( \frac{2ce}{c+e} \Big)

\Rightarrow c( c+e) = e ( a + c)

\Rightarrow c^2 + ce = ce + ea

Hence a, c, e are in G.P.

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Question 20: If a, b, c are in A.P. and, a, x, b and b, y, c are in G.P., show that x^2, b^2, y^2   are in A.P.

Answer:

Given a, b, c are in A.P.  \Rightarrow 2b = a + c       … … … … … i)

Also a, x, b are in G.P.  \Rightarrow x^2 = ab     … … … … … ii)

and b, y, c are in G.P.  \Rightarrow y^2 = bc       … … … … … iii)

\Rightarrow \frac{y^2}{b} = 2b-a

\Rightarrow \frac{y^2}{b} = 2b - \frac{x^2}{b}

\Rightarrow 2b = \frac{x^2}{b} + \frac{y^2}{b}

\Rightarrow 2b^2 = x^2 + y^2

Hence x^2, b^2, y^2   are in A.P.

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Question 21: If a, b, c are in A.P. and a, b, d are in G.P., show that a, (a - b), (d. - c) are in G.P.

Answer:

Given a, b, c are in A.P.  \Rightarrow  2b = a + c

Given a, b, d are in G.P. \Rightarrow  b^2 = ad

To prove: (a-b)^2 = a (d-c)

LHS = ( a-b)^2

= a^2 - 2ab + b^2

= a^2 - a( a+c) + ad

= a^2 - a^2 - ac + ad

= a ( d-c)

Hence a, (a - b), (d. - c) are in G.P.

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Question 22: If a, b, c are three distinct real numbers in G.P. and. a + b + c = xb , then prove that either  x <-1 , or x > 3 .

Answer:

Let r be the common ratio.

a = a ,      b=ar ,      c = ar^2

Given a + b + c = x b

\Rightarrow a + ar + ar^2 = x ( ar)

\Rightarrow ar^2 + ( 1 - x)ar + a = 0

\Rightarrow r^2 + ( 1 - x) r + 1 = 0

We know r is always real

\therefore D \geq 0

\Rightarrow (1-x)^2 - 4 ( 1) (1) \geq 0

\Rightarrow (1-x)^2 - 4 \geq 0

\Rightarrow ( 1 - x - 2)( 1 - x + 2) \geq 0

\Rightarrow (-x-1)(3-x) \geq 0

Case 1:  ( -x -1 ) \geq 0  \text{ and } (3-x) \geq 0 \Rightarrow x \leq -1 \text{ and } x \leq 3 \Rightarrow  x \leq -1

Case 2:  ( -x -1 ) \leq 0 \text{ and } (3-x) \leq 0 \Rightarrow x \geq -1 \text{ and } x \geq 3 \Rightarrow  x \geq 3

But x \neq 3   and x \neq -1 as a, b, c are distinct numbers.

Therefore x < -1 and x > 3

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Question 23: If p^{th}, q^{th} \ and \ r^{th} terms of an A.P. and G.P. are both a, b and c respectively, show that a^{b-c} b^{c-a} c^{a-b} = 1

Answer:

Given p^{th}, q^{th} \ and \ r^{th} terms of an A.P. and G.P. are both a, b and c respectively

Let a be the first term and d be the common difference, then

a_p = A + ( p-1)d = a      … … … … … i)

a_q = A + ( q-1)d = b      … … … … … ii)

a_r = A + ( r-1) d = c      … … … … … iii)

Let A' be the first term and R be the common ratio 

a_p = A'R^{p-1} = a      … … … … … iv)

a_q = A'R^{q-1} = b      … … … … …v)

a_r = A'R^{r-1} = c    … … … … … vi)

Subtracting ii) from i) we get    (p-q)d = ( a- d)    … … … … … vii)

Subtracting iii) from ii) we get     (q-r)d = ( b- c)      … … … … … viii)

Subtracting i) from iii) we get    (r-p)d = ( c- a)      … … … … … xi)

To prove: a^{b-c} \cdot b^{c-a} \cdot c^{a-b} = 1

LHS = [ A'R^{p-1}]^{b-c} \cdot [ A'R^{q-1}]^{c-a} \cdot [ A'R^{r-1}]^{a-b}

= [ A'R^{p-1}]^{(q-r)d} \cdot [ A'R^{q-1}]^{(r-p)d} \cdot [ A'R^{r-1}]^{(p-q)d}

= {A'}^{[ (q-r)d+(r-p)d +(p-q)d]} \cdot R^{[ (p-1)(q-r)d+ (q-1)(r-p)d +(r-1)(p-q)d]  }

= {A'}^{[ (q-r+r-p+p-q) d]} \cdot R^{[ pq-q-r+r+qr-r-pq+p+rp-p-rq+q]d }

= {A'}^0 \cdot R^0 = 1= RHS