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

Answer:

\displaystyle \text{Given } a, b, c \text{are in G.P. }

\displaystyle \therefore b^2 = ac

\displaystyle \text{Taking } \log on both sides we get

\displaystyle \log b^2 = \log ac

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

\displaystyle \text{Hence } \log a, \log b , \log c \text{are in A.P. }

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

Answer:

\displaystyle \text{Given } a, b, c \text{are in G.P. }

\displaystyle \therefore b^2 = ac

\displaystyle \text{Taking } \log on both sides we get

\displaystyle \log b^2 = \log ac

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

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

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

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

Answer:

\displaystyle \text{Given } k + 9, k - 6 \text{ and } 4 form three consecutive terms of a G.P.

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

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

\displaystyle \Rightarrow k^2 - 16k = 0

\displaystyle \Rightarrow k(k-16) = 0

\displaystyle \Rightarrow k = 0 \text{ or } k = 16

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Question 4: Three numbers are in A.P. and their sum is \displaystyle 15 . If \displaystyle 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 \displaystyle a and the common difference be \displaystyle d

\displaystyle \therefore a_1 + a_2 + a_3 = 15

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

\displaystyle \Rightarrow 3a + 3 d = 15

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

\displaystyle \text{Also } \text{Given } a_1 + 1, a_2+3, a_3+9 are in G.P

\displaystyle \Rightarrow a+1, a+d + 3, a+ 2d + 9 \text{are in G.P. }

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

Substituting from i) we get

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

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

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

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

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

\displaystyle \Rightarrow d = 2 \text{ or } d = -10

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

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

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

Answer:

\displaystyle \text{Let the three terms be } a, ( a+d), (a+2d) in A.P.

\displaystyle \text{Given } a+ ( a+d)+ (a+2d) = 21

\displaystyle \Rightarrow 3a + 3d = 21

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

\displaystyle \text{Also } \text{Given } a, ( a+d-1), ( a+2d+1) \text{are in G.P. }

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

Substituting \displaystyle d = 7 - a we get

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

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

\displaystyle \Rightarrow 36 = a ( -a + 15)

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

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

\displaystyle \therefore a = 3 \text{ or } a = 12

\displaystyle \text{When } a = 3, d = 7-3 = 4 \text{ and the numbers are } 3, 7, 11

\displaystyle \text{When } a = 12, d = 7 - 12 = -5 \text{ and the numbers are } 12, 7, 2

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

Answer:

\displaystyle \text{Let the three terms be } a, ( a+d), (a+2d) in A.P.

\displaystyle \text{Given } a+ ( a+d)+ (a+2d) = 18

\displaystyle \Rightarrow 3a + 3d = 18

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

\displaystyle \text{Also } \text{Given } a+4, ( a+d+4), ( a+2d+36) \text{are in G.P. }

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

Substituting \displaystyle a = 6 - d we get

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

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

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

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

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

\displaystyle \therefore d = -40 \text{ or } d = 8

\displaystyle \text{When } d = -40, a = 6-(-40) = 46 \text{ and the numbers are } 46, 6, -34

\displaystyle \text{When } d = 8, a = 6 - 8 = -2 \text{ and the numbers are } -2, 6, 14

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

Answer:

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

\displaystyle \text{Given } a + ar + ar^2 = 56

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

\displaystyle \text{Also } a-1, ar-7, ar^2 - 21 \text{are in A.P. }

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

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

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

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

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

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

From i) and ii) we get

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

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

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

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

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

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

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

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

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

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Question 8: If \displaystyle a, b, c \text{are in G.P., } prove that:

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

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

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

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

Answer:

i) \displaystyle \text{Given } a, b, c \text{are in G.P. }

Therefore \displaystyle b^2 = ac

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

\displaystyle \text{LHS } = a(b^2+c^2) = ab^2 + ac^2 = a(ac) + b^2(c) = c( a^2 + b^2) = \text{ RHS. Hence proved. }

ii) \displaystyle \text{Given } a, b, c \text{are in G.P. }

Therefore \displaystyle b^2 = ac

To prove: \displaystyle 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

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

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

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

 \displaystyle = a^3 + b^3 + c^3 = \text{ RHS. Hence proved. }

iii) \displaystyle \text{Given } a, b, c \text{are in G.P. }

Therefore \displaystyle b^2 = ac

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

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

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

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

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

iv) \displaystyle \text{Given } a, b, c \text{are in G.P. }

Therefore \displaystyle b^2 = ac

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

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

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

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

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

\displaystyle = \frac{1}{ac-c^2}  

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

v) \displaystyle \text{Given } a, b, c \text{are in G.P. }

Therefore \displaystyle b^2 = ac

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

\displaystyle \text{LHS } = (a + 2b + 2c) (a - 2b + 2c)

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

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

\displaystyle = a^2 + 4c^2 = \text{ RHS. Hence proved. }

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Question 9: If \displaystyle a, b, c, d \text{are in G.P., } prove that:

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

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

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

Answer:

i) \displaystyle \text{Given } a, b, c, d \text{are in G.P. } Therefore

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

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

\displaystyle \text{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}  

\displaystyle = \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}  

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

ii) \displaystyle \text{Given } a, b, c, d \text{are in G.P. } Therefore

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

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

\displaystyle \text{LHS } = ( a + b+c+d)^2

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

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

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

\displaystyle = (a+b)^2 + 2 (b+c)^2 + ( c+d)^2 = \text{ RHS. Hence proved. }

iii) \displaystyle \text{Given } a, b, c, d \text{are in G.P. } Therefore

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

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

\displaystyle \text{LHS } = (b + c) (b + d)

\displaystyle = b^2 + bc + bd + cd

\displaystyle = ac + c^2 + ad + cd

\displaystyle = c ( a+c) + d ( a+c)

\displaystyle = ( a+c)(c+d) = \text{ RHS. Hence proved. }

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Question 10: If \displaystyle a, b, c \text{are in G.P., } prove that the following are also in G.P.:

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

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

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

Answer:

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

\displaystyle \text{Given } a, b, c \text{are in G.P. }

Therefore \displaystyle b^2 = ac

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

Therefore \displaystyle a^2 , b^2, c^2 \text{are in G.P. }

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

\displaystyle \text{Given } a, b, c \text{are in G.P. }

Therefore \displaystyle b^2 = ac

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

To prove: \displaystyle a^3, b^3, c^3 \text{are in G.P. }

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

\displaystyle \text{Given } a, b, c \text{are in G.P. }

Therefore \displaystyle b^2 = ac

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

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

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

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

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

Therefore \displaystyle a^2 + b^2 , ab + bc, b^2 +c^2 \text{are in G.P. }

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Question 11: If \displaystyle a, b, c, d \text{are in G.P., } prove that:

i) \displaystyle (a^2 + b^2), (b^2+c^2), (c^2+d^2) \text{are in G.P. }

ii) \displaystyle (a^2 - b^2), (b^2-c^2), (c^2-d^2) \text{are in G.P. }

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

iv) \displaystyle (a^2+b^2+c^2), (ab+bc+cd), (b^2+c^2+d^2) \text{are in G.P. }

Answer:

i) \displaystyle \text{Given } a, b, c, d \text{are in G.P. } Therefore

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

To prove: \displaystyle (a^2 + b^2), (b^2+c^2), (c^2+d^2) \text{are in G.P. }

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

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

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

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

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

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

Therefore \displaystyle (a^2 + b^2), (b^2+c^2), (c^2+d^2) \text{are in G.P. }

ii) \displaystyle \text{Given } a, b, c, d \text{are in G.P. } Therefore

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

To prove: \displaystyle (a^2 - b^2), (b^2-c^2), (c^2-d^2) \text{are in G.P. }

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

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

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

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

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

\displaystyle \text{Hence } (a^2 - b^2), (b^2-c^2), (c^2-d^2) \text{are in G.P. }

iii) \displaystyle \text{Given } a, b, c, d \text{are in G.P. } Let the common ratio be \displaystyle r

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

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

\displaystyle \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]  

\displaystyle \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]  

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

\displaystyle \Rightarrow LHS = RHS . Hence proved.

iv) \displaystyle \text{Given } a, b, c, d \text{are in G.P. } Therefore

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

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

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

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

\displaystyle = 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)

\displaystyle = 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

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

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

\displaystyle \text{Hence } (a^2+b^2+c^2), (ab+bc+cd), (b^2+c^2+d^2) \text{are in G.P. }

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

Answer:

\displaystyle \text{Given } (a -b),(b - c),(c - a) \text{are in G.P. }

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

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

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

\displaystyle \text{LHS } = (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac

\displaystyle = bc + ac + ab + 2 ab + 2bc + 2 ac

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

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

Answer:

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

\displaystyle \text{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 \displaystyle = \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 \displaystyle 4^{th} , \displaystyle 10^{th} \text{ and } 16^{th} terms of a G.P. are \displaystyle x, y \text{ and } z respectively. Prove that \displaystyle x, y , z \text{are in G.P. }

Answer:

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

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

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

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

\displaystyle \Rightarrow y^2 = xz

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

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

Answer:

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

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

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

\displaystyle \text{LHS } = ( a - b)^2 = a^2 - 2ab + b^2

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

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

\displaystyle = ac + ad = a ( d-c)= \text{ RHS. Hence proved. }

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

Answer:

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

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

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

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

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

Subtracting ii) from i) we get

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

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

Subtracting iii) from ii) we get

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

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

Subtracting iv from iii) we get

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

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

From vi) we get

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

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

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

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

\displaystyle \therefore p-q, q-r, r-s \text{are in G.P. }

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

Answer:

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

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

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

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

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

\displaystyle \Rightarrow ac = b^2

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

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

Answer:

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

Take \displaystyle \log of both sides

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

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

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

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

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

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

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

From i) and ii) we get

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

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

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

\displaystyle \Rightarrow 2ac = bc + ab

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

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

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

Answer:

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

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

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

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

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

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

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

\displaystyle \Rightarrow c^2 + ce = ce + ea

\displaystyle \text{Hence } a, c, e \text{are in G.P. }

\displaystyle \\

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

Answer:

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

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

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

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

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

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

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

\displaystyle \text{Hence } x^2, b^2, y^2 \text{are in A.P. }

\displaystyle \\

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

Answer:

\displaystyle \text{Given } a, b, c \text{are in A.P. } \displaystyle \Rightarrow 2b = a + c

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

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

\displaystyle \text{LHS } = ( a-b)^2

\displaystyle = a^2 - 2ab + b^2

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

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

\displaystyle = a ( d-c)

\displaystyle \text{Hence } a, (a - b), (d. - c) \text{are in G.P. }

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

Answer:

Let \displaystyle r be the common ratio.

\displaystyle a = a , \displaystyle b=ar , \displaystyle c = ar^2

\displaystyle \text{Given } a + b + c = x b

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

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

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

We know \displaystyle r is always real

\displaystyle \therefore D \geq 0

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

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

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

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

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

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

Therefore \displaystyle x < -1 \text{ and } x > 3

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

Answer:

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

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

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

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

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

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

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

\displaystyle a_q = A'R^{q-1} = b … … … … …v)

\displaystyle a_r = A'R^{r-1} = c … … … … … vi)

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

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

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

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

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

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

\displaystyle = {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] }  

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

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