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

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

$\displaystyle \therefore b^2 = ac$

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

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

$\displaystyle \therefore b^2 = ac$

$\displaystyle \text{Taking } \log$ 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.

$\displaystyle \text{Given } k + 9, k - 6 \text{ and } 4$ 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.

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.

$\displaystyle \text{Let the three terms be } a, ( a+d), (a+2d)$ 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 $\displaystyle a, b, c$.

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

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

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

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

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

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 $\displaystyle (a+b+c)^2= 3 ( ab + bc+cd)$

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

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

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

$\displaystyle \Rightarrow a^2 + b^2 + c^2 = bc + ac + ab$ … … … … … 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 $\displaystyle \frac{a^2 + ab + b^2}{bc+ca + ab} = \frac{b+a}{c+b}$

$\displaystyle \text{Given } a, b, c \text{are in G.P. }$ $\displaystyle \therefore b^2 = ac$ … … … … … 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 $\displaystyle x, y , z \text{are in G.P. }$

$\displaystyle \text{Given } 4^{th}$ , $\displaystyle 10^{th} \text{ and } 16^{th}$ 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 $\displaystyle a, a-b, d-c \text{are in G.P. }$

$\displaystyle \text{Given } a, b, c \text{are in A.P. }$ $\displaystyle \Rightarrow 2b = a + c$ … … … … … 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 $\displaystyle p-q, q-r, r-s \text{are in G.P. }$

$\displaystyle \text{Given } p^{th}, q^{th}, r^{th} \text{ and } s^{th}$ 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

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

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

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

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

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

$\displaystyle \\$

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.

$\displaystyle \text{Given } \frac{1}{a+b}, \frac{1}{2b}, \frac{1}{b+c}$ 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.

$\displaystyle \\$

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

$\displaystyle \text{Given } a, b, c \text{are in A.P. }$ $\displaystyle \Rightarrow 2b = a + c$ … … … … … 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 $\displaystyle x^2, b^2, y^2 \text{are in A.P. }$

$\displaystyle \text{Given } a, b, c \text{are in A.P. }$ $\displaystyle \Rightarrow 2b = a + c$ … … … … … 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 $\displaystyle a, (a - b), (d. - c) \text{are in G.P. }$

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

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

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

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

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

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

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

$\displaystyle = a ( d-c)$

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

$\displaystyle \\$

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 $\displaystyle x > 3$.

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

$\displaystyle \\$

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 $\displaystyle a^{b-c} b^{c-a} c^{a-b} = 1$

$\displaystyle \text{Given } p^{th}, q^{th} \ and \ r^{th}$ 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