ML Aggarwal Solution Class 10 Chapter 7 Ratio and Proportion Exercise 7.2

 Exercise 7.2

Question 1

Find the value of x in the following proportions :

(i) 10 : 35 = x : 42

(ii) 3 : x = 24 : 2

(iii) 2.5 : 1.5 = x : 3

(iv) x : 50 :: 3 : 2

Sol :

(i) 10 : 35 = x : 42

⇒ 35×x = 10 × 42

$\therefore \quad x=\frac{10 \times 42}{35}=2 \times 6=12$

(ii) $3: x=24: 2$

$\Rightarrow x \times 24=3 \times 2$

$\therefore x=\frac{3 \times 2}{24}=\frac{1}{4}$

(iii) $2 \cdot 5: 1 \cdot 5=x: 3$

$\Rightarrow 1 \cdot 5 \times x=2 \cdot 5 \times 3$

$x=\frac{2 \cdot 5 \times 3}{1 \cdot 5}=5 \cdot 0$

(iv) $x: 50:: 3: 2$

$\Rightarrow x \times 2=50 \times 3$

$x=\frac{50 \times 3}{2}=75$

Question 2

Find the fourth proportional to

(i) 3, 12, 15

(ii) $\frac{1}{3}, \frac{1}{4}, \frac{1}{5}$

(iii) 1.5, 2.5, 4.5

(iv) 9.6 kg, 7.2 kg, 28.8 kg

Sol :

(i) Let fourth proportional to

3, 12, 15 be x.

then 3 : 12 :: 15 : x

$\Rightarrow 3 \times x=12 \times 15$

$x=\frac{12 \times 15}{3}=60$

(ii) Let fourth proportional to $\frac{1}{3}, \frac{1}{4}, \frac{1}{5}$ be x

then $\frac{1}{3}: \frac{1}{4}:: \frac{1}{5}: x$

$\Rightarrow \frac{1}{3} \times x=\frac{1}{4} \times \frac{1}{5}$

$\Rightarrow x=\frac{1}{4} \times \frac{1}{5} \times \frac{3}{1}=\frac{3}{20}$

(iii) Let fourth proportional to

$1 \cdot 5,2 \cdot 5,4 \cdot 5$ be $x$

then $1 \cdot 5: 2 \cdot 5:: 4 \cdot 5: x$

$\therefore 1 \cdot 5 \times x=2 \cdot 5 \times 4 \cdot 5$

$x=\frac{2 \cdot 5 \times 4 \cdot 5}{1 \cdot 5}=7 \cdot 5$

(iv) Let fourth proportional to 9.6kg, 7.2 kg, 28.8 kg be x

then 9.6 : 7.2 :: 28.8 : x

$\Rightarrow \quad 9 \cdot 6 \times x=7 \cdot 2 \times 28 \cdot 8$

$x=\frac{7 \cdot 2 \times 28 \cdot 8}{9 \cdot 6}=21 \cdot 6$

Question 3

Find the third proportional to

(i) 5, 10

(ii) 0.24, 0.6

(iii) Rs. 3, Rs. 12

(iv) $5 \frac{1}{4}$ and 7

Sol :

(i) Let x be the third proportional to 5, 10,

then 5 : 10 :: 10 : x

$\therefore 5 \times x=10 \times 10 \Rightarrow x=\frac{10 \times 10}{5}=20$

∴Third proportional =20

(ii) Let x be the third proportional to 0.24, 0.6

then 0.24 : 0.6 :: 0.6 : x

$\therefore 0 \cdot 24 \times x=0 \cdot 6 \times 0 \cdot 6$

$x=\frac{0 \cdot 6 \times 0 \cdot 6}{0 \cdot 24}=1 \cdot 5$

∴ Third proportional =1.5

(iii) Let x be the third proportional to Rs. 3 and Rs. 12 then Rs. 3: Rs. 12: Rs. 12: x

$\therefore x=\frac{12 \times 12}{3}=48$

$\therefore$ Third proportional Rs 48

(iv) Let x be the third proportional to $5 \frac{1}{4}$ and 7

then $5 \frac{1}{4}: 7:: 7: x$

$\Rightarrow \frac{21}{4}: 7:: 7: x$

$\therefore \frac{21}{4} \times x=7 \times 7$

$x=\frac{7 \times 7 \times 4}{21}$

$=\frac{28}{3}=9 \frac{1}{3}$

∴ Third proportional $=9 \frac{1}{3}$ Ans.

Question 4

Find the mean proportion of:

(i) 5 and 80

(ii) $\frac{1}{12}$ and $\frac{1}{75}$

(iii) 8.1 and 2.5

(iv) (a – b) and (a³ – a²b), a> b

Solution:

(i) Let x be the mean proportion of 5 and 80 ,

then 5 : x : : x : 80

x² = 5 x 80

⇒ x = √5×80=√400= 20

x = 20

Hence mean proportion=20

(ii) Let x be the mean proportion of $\frac{1}{12}$ and $\frac{1}{75}$

then $\frac{1}{12}: x:: x: \frac{1}{75}$

$\therefore x^{2}=\frac{1}{12} \times \frac{1}{75}=\frac{1}{900}$

$\therefore x=\sqrt{\frac{1}{900}}=\frac{1}{30}$

Hence the mean proportion $=\frac{1}{30}$

(iii) Let the x be the mean proportion of 8.1 and 2.5 

∴ 8.1: x:: x: 2.5

$\therefore \quad x^{2}=8 \cdot 1 \times 2 \cdot 5$

$\therefore \quad x=\sqrt{8 \cdot 1 \times 2 \cdot 5}=\sqrt{20 \cdot 25}=4 \cdot 5$

Hence mean proportion =4.5

(iv) Let x be the mean proportion to $(a-b)$ and $\left(a^{3}-a^{2} b\right), a>b$

then $(a-b): x:: x:\left(a^{3}-a^{2} b\right)$

$x^{2}=(a-b)\left(a^{3}-a^{2} b\right)$

$=(a-b) a^{2}(a-b)=a^{2}(a-b)^{2}$

∴ x=a(a-b)

Hence the mean proportion=a(a-b)

Question 5

If a, 12, 16 and b are in continued proportion find a and b.

Sol :

∵ a, 12, 16, b are in continued proportion, then

$\frac{a}{12}=\frac{12}{16}=\frac{16}{b}$

$\Rightarrow \frac{a}{12}=\frac{12}{16} \Rightarrow 16 a=144$

$\Rightarrow a=\frac{144}{16}=$

and $\frac{12}{16}=\frac{16}{b} \Rightarrow 12 b=16 \times 16=256$

$b=\frac{256}{12}=\frac{64}{3}=21 \frac{1}{3}$

Hence $a=9, b=\frac{64}{3}$ or $21 \frac{1}{3}$

Question 6

What number must be added to each of the numbers 5, 11, 19 and 37 so that they are in proportion ? (2009)

Sol :

Let x be added to 5, 11, 19 and 37 to make them in proportion.

5 + x : 11 + x : : 19 + x : 37 + x

⇒ (5+x)(37+x)=(11+x)(19+x)

$\Rightarrow 185+5 x+37 x+x^{2}=209+11 x+19 x+x^{2}$

$\Rightarrow 185+42 x+x^{2}=209+30 x+x^{2}$

$\Rightarrow 42 x-30 x+x^{2}-x^{2}=209-185$

⇒ 12 x=24

⇒ x=2

∴ Least number to be added =2

Question 7

What number should be subtracted from each of the numbers 23, 30, 57 and 78 so that the remainders are in proportion ? (2004)

Sol :

Let x be subtracted from each term, then

23 – x, 30 – x, 57 – x and 78 – x are proportional

23 – x : 30 – x : : 57 – x : 78 – x

$\Rightarrow  \frac{23-x}{30-x}=\frac{57-x}{78-x}$

⇒ (23-x)(78-x)=(30-x)(57-x)

$\Rightarrow 1794-23 x-78 x+x^{2}$

$=1710-30 x-57 x+x^{2}$

$\Rightarrow x^{2}-101 x+1794=x^{2}-87 x+1710$

$\Rightarrow  x^{2}-101 x+1794-x^{2}+87 x-1710=0$

⇒-14 x+84=0

⇒ 14 x=84

$\therefore x=\frac{84}{14}=6$

Hence 6 is to be subtracted Ans.

Question 8

If 2x – 1, 5x – 6, 6x + 2 and 15x – 9 are in proportion, find the value of x.

Sol :

∵ 2x – 1, 5x – 6, 6x + 2 and 15x – 9 are in proportion.

then (2x – 1) (15x – 9) = (5x – 6) (6x + 2)

$\Rightarrow 30 x^{2}-18 x-15 x+9 \Rightarrow 30 x^{2}+10 x-36 x-12$

$\Rightarrow 30 x^{2}-33 x+9=30 x^{2}-26 x-12$

$\Rightarrow 30 x^{2}-33 x-30 x^{2}+26 x=-12-9$

$\Rightarrow-7x=-21$

$\therefore \quad x=\frac{-21}{-7}=3$

Hence x=3

Question 9

If 2x – 1, 5x – 6, 6x + 2 and 15x – 9 are in proportion, find the value of x.

Sol :

∵ 2x – 1, 5x – 6, 6x + 2 and 15x – 9 are in proportion.

then (2x – 1) (15x – 9) = (5x – 6) (6x + 2)

$\Rightarrow 30 x^{2}-18 x-15 x+9 \Rightarrow 30 x^{2}+10 x-36 x-12$

$\Rightarrow 30 x^{2}-33 x+9=30 x^{2}-26 x-12$

$\Rightarrow 30 x^{2}-33 x-30 x^{2}+26 x=-12-9$

⇒-7 x=-21

$\therefore x=\frac{-21}{-7}=3$

Hence x=3

Question 10

What number must be added to each of the numbers 16, 26 and 40 so that the resulting numbers may be in continued proportion?

Sol :

Let x be added to each number then

16 + x, 26 + x and 40 + x

are in continued proportion.

$\Rightarrow \frac{16+x}{26+x}=\frac{26+x}{40+x}$

Cross Multiplying,

(16+x)(40+x)=(26+x)(26+x)

$\Rightarrow 640+16 x+40 x+x^{2}=676+26 x+26 x+x^{2}$

$\Rightarrow 640+56 x+x^{2}=676+52 x+x^{2}$

$\Rightarrow 56 x+x^{2}-52 x-x^{2}=676-640$

⇒4x=36

$\Rightarrow x=\frac{36}{4}=9$

∴9 is to be added.

Question 11

Find two numbers such that the mean proportional between them is 28 and the third proportional to them is 224.

Sol :

Let the two numbers are a and b.

∵ 28 is the mean proportional

∵ a : 28 : : 28 : b

$\therefore a b=(28)^{2}=784$

$\Rightarrow a=\frac{784}{b} \quad \ldots(i)$

∵ 224 is the third proportional

∵ a: b:: b: 224

$\Rightarrow \quad b^{2}=224 a$

Substituting the value of a in (ii)

$b^{2}=224 \times \frac{784}{b} \Rightarrow b^{3}=224 \times 784$

$\Rightarrow b^{3}=175616=(56)^{3}$

∴ b=56

Now substituting the value of b in (i) 

$a=\frac{784}{56}=14$

Hence numbers are 14,56

Question 12

If b is the mean proportional between a and c, prove that a, c, a² + b², and b² + c² are proportional.

Sol :

∵ b is the mean proportional between a and c, then,

b² = a × c ⇒ b² = ac …(i)

Now $a, c, a^{2}+b^{2}$ and $b^{2}+c^{2}$ are in proportion

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

if $a\left(b^{2}+c^{2}\right)=c\left(a^{2}+b^{2}\right)$

if $a\left(a c+c^{2}\right)=c\left(a^{2}+a c\right)[$ from $(i)]$ if $\quad a c(a+c)=a^{2} c+a c^{2}$

if ac(a+c)=ac(a+c) which is true. Hence proved.

Question 13

If b is the mean proportional between a and c, prove that (ab + bc) is the mean proportional between (a² + b²) and (b² + c²).

Sol :

b is the mean proportional between a and c then

b² = ac …(i)

Now if (ab + bc) is the mean proportional

between $\left(a^{2}+b^{2}\right)$ and $\left(b^{2}+c^{2}\right),$ then $(a b+b c)^{2}=\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)$

Now L.H.S. $=(a b+b c)^{2}=a^{2} b^{2}+b^{2} c^{2}+2 a b^{2} c$

$=a^{2}(a c)+a c(c)^{2}+2 a$ ac. $c \quad[$ from $(i)]$

$=a^{3} c+a c^{3}+2 a^{2} c^{2}$

$=a c\left(a^{2}+c^{2}+2 a c\right)=a c(a+c)^{2}$

$\mathrm{R.H.S.}=\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)$

$=\left(a^{2}+a c\right)\left(a c+c^{2}\right) \quad[$ from $(i)]$

$=a(a+c) c(a+c)=a c(a+c)^{2}$

$\therefore L.H.S=R.H.S$

Question 14

If y is mean proportional between x and z, prove that

xyz (x + y + z)³ = (xy + yz + zx)³.

Sol :

∵ y is the mean proportional between

x and z, then

y² = xz …(i)

L.H.S. $=x y z(x+y+z)^{3}$

$=x z, y(x+y+z)^{3}$

$=y^{2} y(x+y+z)^{3}$ [from (i)]

$=y^{3}(x+y+z)^{3}$

$=[y(x+y+z)]^{3}$

$=\left[x y+y^{2}+y z\right]^{3}$

$=(x y+y z+z x)^{3}$ (from i)

=R.H.S

Question 15

If a + c = mb and  $\frac{1}{b}+\frac{1}{d}=\frac{m}{c}$ , prove that a, b, c and d are in proportion.

Sol :

a + c = mb and $\frac{1}{b}+\frac{1}{d}=\frac{m}{c}$

a + c = mb

$\frac{a}{b}+\frac{c}{d}=m$ (Dividing by b)...(i)

and $\frac{1}{b}+\frac{1}{d}=\frac{m}{c}$

$\frac{c}{b}+\frac{c}{d}=m$ (Multiplying by c)...(ii)

From (i) and (ii)

$\frac{a}{b}+\frac{c}{b}=\frac{c}{b}+\frac{c}{d} \Rightarrow \frac{a}{b}=\frac{c}{d}$

Hence, a, b, c and d are proportional.

Question 16

If $\frac{x}{a}=\frac{y}{b}=\frac{z}{c},$ prove that

(i) $\frac{x^{3}}{a^{2}}+\frac{y^{3}}{b^{2}}+\frac{z^{3}}{c^{2}}=\frac{(x+y+z)^{3}}{(a+b+c)^{2}}$

(ii) $\left[\frac{a^{2} x^{2}+b^{2} y^{2}+c^{2} z^{2}}{a^{3} x+b^{3} y+c^{3} z}\right]^{3}=\frac{x y z}{a b c}$

(iii) $\frac{a x-b y}{(a+b)(x-y)}+\frac{b y-c z}{(b+c)(y-z)}+\frac{c z-a x}{(c+a)(z-x)}=3$

Sol :

$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$

$\therefore x=a k, y=b k, z=c k$

(i) L.H.S. $=\frac{x^{3}}{a^{2}}+\frac{y^{3}}{b^{2}}+\frac{z^{3}}{c^{2}}$

$=\frac{a^{3} k^{3}}{a^{2}}+\frac{b^{3} k^{3}}{b^{2}}+\frac{c^{3} k^{3}}{c^{2}}$

$=a k^{3}+b k^{3}+c k^{3}=k^{3}(a+b+c)$

R.H.S. $=\frac{(x+y+z)^{3}}{(a+b+c)^{2}}$

$=\frac{(a k+b k+c k)^{3}}{(a+b+c)^{2}}=\frac{k^{3}(a+b+c)^{3}}{(a+b+c)^{2}}$

$=k^{3}(a+b+c)$

Hence L.H.S. = R.H.S.

(ii) L.H.S $=\left[\frac{a^{2} x^{2}+b^{2} y^{2}+c^{2} z^{2}}{a^{3} x+b^{3} y+c^{3} z}\right]^{3}$

$=\left[\frac{a^{2} \cdot a^{2} k^{2}+b^{2} \cdot b^{2} k^{2}+c^{2} \cdot c^{2} k^{2}}{a^{3} \cdot a \cdot k+b^{3} \cdot b k+c^{3} \cdot c k}\right]^{3}$

$=\left[\frac{a^{4} k^{2}+b^{4} k^{2}+c^{4} k^{2}}{a^{4} k+b^{4} k+c^{4} k}\right]^{3}$

$=\left[\frac{k^{2}\left(a^{4}+b^{4}+c^{4}\right)}{k\left(a^{4}+b^{4}+c^{4}\right)}\right]^{3}=k^{3}$

$\mathrm{R.H.S}=\frac{x y z}{a b c}=\frac{a k \cdot b k \cdot c k}{a b c}=k^{3}$

∴L.H.S=R.H.S

(iii) L.H.S $\frac{a x-b y}{(a+b)(x-y)}+\frac{b y-c z}{(b+c)(y-z)}+\frac{c z-a x}{(c+a)(z-x)}$

$=\frac{a \cdot a k-b \cdot b k}{(a+b)(a k-b k)}+\frac{b \cdot b k-c \cdot c k}{(b+c)(b k-c k)}+\frac{c . c k-a . a k}{(c+a)(c k-a k)}$

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

$=\frac{k\left(a^{2}-b^{2}\right)}{k\left(a^{2}-b^{2}\right)}+\frac{k\left(b^{2}-c^{2}\right)}{k\left(b^{2}-c^{2}\right)}+\frac{k\left(c^{2}-a^{2}\right)}{k\left(c^{2}-a^{2}\right)}$

=1+1+1=3=R.H.S

Question 17

If $\frac{a}{b}=\frac{c}{d}=\frac{e}{t}$ prove that

(i) (b² + d² + f²) (a² + c² + e²) = (ab + cd + ef)²

(ii) $\frac{\left(a^{3}+c^{3}\right)^{2}}{\left(b^{3}+d^{3}\right)^{2}}=\frac{e^{6}}{f^{6}}$

(iii) $\frac{a^{2}}{b^{2}}+\frac{c^{2}}{d^{2}}+\frac{e^{2}}{f^{2}}=\frac{a c}{b d}+\frac{c e}{d f}+\frac{a e}{d f}$

(iv) $b d f\left[\frac{a+b}{b}+\frac{c+d}{d}+\frac{c+f}{f}\right]^{3}$=27(a+b)(c+d)(e+f)

Sol :

$\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=k(\mathrm{say})$

∴ a = bk, c = dk, e =fk

(i) L.H.S. $=\left(b^{2}+d^{2}+f^{2}\right)\left(a^{2}+c^{2}+e^{2}\right)$

$=\left(b^{2}+d^{2}+f^{2}\right)\left(b^{2} k^{2}+d^{2} k^{2}+f^{2} k^{2}\right)$

$=\left(b^{2}+d^{2}+f^{2}\right) k^{2}\left(b^{2}+d^{2}+f^{2}\right)$

$=k^{2}\left(b^{2}+d^{2}+f^{2}\right)$

$\mathrm{R} \cdot \mathrm{H} \cdot \mathrm{S}=(a b+c d+e f)^{2}$

$=(b . k b .+d k . d+f k . f)^{2}$

$=\left(k b^{2}+k d^{2}+k f^{2}\right)=k^{2}\left(b^{2}+d^{2}+f^{2}\right)^{2}$

$\therefore \mathrm{L} \cdot \mathrm{H.S}=\mathrm{R} \cdot \mathrm{H.S}$

(ii) $\frac{\left(a^{3}+c^{3}\right)^{2}}{\left(b^{3}+d^{3}\right)^{2}}=\frac{\left(b^{3} k^{3}+d^{3} k^{3}\right)^{2}}{\left(b^{3}+d^{3}\right)^{2}}$

$=\frac{\left[k^{3}\left(b^{3}+d^{3}\right)\right]^{2}}{\left(b^{3}+a^{3}\right)^{2}}=\frac{k^{6}\left(b^{3}+d^{3}\right)^{2}}{\left(b^{3}+d^{3}\right)^{2}}=k^{6}$

$\mathrm{R} . \mathrm{H.S}=\frac{e^{6}}{f^{6}}=\frac{f^{6} k^{6}}{f^{6}}=k^{6}$

$\therefore \mathrm{L} \cdot \mathrm{H.S}=\mathrm{R} . \mathrm{H.S}$

(iii) $\mathrm{L.H.S}=\frac{a^{2}}{b^{2}}+\frac{c^{2}}{d^{2}}+\frac{e^{2}}{f^{2}}=\frac{b^{2} k^{2}}{b^{2}}$

$+\frac{d^{2} k^{2}}{d^{2}}+\frac{f^{2} k^{2}}{f^{2}}=k^{2}+k^{2}+k^{2}=3 k^{2}$

$\mathrm{R} \cdot \mathrm{H.S}=\frac{a c}{b d}+\frac{c e}{d f}+\frac{a e}{b f}$

$=\frac{b k \cdot d k}{b \cdot d}+\frac{d k \cdot f k}{d \cdot f}+\frac{b k \cdot f k}{b \cdot f}$

$=k^{2}+k^{2}+k^{2}=3 k^{2}$

$\therefore \mathrm{L} \cdot \mathrm{H.S}=\mathrm{R} . \mathrm{H.S}$

(iv) L.H.S $=b d f\left[\frac{a+b}{b}+\frac{c+d}{d}+\frac{e+f}{f}\right]^{3}$

$=b d f\left[\frac{b k+b}{b}+\frac{d k+d}{d}+\frac{f k+f}{f}\right]^{3}$

$=b d f\left[\frac{b(k+1)}{b}+\frac{d(k+1)}{d}+\frac{f(k+1)}{f}\right]$

$=b d f(k+1+k+1+k+1)^{3}$

$=b d f(3 k+3)^{3}=27 b d f(k+1)^{3}$

R.H.S $=27(a+b)(c+d)(e+f)$

$=27(b k+b)(d k+d)(f k+f)$

$=27 b(k+1) d(k+1) f(k+1)$

$=27 b d f(k+1)^{3}$

$\therefore \mathrm{L} \cdot \mathrm{H.S}=\mathrm{R} . \mathrm{H.S}$

Question 18

If ax = by = cz; prove that

$\frac{x^{2}}{y z}+\frac{y^{2}}{z x}+\frac{z^{2}}{x y}=\frac{b c}{a^{2}}+\frac{c a}{b^{2}}+\frac{a b}{c^{2}}$

Sol :

Let ax = by = cz = k

$\therefore  x=\frac{k}{a}, y=\frac{k}{b}, z=\frac{k}{c}$

L.H.S. $=\frac{x^{2}}{y z}+\frac{y^{2}}{z x}+\frac{z^{2}}{x y}$

$=\frac{\frac{k^{2}}{a^{2}}}{\frac{k}{b} \cdot \frac{k}{c}}+\frac{\frac{k^{2}}{b^{2}}}{\frac{k}{c} \cdot \frac{k}{a}}+\frac{\frac{k^{2}}{c^{2}}}{\frac{k}{a} \cdot \frac{k}{b}}$

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

$=\frac{k^{2}}{a^{2}} \times \frac{b c}{k^{2}}+\frac{k^{2}}{b^{2}} \times \frac{c a}{k^{2}}+\frac{k^{2}}{c^{2}} \times \frac{a b}{k^{2}}$

$=\frac{b c}{a^{2}}+\frac{c a}{b^{2}}+\frac{a b}{c^{2}}=\mathrm{R} \cdot \mathrm{H.S}$

Question 19

If a, b, c and d are in proportion, prove that:

(i) (5a + 7b) (2c – 3d) = (5c + 7d) (2a – 3b)

(ii) (ma + nb) : b = (mc + nd) : d

(iii) (a4 + c4) : (b4 + d4) = a2 c2 : b2 d2.

(iii) $\left(a^{4}+c^{4}\right):\left(b^{4}+d^{4}\right)=a^{2} c^{2}: b^{2} d^{2}$

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

(v) $\frac{(a+c)^{3}}{(b+d)^{3}}=\frac{a(a-c)^{2}}{b(b-d)^{2}}$

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

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

(viii) abcd $\left[\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}\right.=a^{2}+b^{2}+c^{2}+d^{2}$

Sol :

∵ a, b, c, d are in proportion

$\frac{a}{b}=\frac{c}{d}=k($ say $)$

a=bk, c=dk

(i) L.H.S. $=(5 a+7 b)(2 c-3 d)$

$=(5 \cdot b k+7 b)(2 d k-3 d)$

$=k(5 b+7 b) k(2 d-3 d)$

$=k^{2}(12 b) \times(-d)=-12 b d k^{2}$

$\mathrm{R} . \mathrm{H.S.}=(5 c+7 d)(2 a-3 b)$

$=(5 d k+7 d)(2 k \cdot b-3 b)$

$=k(5 d+7 d) k(2 b-3 b)$

$=k^{2}(12 d)(-b)=-12 k^{2} b d=-12 b d k^{2}$

$\therefore \mathrm{L} \cdot \mathrm{H.S}=\mathrm{R} \cdot \mathrm{H.S}$

(ii) $(m a+n b): b=(m c+n d)^{*}: d$

$\Rightarrow \frac{m a+n b}{b}=\frac{m c+n d}{d}$

L.H.S. $=\frac{m b k+n b}{b}=\frac{b(m k+n)}{b}$

=m k+n

$\mathrm{R} . \mathrm{H.S.}=\frac{m c+n d}{d}=\frac{m d k+n d}{d}$

$=\frac{d(m k+n)}{d}=m k+n$

$\therefore \mathrm{L} \cdot \mathrm{H.S}=\mathrm{R} . \mathrm{H.S}$

(iii) $\left(a^{4}+c^{4}\right):\left(b^{4}+d^{4}\right)=a^{2} c^{2}: b^{2} d^{2}$

$\frac{a^{4}+c^{4}}{b^{4}+d^{4}}=\frac{a^{2} c^{2}}{b^{2} d^{2}}$

L. H.S. $=\frac{a^{4}+c^{4}}{b^{4}+d^{4}}=\frac{b^{4} k^{4}+d^{4} k^{4}}{b^{4}+d^{4}}$

$=\frac{k^{4}\left(b^{4}+d^{4}\right)}{\left(b^{4}+d^{4}\right)}=k^{4}$

R.H.S. $=\frac{a^{2} c^{2}}{b^{2} d^{2}}=\frac{k^{2} b^{2} \cdot k^{2} d^{2}}{b^{2} \cdot d^{2}}=k^{4}$

Hence L.H.S.=R.H.S.

(iv) $\quad$ L.H.S $=\frac{a^{2}+a b}{c^{2}+c d}=\frac{k^{2} b^{2}+b k \cdot b}{k^{2} d^{2}+d k \cdot d}$

$=\frac{k b^{2}(k+1)}{d^{2} k(k+1)}=\frac{b^{2}}{d^{2}}$

$\mathrm{R.H.S}=\frac{b^{2}-2 a b}{d^{2}-2 c d}=\frac{b^{2}-2 \cdot b k b}{d^{2}-2 d k d}$

$=\frac{b^{2}(1-2 k)}{d^{2}(1-2 k)}=\frac{b^{2}}{d^{2}}$

$\therefore \mathrm{L} \cdot \mathrm{H.S}=\mathrm{R} \cdot \mathrm{H.S}$

(v) L.H.S. $=\frac{(a+c)^{3}}{(b+d)^{3}}=\frac{(b k+d k)^{3}}{(b+d)^{3}}$

$=\frac{k^{3}(b+d)^{3}}{(b+d)^{3}}=k^{3}$

R.H.S. $=\frac{a(a-c)^{2}}{b(b-d)^{2}}=\frac{b k(b k-d k)^{2}}{b(b-d)^{2}}$

$=\frac{b k \cdot k^{2}(b-d)^{2}}{b(b-d)^{2}}=k^{3}$

$\therefore \mathrm{L} . \mathrm{H.S}=\mathrm{R} . \mathrm{H.S}$

$(v i) \quad L . H . S=\frac{a^{2}+a b+b^{2}}{a^{2}-a b+b^{2}}$

$=\frac{b^{2} k^{2}+b k \cdot b+b^{2}}{b^{2} k^{2}-b k \cdot b+b^{2}}$

$=\frac{b^{2}\left(k^{2}+k+1\right)}{b^{2}\left(k^{2}-k+1\right)}=\frac{k^{2}+k+1}{k^{2}-k+1}$

$\mathrm{R} \cdot \mathrm{H.S}=\frac{c^{2}+c d+d^{2}}{c^{2}-c d+d^{2}}$

$=\frac{d^{2} k^{2}+d k d+d^{2}}{d^{2} k^{2}-d k \cdot d+d^{2}}$

$=\frac{d^{2}\left(k^{2}+k+1\right)}{d^{2}\left(k^{2}-k+1\right)}=\frac{k^{2}+k+1}{k^{2}-k+1}$

$\therefore \mathrm{L} \cdot \mathrm{H.S}=\mathrm{R} \cdot \mathrm{H.S}$

$(vi) L . H . S=\frac{a^{2}+a b+b^{2}}{a^{2}-a b+b^{2}}$

$=\frac{b^{2} k^{2}+b k \cdot b+b^{2}}{b^{2} k^{2}-b k \cdot b+b^{2}}$

$=\frac{b^{2}\left(k^{2}+k+1\right)}{b^{2}\left(k^{2}-k+1\right)}=\frac{k^{2}+k+1}{k^{2}-k+1}$

$\mathrm{R} \cdot \mathrm{H.S}=\frac{c^{2}+c d+d^{2}}{c^{2}-c d+d^{2}}$

$=\frac{d^{2} k^{2}+d k d+d^{2}}{d^{2} k^{2}-d k \cdot d+d^{2}}$

$=\frac{d^{2}\left(k^{2}+k+1\right)}{d^{2}\left(k^{2}-k+1\right)}=\frac{k^{2}+k+1}{k^{2}-k+1}$

$\therefore \mathrm{L} \cdot \mathrm{H.S}=\mathrm{R} \cdot \mathrm{H.S}$

(viii) $\mathrm{L.H.S.}=a b c d\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}\right)$

(viii) L.H.S. $=a b c d\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}\right)$

$=b k$.b. dk.d $\left[\frac{1}{b^{2} k^{2}}+\frac{1}{b^{2}}+\frac{1}{d^{2} k^{2}}+\frac{1}{d^{2}}\right]$

$=k^{2} b^{2} d^{2}\left[\frac{d^{2}+d^{2} k^{2}+b^{2}+b^{2} k^{2}}{b^{2} d^{2} k^{2}}\right]$

$=d^{2}\left(1+k^{2}\right)+b^{2}\left(1+k^{2}\right)=\left(1+k^{2}\right)\left(b^{2}+d^{2}\right)$

R.H.S $=a^{2}+b^{2}+c^{2}+d^{2}$

$=b^{2} k^{2}+b^{2}+d^{2} k^{2}+d^{2}$

$=b^{2}\left(k^{2}+1\right)+d^{2}\left(k^{2}+1\right)=\left(k^{2}+1\right)\left(b^{2}+d^{2}\right)$

$\therefore$ L.H.S = R.H.S.

Question 20

If x, y, z are in continued proportion, prove that: $\frac{(x+y)^{2}}{(y+z)^{2}}=\frac{x}{z}$. (2010)

Sol :

x, y, z are in continued proportion

Let $\frac{x}{y}=\frac{y}{z}=k$

Then y=k z

$x=y k=k z \times k=k^{2} z$

Now L.H.S. $=\frac{(x+y)^{2}}{(y+z)^{2}}$

$=\frac{\left(k^{2} z+k z\right)^{2}}{(k z+z)^{2}}=\frac{\{k z(k+1)\}^{2}}{\{z(k+1)\}^{2}}$

$=\frac{k^{2} z^{2}(k+1)^{2}}{z^{2}(k+1)^{2}}=k^{2}$

$\mathrm{R.H.S.}=\frac{x}{z}=\frac{k^{2} z}{z}=k^{3}$

∴L.H.S=R.H.S

Question 21

If a, b, c are in continued proportion, prove that:

$\frac{p \alpha^{2}+q a b+r b^{2}}{p b^{2}+q b c+r c^{2}}=\frac{\alpha}{c}$

Sol :

Given a, b, c are in continued proportion

$\frac{p \alpha^{2}+q \alpha b+r b^{2}}{p b^{2}+q b c+r c^{2}}=\frac{a}{c}$

Let $\frac{a}{b}=\frac{b}{c}=k$

⇒a=bk and b=ck...(i)

⇒a=(ck)k$=c k^{2}$  [using (i)]

b=ck

$\mathrm{L.H.S.}=\frac{a}{c}=\frac{c k^{2}}{c}=k^{2}$

R.H.S. $=\frac{p\left(c k^{2}\right)^{2}+q\left(c k^{2}\right) c k+r(c k)^{2}}{p(c k)^{2}+q(c k) c+r c^{2}}$

$=\frac{p c^{2} k^{4}+q c^{2} k^{3}+r c^{2} k^{2}}{p c^{2} k^{2}+q c^{2} k+r c^{2}}$

$=\frac{c^{2} k^{2}}{c^{2}}\left[\frac{p k^{2}+q k+r}{p k^{2}+q k+r}\right]=k^{2}$...(iii)

From (ii) and (iii) L.H.S=R.H.S

∴b=ck, a=bk=ckk$=c k^{2}$

(i) L.H.S

$=\frac{a+b}{b+c}=\frac{c k^{2}+c k}{c k+c}=\frac{c k(k+1)}{c(k+1)}=k$

$\mathrm{R} . \mathrm{H.S}=\frac{a^{2}(b-c)}{b^{2}(a-b)}$

$=\frac{\left(c k^{2}\right)^{2}(c k-c)}{(c k)^{2}\left(c k^{2}-c k\right)}$

$=\frac{c^{2} k^{4} c(k-1)}{c^{2} k^{2} c k(k-1)}$

Question 22

If a, b, c are in continued proportion, prove that:

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

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

(iii) $a: c=\left(a^{2}+b^{2}\right):\left(b^{2}+c^{2}\right)$

(iv) $a^{2} b^{2} c^{2}\left(a^{-4}+b^{-4}+c^{-4}\right)=b^{-2}\left(a^{4}+b^{4}+c^{4}\right)$

(v) $a b c(a+b+c)^{3}=(a b+b c+c a)^{3}$

(vi) $(a+b+c)(a-b+c)=a^{2}+b^{2}+c^{2}$

Sol :

As a, b, c, are in continued proportion

Let $\frac{a}{b}=\frac{b}{c}=k$

$=\frac{c^{3} k^{4}(k-1)}{c^{3} k^{3}(k-1)}=k$

∴L.H.S=R.H.S

(ii) L.H.S. $=\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}$

$=\frac{1}{\left(c k^{2}\right)^{3}}+\frac{1}{(c k)^{3}}+\frac{1}{c^{3}}$

$=\frac{1}{c^{3} k^{6}}+\frac{1}{c^{3} k^{3}}+\frac{1}{c^{3}}$

$=\frac{1}{c^{3}}\left[\frac{1}{k^{6}}+\frac{1}{k^{3}}+\frac{1}{1}\right]$

R.H.S. $=\frac{a}{b^{2} c^{2}}+\frac{b}{c^{2} a^{2}}+\frac{c}{a^{2} b^{2}}$

$=\frac{c k^{2}}{(c k)^{2} c^{2}}+\frac{c k}{c^{2}\left(c k^{2}\right)^{2}}+\frac{c}{\left(c k^{2}\right)^{2}(c k)^{2}}$

$=\frac{c k^{2}}{c^{4} k^{2}}+\frac{c k}{c^{4} k^{4}}+\frac{c}{c^{4} k^{6}}$

$=\frac{1}{c^{3}}+\frac{1}{c^{3} k^{3}}+\frac{1}{c^{3} k^{6}}$

$=\frac{1}{c^{3}}\left[1+\frac{1}{k^{3}}+\frac{1}{k^{6}}\right]$

$=\frac{1}{c^{3}}\left[\frac{1}{k^{6}}+\frac{1}{k^{3}}+1\right]$

∴ L.H.S = R.H.S.

(iii) $a: c=\left(a^{2}+b^{2}\right):\left(b^{2}+c^{2}\right)$

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

L.H.S. $\frac{a}{c}=\frac{c k^{2}}{c}=k^{2}$

R.H.S. $=\frac{\left(c k^{2}\right)^{2}+(c k)^{2}}{(c k)^{2}+c^{2}}$

$=\frac{c^{2} k^{4}+c^{2} k^{2}}{c^{2} k^{2}+c^{2}}$

$=\frac{c^{2} k^{2}\left(k^{2}+1\right)}{c^{2}\left(k^{2}+1\right)}=k^{2}$

∴ L.H.S = R.H.S.

(iv) L.H.S. $=a^{2} b^{2} c^{2}\left(a^{-4}+b^{-4}+c^{-4}\right)$

$=a^{2} b^{2} c^{2}\left[\frac{1}{a^{4}}+\frac{1}{b^{4}}+\frac{1}{c^{4}}\right]$

$=\frac{a^{2} b^{2} c^{2}}{a^{4}}+\frac{a^{2} b^{2} c^{2}}{b^{4}}+\frac{a^{2} b^{2} c^{2}}{c^{4}}$

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

$=\frac{(c k)^{2} \cdot c^{2}}{\left(c k^{2}\right)^{2}}+\frac{c^{2}\left(c k^{2}\right)^{2}}{(c k)^{2}}+\frac{\left(c k^{2}\right)^{2}(c k)^{2}}{c^{2}}$

$=\frac{c^{2} k^{2} \cdot c^{2}}{c^{2} k^{4}}+\frac{c^{2} \cdot c^{2} k^{4}}{c^{2} k^{2}}+\frac{c^{2} k^{4} \cdot c^{2} k^{2}}{c^{2}}$

$=\frac{c^{2}}{k^{2}}+\frac{c^{2} k^{2}}{1}+\frac{c^{2} k^{6}}{1}$

$=c^{2}\left[\frac{1}{k^{2}}+k^{2}+k^{6}\right]$

$=\frac{c^{2}}{k^{2}}\left[1+k^{4}+k^{8}\right]$

R.H.S. $=b^{-2}\left[a^{4}+b^{4}+c^{4}\right]$

$=\frac{1}{b^{2}}\left[a^{4}+b^{4}+c^{4}\right]$

$=\frac{1}{(c k)^{2}}\left[\left(c k^{2}\right)^{4}+(c k)^{4}+c^{4}\right]$

$=\frac{1}{c^{2} k^{2}}\left[c^{4} k^{8}+c^{4} k^{4}+c^{4}\right]$

$=\frac{c^{4}}{c^{2} k^{2}}\left[k^{8}+k^{4}+1\right]$

$=\frac{c^{2}}{k^{2}}\left[1+k^{4}+k^{8}\right]$

∴ L.H.S = R.H.S.

(v) L.H.S. $=a b c(a+b+c)^{3}$

$=c k^{2}$.ck.c $\left[c k^{2}+c k+c\right]^{3}$

$=c^{3} k^{3}\left\{c\left(k^{2}+k+1\right)\right]^{3}$

$=c^{3} k^{3} \cdot c^{3} \cdot\left(k^{2}+k+1\right)^{3}=c^{6} k^{3}\left(k^{2}+k+1\right)^{3}$

$\mathrm{R} . \mathrm{H.S.}=(a b+b c+c a)^{3}$

$=\left(c k^{2} \cdot c k+c k \cdot c+c \cdot c k^{2}\right)^{3}$

$=\left(c^{2} k^{3}+c^{2} k+c^{2} k^{2}\right)^{3}=\left(c^{2} k^{3}+c^{2} k^{2}+c^{2} k\right)^{3}$

$=\left[c^{2} k\left(k^{2}+k+1\right)\right]^{3}=c^{6} k^{3}\left(k^{2}+k+1\right)^{3}$

∴ L.H.S = R.H.S.

(vi) L.H.S. $=(a+b+c)(a-b+c)$

$=\left(c k^{2}+c k+c\right)\left(c k^{2}-c k+c\right)$

$=c\left(k^{2}+k+1\right) c\left(k^{2}-k+1\right)$

$=c^{2}\left(k^{2}+k+1\right)\left(k^{2}-k+1\right)$

$=c^{2}\left(k^{4}+k^{2}+1\right)$

R.H.S. $=a^{2}+b^{2}+c^{2}$

$=\left(c k^{2}\right)^{2}+(c k)^{2}+(c)^{2}$

$=c^{2} k^{4}+c^{2} k^{2}+c^{2}=c^{2}\left(k^{4}+k^{2}+1\right)$

∴ L.H.S = R.H.S.

Question 23

If a, b, c, d are in continued proportion, prove that:

(i) $\frac{a^{3}+b^{3}+c^{3}}{b^{3}+c^{3}+3^{3}}=\frac{a}{d}$

(ii) $\left(a^{2}-b^{2}\right)\left(c^{2}-d^{2}\right)=\left(b^{2}-c^{2}\right)^{2}$

(iii) $(a+d)(b+c)-(a+c)(b+d)=(b-c)^{2}$

(iv) $a: d=$ triplicate ratio of $(a-b):(b-c)$

(v) $\left(\frac{a-b}{c}+\frac{a-c}{b}\right)^{2}-\left(\frac{d-b}{c}+\frac{d-c}{b}\right)^{2}$.

$=(a-d)^{2}\left(\frac{1}{c^{2}}-\frac{1}{b^{2}}\right)$

Sol :

a, b, c, d are in continued proportion

$\therefore \frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k(s a y)$

$\therefore c=d k, b=c k=d k . k=d k^{2}$

$a=b k=d k^{2} \cdot k=d k^{3}$

(i) L.H.S. $=\frac{a^{3}+b^{3}+c^{3}}{b^{3}+c^{3}+d^{3}}$

$=\frac{\left(d k^{3}\right)^{3}+\left(d k^{2}\right)^{3}+(d k)^{3}}{\left(d k^{2}\right)^{3}+(d k)^{3}+d^{3}}$

$=\frac{d^{3} k^{9}+d^{3} k^{6}+d^{3} k^{3}}{d^{3} k^{6}+d^{3} k^{3}+d^{3}}$

$=\frac{d^{3} k^{3}\left(k^{6}+k^{3}+1\right)}{d^{3}\left(k^{6}+k^{3}+1\right)}=k^{3}$

$\mathrm{R.H.S.}=\frac{a}{d}=\frac{d k^{3}}{d}=k^{3}$

∴ L.H.S = R.H.S.

(ii) L.H.S. $=\left(a^{2}-b^{2}\right)\left(c^{2}-d^{2}\right)$

$=\left[\left(d k^{3}\right)^{2}-\left(d k^{2}\right)^{2}\right]\left[(d k)^{2}-d^{2}\right]$

$=\left(d^{2} k^{6}-d^{2} k^{4}\right)\left(d^{2} k^{2}-d^{2}\right)$

$=d^{2} k^{4}\left(k^{2}-1\right) d^{2}\left(k^{2}-1\right)=d^{4} k^{4}\left(k^{2}-1\right)^{2}$

$\mathrm{R} \cdot \mathrm{H} . \mathrm{S} .=\left(b^{2}-c^{2}\right)^{2}=\left[\left(d k^{2}\right)^{2}-(d k)^{2}\right]^{2}$

$=\left[d^{2} k^{4}-d^{2} k^{2}\right]^{2}=\left[d^{2} k^{2}\left(k^{2}-1\right)\right]^{2}$

$=d^{4} k^{4}\left(k^{2}-1\right)^{2}$

∴ L.H.S = R.H.S.

(iii) L.H.S. $=(a+d)(b+c)-(a+c)(b+d)$

$=\left(d k^{3}+d\right)\left(d k^{2}+d k\right)-\left(d k^{3}+d k\right)\left(d k^{2}+d\right)$

$=d\left(k^{3}+1\right) d k(k+1)-d k\left(k^{2}+1\right) d\left(k^{2}+1\right)$

$=d^{2} k(k+1)\left(k^{3}+1\right)-d^{2} k\left(k^{2}+1\right)\left(k^{2}+1\right)$

$=d^{2} k\left[k^{4}+k^{3}+k+1-k^{4}-2 k^{2}-1\right]$

$=d^{2} k\left[k^{3}-2 k^{2}+k\right]=d^{2} k^{2}\left[k^{2}-2 k+1\right]$

$=d^{2} k^{2}(k-1)^{2}$

$\mathrm{R} . \mathrm{H.S.}=(b-c)^{2}=\left(d k^{2}-d k\right)^{2}=d^{2} k^{2}(k-1)^{2}$

∴ L.H.S = R.H.S.

(iv) a : d= triplicate ratio of (a-b) : (b-c)

$=(a-b)^{3}:(b-c)^{3}$

L.H.S. $=a: d=\frac{a}{d}=\frac{d k^{3}}{d}=k^{3}$

$\mathrm{R} . \mathrm{H.S.}=\frac{(a-b)^{3}}{(b-c)^{3}}$

$=\frac{\left(d k^{3}-d k^{2}\right)^{3}}{\left(d k^{2}-d k\right)^{3}}=\frac{d^{3} k^{6}(k-1)^{3}}{d^{3} k^{3}(k-1)^{3}}=k^{3}$

∴ L.H.S = R.H.S.

(v) L.H.S=

$\left(\frac{a-b}{c}+\frac{a-c}{b}\right)^{2}-\left(\frac{d-b}{c}+\frac{d-c}{b}\right)^{2}$

$=\left(\frac{d k^{3}-d k^{2}}{d k}+\frac{d k^{3}-d k}{d k^{2}}\right)^{2}-\cdot\left(\frac{d-d k^{2}}{d k}+\frac{d-d k}{d k^{2}}\right)^{2}$

$=\left(\frac{d k^{2}(k-1)}{d k}+\frac{d k\left(k^{2}-1\right)}{d k^{2}}\right)^{2}-\left(\frac{d\left(1-k^{2}\right)}{d k}+\frac{d(1-k)}{d k^{2}}\right)^{2}$

$=\left(k(k-1)+\frac{k^{2}-1}{k}\right)^{2}-\left(\frac{1-k^{2}}{k}+\frac{1-k}{k^{2}}\right)^{2}$

$=\left(\frac{k^{2}(k-1)+\left(k^{2}-1\right)}{k}\right)^{2}-\left(\frac{k\left(1-k^{2}\right)+1-k}{k^{2}}\right)^{2}$

$=\left(\frac{k^{3}-k^{2}+k^{2}-1}{k}\right)^{2}-\left(\frac{k-k^{3}+1-k}{k^{2}}\right)^{2}$

$=\frac{\left(k^{3}-1\right)^{2}}{k^{2}}-\frac{\left(-k^{3}+1\right)^{2}}{k^{4}}$

$=\frac{\left(k^{3}-1\right)^{2}}{k^{2}}-\frac{\left(1-k^{3}\right)^{2}}{k^{4}}$

$=\left(\frac{\left(k^{3}-1\right)}{k^{2}}\right)^{2}\left(1-\frac{1}{k^{2}}\right)=\frac{\left(k^{3}-1\right)^{2}\left(k^{2}-1\right)}{k^{4}}$

$=\frac{\left(k^{3}-1\right)^{2}\left(k^{2}-1\right)}{k^{4}}$

R.H.S. $=(a-d)^{2}\left(\frac{1}{c^{2}}-\frac{1}{b^{2}}\right)$

$=\left(d k^{3}-d\right)^{2}\left(\frac{1}{d^{2} k^{2}}-\frac{1}{d^{2} k^{4}}\right)$

$=d^{2}\left(k^{3}-1\right)^{2}\left(\frac{k^{2}-1}{d^{2} k^{4}}\right)=\frac{\left(k^{3}-1\right)^{2}\left(k^{2}-1\right)}{k^{4}}$

∴ L.H.S = R.H.S.