Question

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

1 ) 2a + 3d : 3a - 4d = 2a³ + 3b³ : 3a³ - 4b³


2) (a² + b² + c²) (b² + c² + d² ) = (ab + bc + cd)²


Irrelevant answers will be deleted.​

Answer 1

Given -

• a, b and c are in continued proportion.

⠀

To prove -

1. 2a + 3d : 3d - 4d = 2a³ + 3b³ : 3a³ - 4b³
2. (a² + b² + c²) (b² + c² + d²) = (ab + bc + cd)²

⠀

Solution 1 -

⠀

We know $\dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} = k$, then

• a = bk = ck² = dk³
• b = ck = dk²
• c = dk

⠀

To prove :- $\bf{\dfrac{2a + 3d}{3a - 4d} = \dfrac{2a^3 + 3b^3}{3a^3 - 4b^3}}$

⠀

Taking L.H.S

$\sf{\dashrightarrow{\dfrac{2a + 3d}{3a - 4d}}}$

⠀

$\sf{\dashrightarrow{\dfrac{2dk^3 + 3d}{3dk^3 - 4d}}}$

⠀

$\sf{\dashrightarrow{\dfrac{2k^3 + 3}{3k^3 - 4}}}$

⠀

Now, solving R.H.S

$\sf{\dashrightarrow{\dfrac{2a^3 + 3b^3}{3a^3 - 4b^3}}}$

⠀

$\sf{\dashrightarrow{\dfrac{2d^3 k^9 + 3d^3 k^6}{3d^3 k^9 - 4d^3 k^6}}}$

⠀

$\sf{\dashrightarrow{\dfrac{d^3 k^6 (2k^3 + 3)}{d^3 k^6 (3k^3 - 4)}}}$

⠀

$\sf{\dashrightarrow{\dfrac{2k^3 + 3}{3k^3 - 4}}}$

⠀

•°• L.H.S = R.H.S

➛ Hence proved

⠀━━━━━━━━━━━━━━━━━━━━

⠀

Solution 2 -

⠀

We know that,

• a = bk
• b = ck
• c = dk

⠀

To prove :- (a² + b² + c²) (b² + c² + d²) = (ab + bc + cd)²

⠀

Substituting the values of a, b and c

$\sf{\dashrightarrow{[ (bk)^2 + (ck)^2+ (dk)^2 ] (b^2 + c^2 + d^2) = (b^2 k + c^2 k + d^2 k)^2}}$

⠀

$\sf{\dashrightarrow{[b^2 k^2 + c^2 k^2 + d^2 k^2] (b^2 + c^2 + d^2) = k^2(b^2 + c^2 + d^2)^2}}$

⠀

$\sf{\dashrightarrow{k^2(b^2 + c^2 + d^2) (b^2 + c^2 + d^2) = k^2 (b^2 + c^2 + d^2)^2}}$

⠀

$\sf{\dashrightarrow{(b^2+ c^2 + d^2)^2 = (b^2 + c^2 + d^2)^2}}$

⠀

•°• L.H.S = R.H.S

➛ Hence proved

━━━━━━━━━━━━━━━━━━━━━━

Answer 2

Answer:

Given -

a, b and c are in continued proportion.

⠀

To prove -

2a + 3d : 3d - 4d = 2a³ + 3b³ : 3a³ - 4b³

(a² + b² + c²) (b² + c² + d²) = (ab + bc + cd)²

⠀

Solution 1 -

⠀

We know \dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} = k

b

a

=

c

b

=

d

c

=k , then

a = bk = ck² = dk³

b = ck = dk²

c = dk

⠀

To prove :- \bf{\dfrac{2a + 3d}{3a - 4d} = \dfrac{2a^3 + 3b^3}{3a^3 - 4b^3}}

3a−4d

2a+3d

=

3a

3

−4b

3

2a

3

+3b

3

⠀

Taking L.H.S

\sf{\dashrightarrow{\dfrac{2a + 3d}{3a - 4d}}}⇢

3a−4d

2a+3d

⠀

\sf{\dashrightarrow{\dfrac{2dk^3 + 3d}{3dk^3 - 4d}}}⇢

3dk

3

−4d

2dk

3

+3d

⠀

\sf{\dashrightarrow{\dfrac{2k^3 + 3}{3k^3 - 4}}}⇢

3k

3

−4

2k

3

+3

⠀

Now, solving R.H.S

\sf{\dashrightarrow{\dfrac{2a^3 + 3b^3}{3a^3 - 4b^3}}}⇢

3a

3

−4b

3

2a

3

+3b

3

⠀

\sf{\dashrightarrow{\dfrac{2d^3 k^9 + 3d^3 k^6}{3d^3 k^9 - 4d^3 k^6}}}⇢

3d

3

k

9

−4d

3

k

6

2d

3

k

9

+3d

3

k

6

⠀

\sf{\dashrightarrow{\dfrac{d^3 k^6 (2k^3 + 3)}{d^3 k^6 (3k^3 - 4)}}}⇢

d

3

k

6

(3k

3

−4)

d

3

k

6

(2k

3

+3)

⠀

\sf{\dashrightarrow{\dfrac{2k^3 + 3}{3k^3 - 4}}}⇢

3k

3

−4

2k

3

+3

⠀

•°• L.H.S = R.H.S

➛ Hence proved

⠀━━━━━━━━━━━━━━━━━━━━

⠀

Solution 2 -

⠀

We know that,

a = bk

b = ck

c = dk

⠀

To prove :- (a² + b² + c²) (b² + c² + d²) = (ab + bc + cd)²

⠀

Substituting the values of a, b and c

\sf{\dashrightarrow{[ (bk)^2 + (ck)^2+ (dk)^2 ] (b^2 + c^2 + d^2) = (b^2 k + c^2 k + d^2 k)^2}}⇢[(bk)

2

+(ck)

2

+(dk)

2

](b

2

+c

2

+d

2

)=(b

2

k+c

2

k+d

2

k)

2

⠀

\sf{\dashrightarrow{[b^2 k^2 + c^2 k^2 + d^2 k^2] (b^2 + c^2 + d^2) = k^2(b^2 + c^2 + d^2)^2}}⇢[b

2

k

2

+c

2

k

2

+d

2

k

2

](b

2

+c

2

+d

2

)=k

2

(b

2

+c

2

+d

2

)

2

⠀

\sf{\dashrightarrow{k^2(b^2 + c^2 + d^2) (b^2 + c^2 + d^2) = k^2 (b^2 + c^2 + d^2)^2}}⇢k

2

(b

2

+c

2

+d

2

)(b

2

+c

2

+d

2

)=k

2

(b

2

+c

2

+d

2

)

2

⠀

\sf{\dashrightarrow{(b^2+ c^2 + d^2)^2 = (b^2 + c^2 + d^2)^2}}⇢(b

2

+c

2

+d

2

)

2

=(b

2

+c

2

+d

2

)

2

⠀

•°• L.H.S = R.H.S

➛ Hence proved

━━━━━━━━━━━━━━━━━━━━━━

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