Question

1) If a∝b, b∝c then p.t a + b + c ∝ √bc + √ca + √ab
2) a,b,c,d are in continued proportion. Prove that a² + b² + c² : b² + c² + d² = b:d​

Answer 1

SOLUTION :-

QUESTION :-

1) If a∝b, b∝c

then p.t a + b + c ∝ √bc + √ca + √ab

2) a,b,c,d are in continued proportion.

Prove that a² + b² + c² : b² + c² + d² = b:d

EVALUATION :-

1.

a ∝ b implies a = kb .......... (1)

b ∝ c implies b = mc ..........(2)

Where k and m are non zero constants

Equation (1) & (2) gives

a = kmc ........... (3)

Now

$\displaystyle \sf{ \frac{a + b + c}{ \sqrt{bc} + \sqrt{ca} + \sqrt{ab} } \: }$

$= \displaystyle \sf{ \frac{kmc + mc + c}{ \sqrt{mc.c} + \sqrt{c.kmc} + \sqrt{kmc.mc} } \: }$

$= \displaystyle \sf{ \frac{km+ m+ 1}{ \sqrt{m} + \sqrt{km} + \sqrt{k{m}^{2} } } \: }$

= constant

Hence a + b + c ∝ √bc + √ca + √ab

2.

a,b,c,d are in continued proportion

Therefore

a : b = b : c = c : d

a = kb

b = kc

c = kd

Where k is a non zero constant

Above three equations give

$\sf{a = {k}^{3}d \: }$

$\sf{b = {k}^{2}d }$

c = kd

Now

LHS

= a² + b² + c² : b² + c² + d²

$= \sf{ ({k}^{6} {d}^{2} +{k}^{4} {d}^{2} +{k}^{2} {d}^{2} ) : ({k}^{4} {d}^{2} +{k}^{2} {d}^{2} + {d}^{2} )}$

$= \sf{ {k}^{2} {d}^{2} ({k}^{4} +{k}^{2} +1 ) : {d}^{2} ({k}^{4} +{k}^{2} + 1 )}$

$= \sf{ {k}^{2} :1 }$

RHS

$\sf{b : d}$

$= \sf{ {k}^{2} d : d}$

$\sf{ = {k}^{2} :1 }$

Hence LHS = RHS

Hence proved

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