Question

If a , b , c , d are in continued proportion , prove that : -
1 ) 2a + 3d : 3a - 4d = 2a³ + 3b³ : 3a³ - 4b³

Answer 1

QuestioN :

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

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

GiveN :

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

To FiNd :

• We have to prove.

ANswer :

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

SolutioN :

$\sf \bigstar \huge\underline{We \:know\: that}:-$

$\sf \dfrac{a}{b}=\dfrac dc=\dfrac cd=k, \:then$

$\sf a=bk = ck^2=dk^3$

      $\sf And$

$\sf b=ck=dk^2\:and\:c=dk$

$\sf \bigstar \underline{L.H.S}:-$

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

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

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

$\sf \bigstar \underline{R.H.S}:-$

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

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

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

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

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

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Answer 2

Answer:

Step-by-step explanation: