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if `(2a+3b+4c+5d)/(2a-3b-4c-5d)=(2a-3b+4c-5d)/(2a-3b-4c+5d)`, then show that a,3b,2c and 5d are in proportion. |
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Answer» Given, `(a+3b+4c+5d)/(2a+3b-4c-5d)=(2a-3b+4c+5d)/(2a-3b-4c+5d)` `implies((2a+3b)+(4c+5d))/((2a+3b)-(4c+5d))=((2a-3b)+(4c-5d))/((2a-3b)-(4c-5d))` `implies(2a+3b)/(4c+5d)=(2a-3b)/(4c-5d)implies(2a+3b)/(2a-3b)=(4c+5d)/(4c-5d)` Using componendo and dividendo rule, `((2a+3b)+(2a-3b))/((2a+3b)-(2a-3b))=((4c+5d)+(4c-5d))/((4c+5d)-(4c-5d))` `implies(4a)/(6b)=(8c)/(10d)implies(a)/(3b)=(2c)/(5d)` `thereforea:3b: :2c:5d` So, a,3b, 2c and 5d are in proportion |
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