if a,b,c are in continued proportion and a(b-c) =2b prove that a-c= 2(a+b)/a
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given a , b , c are in continued proportion
⇒ a : b : : b : c ⇒ a x c = b x b
ac=b² .............(1)
now A/Q a(b-c)=2b ⇒ ab - ac -2b = 0
⇒ ab - b² -2b = 0
⇒ b(a - b -2) = 0
⇒ a - b -2 =0 as b≠0
⇒ a - b = 2 ............(2)
now to prove a-c=2(a+b)/a
LHS = a-c ⇒ multipying a to nr. and dr.
⇒ a(a-c)/a ⇒ a² - ac/ a ⇒ a² - b² / a
⇒ (a-b)(a+b)/a ⇒ 2(a+b)/a from (1) and (2)
hence proved
hope this helps
⇒ a : b : : b : c ⇒ a x c = b x b
ac=b² .............(1)
now A/Q a(b-c)=2b ⇒ ab - ac -2b = 0
⇒ ab - b² -2b = 0
⇒ b(a - b -2) = 0
⇒ a - b -2 =0 as b≠0
⇒ a - b = 2 ............(2)
now to prove a-c=2(a+b)/a
LHS = a-c ⇒ multipying a to nr. and dr.
⇒ a(a-c)/a ⇒ a² - ac/ a ⇒ a² - b² / a
⇒ (a-b)(a+b)/a ⇒ 2(a+b)/a from (1) and (2)
hence proved
hope this helps
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