Question

If a, b, c and dare in continued proportion, then prove that 

(a+ d)(b+ c)-(a+ c)(b+ d)= (b-c)²  

Answer 1

`"a"/"b" = "b"/"c" = "c"/"d" = "k"`

⇒ c = kd 

b =kc= k²d 

a= kb= k³d 

(a+ d)(b + c)-(a+ c)(b + d) = (b- c)² 

LHS

(a+ d)(b + c)-(a+ c)(b + d) 

= ab + bd + ac + cd - ab - bc - ad - cd 

= bd+ ca - bc - ad 

= k²d² + k⁴d⁴ - k³d² - k³d²

=  k²d² + k⁴d⁴ - 2k³d²

= k²d² (1+ k² - 2k) 

RHS

(b - c)² = (b - c)(b - c) 

= b² - 2bc + c²

= k⁴d⁴ - 2k³d² + k²d²

= k²d² (k² - 2k + 1) 

LHS = RHS. Hence, proved.