Question

if a,b,c,d are in continued proportion, prove that (ab)^1/2-(bc)^1/2+(cd)^1/2={(a-b+c)(b-c+d)}^1/2

Answer 1

Given : a,b,c,d are in continued proportion,

To  find : prove that √ab  - √bc  + √cd   =   √(a - b + c)(b - c + d)

Solution:

a,b,c,d are in continued proportion,

=> a/b = b/c = c/d  

let say  a/b = b/c = c/d   = k

=> c = dk

=> b = ck = (dk)k  = dk²

& a = bk = dk²k = dk³

√ab  - √bc  + √cd   =   √(a - b + c)(b - c + d)

LHS = √ab  - √bc  + √cd  

= √dk³dk²  - √dk²dk  + √dkd  

= dk²√k  - dk√k + d √k

= d √k ( k² - k + 1)

RHS = √(a - b + c)(b - c + d)

= √(dk³ - dk² + dk)(dk² - dk + d)

= √dk( k² - k + 1)d( k² - k + 1)

= √d²k( k² - k + 1)²

= d√k ( k² - k + 1)

LHS = RHS

QED

Hence proved

√ab  - √bc  + √cd   =   √(a - b + c)(b - c + d)

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