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Question
If a, b, c are in continued proportion, then prove that
`b/[b+c] = [a-b]/[a-c]`
Solution
a, b, c are in continued proportion.
`therefore a/b = b/c = k`
⇒ `a = bk, b = ck`
⇒ `a = bk = ck xx k = ck^2`
LHS = `b/[ b + c] = [ck]/[ ck + c] `
= `[ck]/[ c( k + 1)]`
= `k/( k + 1)` ..(1)
RHS = `[ a - b]/[ a - c ] = [ ck^2 - ck]/[ ck^2 - c ]`
= `[ck( k -1)]/[c(k^2 - 1)]`
= `[k( k -1)]/[(k-1)(k+1)]`
= `k/(k + 1)` ...(2)
From (1) and (2), we get
`b/[b+c] = [a-b]/[a-c]`
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