Sum

If a, b, c are in continued proportion , then prove that

`b/[b+c] = [a-b]/[a-c]`

Advertisement Remove all ads

#### 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`

`b/[ b + c] = [ck]/[ ck + c ] = [ck]/[ c( k + 1)] = k/( k + 1)` ...........(1)

`[ 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]`

Concept: Concept of Proportion

Is there an error in this question or solution?

Advertisement Remove all ads

#### APPEARS IN

Advertisement Remove all ads

Advertisement Remove all ads