Question

If a, b, c are in continued proportion, prove that: `(a + b)/(b + c) = (a^2(b - c))/(b^2(a - b)`.

Answer 1

`(a + b)/(b + c) = (a^2 (b - c))/(b^2 (a - b)`

Since, a, b, c are in continued proportion, `a/b = b/c`.

Let, `a/b = b/c = k`.

Then, a = bk and b = ck

Hence, a = bk = (ck). k = ck²

LHS = `(a + b)/(b + c)`

LHS = `(ck^2 + ck)/(ck + c)`

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

LHS = `[cancelck(cancel(k + 1))]/[cancelc(cancel(k + 1))]`

LHS = k ...(I)

RHS = `(a^2(b - c))/(b^2(a - b)`

RHS = `((ck^2)^2(ck - c))/((ck)^2(ck^2 - ck)`

RHS = `(c^2k^4(ck - c))/(c^2k^2(ck^2 - ck))`

RHS = `(c^3k^4(k - 1))/(c^3k^3(k - 1))`

RHS = `[cancel(c^3)k^(cancel4)(cancel(k - 1))]/[cancel(c^3)(cancel(k - 1))]`

RHS = k  ...(II)

From (I) and (II), 

LHS = RHS