Question

5. Ifa, b, c, d are continued proportion, let us prove that
(b-c)2+(c–a)2+(b-d)2 = (a-d)2?​

Answer 1

Answer:

Given :-

• If a, b, c, d are in continued proportion.

Prove That :-

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

Solution :-

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

Let, a/b = b/c = c/d = k [where, k ≠ 0]

where, a/b = k; b/c = k; c/d = k

Now, we have to find the value of a, b, c

➔ c/d = k = c = dk

➔ b/c = k = b = dk . k = dk²

➔ a/b = k = a = dk² . k = dk³

Hence, a = dk³, b = dk² and c = dk

Now,

➤ L.H.S = (b - c)² + (c - a)² + (b - d)²

Put, a = dk³, b = dk² and c = dk we get,

⇒ (dk² - dk)² + (dk - dk³)² + (dk² - d)²

⇒ {dk(k + 1)}² + {dk(1 - k²)}² + {d(k² - 1)}²

⇒ d²k²(k² - 2k + 1) + d²k²(1 - 2k² + k⁴) + d²(k⁴ - 2k² + 1)

⇒ d²{k⁴ - 2k³ + k² + k² - 2k⁴ + k⁶ + k⁴ + 2k² + 1)

⇒ d²(k⁶ - 2k³ + 1)

➠ d²(k³ - 1)²

➤ R.H.S = (a - d)²

Put a = dk³ we get,

↦ (dk³ - d)²

↦ {d(k³ - 1)}²

➦ d²(k³ - 1)²

∴ L.H.S = R.H.S. (PROVED)