Question

If a, b, c are in continued proportion, prove that: a² b² c² (a^-4 + b^-4 + c^-4) = b^-2(a⁴ + b⁴ + c⁴)

Answer 1

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

`a/b = b/c` = k

`a/b` = k  ∴  a = bk

`b/c` = k  ∴ b = ck

L.H.S. = a² b² c² (a^-4 + b^-4 + c^-4)

L.H.S. = `a^2b^2c^2[1/a^4 + 1/b^4 + 1/c^4]`

L.H.S. = `(a^2b^2c^2)/a^4 + (a^2b^2c^2)/b^4 + (a^2b^2c^2)/c^4`

L.H.S. = `(b^2c^2)/a^2 + (c^2a^2)/b^2 + (a^2b^2)/c^2`

L.H.S. = `((ck)^2.c^2)/((ck^2)^2) + (c^2(ck^2)^2)/(ck)^2 + ((ck^2)^2(ck)^2)/(c^2)`

L.H.S. = `(c^2k^2.c^2)/(c^2k^4) + (c^2.c^2k^4)/(c^2k^2) + (c^2k^4.c^2k^2)/(c^2)`

L.H.S. = `c^2/k^2 + (c^2k^2)/(1) + (c^2k^6)/(1)`

L.H.S. = `c^2[1/k^2 + k^2 + k^6]`

L.H.S. = `c^2/k^2[ 1 + k^4 + k^8]`

R.H.S. = b^{- 2} [a⁴ + b⁴ + c⁴]

R.H.S. = `(1)/b^2[a^4 + b^4 + c^4]`

R.H.S. = `(1)/(ck)^2[(ck^2)^4 + (ck)^4 + c^4]`

R.H.S. = `(1)/(c^2k^2)[c^4k^8 + c^4k^4 + c^4]`

R.H.S. = `c^4/(c^2k^2)[k^8 + k^4 + 1]`

R.H.S. = `c^2/k^2[1 + k^4 + k^8]`

∴ L.H.S. = R.H.S.

Hence proved.