Question

if a,b,c and d are in continuous proportion,prove that (a2+b2+c2) (b2+c2+d2)=(ab+bc+cd)2​

Answer 1

Question:-

• If a,b,c and d are in continuous proportion,prove that (a² + b² + c²) (b² + c² + d²) = (ab + bc + cd)²

Proof :-

Let –

a/b = b/c = c/d = k

Then –

• a = bk
• b = ck
• c = dk

L.H.S –

$\qquad \longrightarrow {\purple{\underline{\underline{\pmb{\sf{ (a² + b² + c²) (b² + c² + d²)}}}}}}$

$\qquad \sf\longrightarrow (b²k² + c²k² + d²k²) (c²k² + d²k² + d²)$

$\qquad \sf\longrightarrow(c²k⁴ + c²k² + d²k²) (d²k²k² + d²k² + d²)$

$\qquad \sf\longrightarrow (d²k⁶ + d²k⁴ + d²k²) (d²k⁴ + d²k² + d²)$

$\qquad \sf\longrightarrow d²k²(k⁴ + k² + 1) d²(k⁴ + k² + 1)$

$\qquad \longrightarrow {\pink{\underline{\underline{\pmb{\sf{ d⁴k²(k⁴ + k² + 1)²}}}}}}$

R.H.S–

$\qquad \longrightarrow {\purple{\underline{\underline{\pmb{\sf{ (ab + bc + cd)²}}}}}}$

$\qquad \sf\longrightarrow (bk.b + ck.c + dk.d)²$

$\qquad \sf\longrightarrow (b²k + c²k + d²k)²$

$\qquad \sf\longrightarrow (c²k²k + d²k²k + d²k)²$

$\qquad \sf\longrightarrow (d²k²k² × k + d²k³ + d²k)²$

$\qquad \sf\longrightarrow (d²k⁵ + d²k³ + d²k)²$

$\qquad \sf\longrightarrow{d²k(k⁴ + k² + 1)}²$

$\qquad \longrightarrow {\pink{\underline{\underline{\pmb{\sf{ d⁴k²(k⁴ + k² + 1)²}}}}}}$

Henceforth –

• L.H.S = R.H.S (Proved)

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