Mathematics High School

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

Answers

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 = bkb = ckc = dk

L.H.S –

[tex]\qquad \longrightarrow {\purple{\underline{\underline{\pmb{\sf{ (a² + b² + c²) (b² + c² + d²)}}}}}}[/tex]

[tex]\qquad \sf\longrightarrow (b²k² + c²k² + d²k²) (c²k² + d²k² + d²)[/tex]

[tex]\qquad \sf\longrightarrow(c²k⁴ + c²k² + d²k²) (d²k²k² + d²k² + d²)[/tex]

[tex]\qquad \sf\longrightarrow (d²k⁶ + d²k⁴ + d²k²) (d²k⁴ + d²k² + d²)[/tex]

[tex]\qquad \sf\longrightarrow d²k²(k⁴ + k² + 1) d²(k⁴ + k² + 1)[/tex]

[tex]\qquad \longrightarrow {\pink{\underline{\underline{\pmb{\sf{ d⁴k²(k⁴ + k² + 1)²}}}}}}[/tex]

R.H.S–

[tex]\qquad \longrightarrow {\purple{\underline{\underline{\pmb{\sf{ (ab + bc + cd)²}}}}}}[/tex]

[tex]\qquad \sf\longrightarrow (bk.b + ck.c + dk.d)²[/tex]

[tex]\qquad \sf\longrightarrow (b²k + c²k + d²k)²[/tex]

[tex]\qquad \sf\longrightarrow (c²k²k + d²k²k + d²k)²[/tex]

[tex]\qquad \sf\longrightarrow (d²k²k² × k + d²k³ + d²k)²[/tex]

[tex]\qquad \sf\longrightarrow (d²k⁵ + d²k³ + d²k)²[/tex]

[tex]\qquad \sf\longrightarrow{d²k(k⁴ + k² + 1)}²[/tex]

[tex]\qquad \longrightarrow {\pink{\underline{\underline{\pmb{\sf{ d⁴k²(k⁴ + k² + 1)²}}}}}}[/tex]

Henceforth –

L.H.S = R.H.S (Proved)

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Answers

Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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  -8          -8

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in pic

(Credits: Symbolab)

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Step-by-step explanation:

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Step-by-step explanation:

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Guided Practice


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A.
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