Math, asked by omshubham589, 9 months ago

4. If a, b, c, d are in continued proportion, prove
that:
(4 (6-c)² + (c-a)² + (d-6)² = (a - a)? plz give me answer fast

Answers

Answered by mayajakhar79

0

Answer:

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(i) Since a, b, c, d are in continued proportion then

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

⇒ a = bk, b = ck , c = dk

⇒ a = ck2

⇒ a = dk3, b = dk2 and c = dk

(ii) L.H.S. = (d2k6 + d2k4 + d2k2)(d2k4 + d2k2 + d2)

= d2k2 (k4 + k2 + 1).d2 (k4 + k2 + 1)

= d4k2(k4 + k2 + 1)2

R.H.S. = (ab + bc + cd)2

= (dk3.dk2 + dk2.dk + dk.d)2

= d4.k2(k4 + k2 + 1)2

L.H.S. = R.H.S.

Hence proved.

Answered by Anonymous

0

(i) Since a, b, c, d are in continued proportion then

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

⇒ a = bk, b = ck , c = dk

⇒ a = ck2

⇒ a = dk3, b = dk2 and c = dk

(ii) L.H.S. = (d2k6 + d2k4 + d2k2)(d2k4 + d2k2 + d2)

= d2k2 (k4 + k2 + 1).d2 (k4 + k2 + 1)

= d4k2(k4 + k2 + 1)2

R.H.S. = (ab + bc + cd)2

= (dk3.dk2 + dk2.dk + dk.d)2

= d4.k2(k4 + k2 + 1)2

L.H.S. = R.H.S.

Hence proved.

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