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If a, b, c are in continued proportion, prove that: (a+b)/(b+c) =a2(b-c) /b2(a-b) (ii) 1/a3+1/b3+1/c3=a/b2c2+ b/c2a2+c/a2b2 (iii) a: c = (a2 + b2): (b2 + c2) (iv) a2b2c2 (a-4 + b-4 + c-4) = b-2 (a4 + b4 + c4) (v) abc (a + b + c)3 = (ab + bc + ca)3 (vi) (a + b + c) (a – b + c) = a2 + b2 + c2

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exam oriented an important question from ML aggarwal, class 10th, chapter 7, ratio and proportion, Avichal publication

This question has been asked in 2015

It is given that a, b, c are in continued proportion ans we are asked to prove the equalities using the condition given in the equation.

Question 21, 7.2

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1 Answer

  1. Solution:

    It is given that

    a, b, c are in continued proportion

    So we get

    a/b = b/c = k

    ML Aggarwal Solutions for Class 10 Chapter 7 Image 32

    ML Aggarwal Solutions for Class 10 Chapter 7 Image 33

    Therefore, LHS = RHS.

    ML Aggarwal Solutions for Class 10 Chapter 7 Image 34

    ML Aggarwal Solutions for Class 10 Chapter 7 Image 35

    Therefore, LHS = RHS.

    (iii) a: c = (a2 + b2): (b2 + c2)

    We can write it as

    ML Aggarwal Solutions for Class 10 Chapter 7 Image 36

    ML Aggarwal Solutions for Class 10 Chapter 7 Image 37

    Therefore, LHS = RHS.

    (iv) a2b2c2 (a-4 + b-4 + c-4) = b-2 (a4 + b4 + c4)

    ML Aggarwal Solutions for Class 10 Chapter 7 Image 38

    ML Aggarwal Solutions for Class 10 Chapter 7 Image 39

    ML Aggarwal Solutions for Class 10 Chapter 7 Image 40

    Therefore, LHS = RHS.

    (v) LHS = abc (a + b + c)3

    We can write it as

    = ck2. ck. c [ck2 + ck + c]3

    Taking out the common terms

    = c3 k3 [c (k2 + k + 1)]3

    So we get

    = c3 k3. c3 (k2 + k + 1)3

    = c6 k3 (k2 + k + 1)3

    RHS = (ab + bc + ca)3

    We can write it as

    = (ck2. ck + ck. c + c. ck2)3

    So we get

    = (c2k3 + c2k + c2k2)3

    = (c2k3 + c2k2 + c2k)3

    Taking out the common terms

    = [c2k (k2 + k + 1)]3

    = c6k3 (k2 + k + 1)3

    Therefore, LHS = RHS.

    (vi) LHS = (a + b + c) (a – b + c)

    We can write it as

    = (ck2 + ck + c) (ck2 – ck + c)

    Taking out the common terms

    = c (k2 + k + 1) c (k2 – k + 1)

    = c2 (k2 + k + 1) (k2 – k + 1)

    So we get

    = c2 (k4 + k2 + 1)

    RHS = a2 + b2 + c2

    We can write it as

    = (ck2)2 + (ck)2 + (c)2

    So we get

    = c2k4 + c2k2 + c2

    Taking out the common terms

    = c2 (k4 + k2 + 1)

    Therefore, LHS = RHS.

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