This is an important ques from the Book ML Aggarwal class 10th,, chapter – 7, ratio and proportion.
It is given that a, b, c, d are in continued proportion and we have to prove the equalities based on the fact given in the question
Question 22, 7.2
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If a, b, c, d are in continued proportion, prove that: (i) (a3+b3+c3) /(b3+c3+d3) =a/d (ii) (a2 – b2) (c2 – d2) = (b2 – c2)2 (iii) (a + d) (b + c) – (a + c) (b + d) = (b – c)2 (iv) a: d = triplicate ratio of (a – b): (b – c) (v) [ (a-b) /c+(a-c) /b]2 – [(d-b) /c +(d-c) /b)] 2 =(a-d) 2(1/c2-1/b2)
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Solution:
It is given that
a, b, c, d are in continued proportion
Here we get
a/b = b/c = c/d = k
c = dk, b = ck = dk . k = dk2
a = bk = dk2 . k = dk3
Therefore, LHS = RHS.
(ii) LHS = (a2 – b2) (c2 – d2)
We can write it as
= [(dk3)2 – (dk2)2] [(dk)2 – d2]
By further calculation
= (d2k6 – d2k4) (d2k2 – d2)
Taking out the common terms
= d2k4 (k2 – 1) d2 (k2 – 1)
= d4k4 (k2 – 1)2
RHS = (b2 – c2)2
We can write it as
= [(dk2)2 – (dk)2]2
By further calculation
= [d2k4 – d2k2]2
Taking out the common terms
= [d2k2 (k2 – 1)]2
= d4 k4 (k2 – 1)2
Therefore, LHS = RHS.
(iii) LHS = (a + d) (b + c) – (a + c) (b + d)
We can write it as
= (dk3 + d) (dk2 + dk) – (dk3 + dk) (dk2 + d)
Taking out the common terms
= d (k3 + 1) dk (k + 1) – dk (k2 + 1) d (k2 + 1)
By further simplification
= d2k (k + 1) (k3 + 1) – d2k (k2 + 1) (k2 + 1)
So we get
= d2k (k4 + k3 + k + 1 – k4 – 2k2 – 1)
= d2k (k3 – 2k2 + k)
Taking k as common
= d2k2 (k2 – 2k + 1)
= d2k2 (k – 1)2
RHS = (b – c)2
We can write it as
= (dk2 – dk)2
Taking out the common terms
= d2k2 (k – 1)2
Therefore, LHS = RHS.
(iv) a: d = triplicate ratio of (a – b): (b – c) = (a – b)3: (b – c)3
We know that
Therefore, LHS = RHS.
(v)
Therefore, LHS = RHS.