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

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 = dk

^{2}a = bk = dk

^{2}. k = dk^{3}Therefore, LHS = RHS.

(ii) LHS = (a

^{2}– b^{2}) (c^{2}– d^{2})We can write it as

= [(dk

^{3})^{2}– (dk^{2})^{2}] [(dk)^{2}– d^{2}]By further calculation

= (d

^{2}k^{6}– d^{2}k^{4}) (d^{2}k^{2}– d^{2})Taking out the common terms

= d

^{2}k^{4}(k^{2}– 1) d^{2}(k^{2}– 1)= d

^{4}k^{4}(k^{2}– 1)^{2}RHS = (b

^{2}– c^{2})^{2}We can write it as

= [(dk

^{2})^{2}– (dk)^{2}]^{2}By further calculation

= [d

^{2}k^{4}– d^{2}k^{2}]^{2}Taking out the common terms

= [d

^{2}k^{2}(k^{2}– 1)]^{2}= d

^{4}k^{4}(k^{2}– 1)^{2}Therefore, LHS = RHS.

(iii) LHS = (a + d) (b + c) – (a + c) (b + d)

We can write it as

= (dk

^{3}+ d) (dk^{2}+ dk) – (dk^{3}+ dk) (dk^{2}+ d)Taking out the common terms

= d (k

^{3}+ 1) dk (k + 1) – dk (k^{2}+ 1) d (k^{2}+ 1)By further simplification

= d

^{2}k (k + 1) (k^{3}+ 1) – d^{2}k (k^{2}+ 1) (k^{2}+ 1)So we get

= d

^{2}k (k^{4}+ k^{3}+ k + 1 – k^{4}– 2k^{2}– 1)= d

^{2}k (k^{3}– 2k^{2}+ k)Taking k as common

= d

^{2}k^{2}(k^{2}– 2k + 1)= d

^{2}k^{2}(k – 1)^{2}RHS = (b – c)

^{2}We can write it as

= (dk

^{2}– dk)^{2}Taking out the common terms

= d

^{2}k^{2}(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.