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

Solution:It is given that

a, b, c are in continued proportion

So we get

a/b = b/c = k

Therefore, LHS = RHS.

Therefore, LHS = RHS.

(iii) a: c = (a

^{2}+ b^{2}): (b^{2}+ c^{2})We can write it as

Therefore, LHS = RHS.

(iv) a

^{2}b^{2}c^{2}(a^{-4}+ b^{-4}+ c^{-4}) = b^{-2}(a^{4}+ b^{4}+ c^{4})Therefore, LHS = RHS.

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

^{3}We can write it as

= ck

^{2}. ck. c [ck^{2}+ ck + c]^{3}Taking out the common terms

= c

^{3}k^{3}[c (k^{2}+ k + 1)]^{3}So we get

= c

^{3}k^{3}. c^{3}(k^{2}+ k + 1)^{3}= c

^{6}k^{3}(k^{2}+ k + 1)^{3}RHS = (ab + bc + ca)

^{3}We can write it as

= (ck

^{2}. ck + ck. c + c. ck^{2})^{3}So we get

= (c

^{2}k^{3}+ c^{2}k + c^{2}k^{2})^{3}= (c

^{2}k^{3}+ c^{2}k^{2}+ c^{2}k)^{3}Taking out the common terms

= [c

^{2}k (k^{2}+ k + 1)]^{3}= c

^{6}k^{3}(k^{2}+ k + 1)^{3}Therefore, LHS = RHS.

(vi) LHS = (a + b + c) (a – b + c)

We can write it as

= (ck

^{2}+ ck + c) (ck^{2}– ck + c)Taking out the common terms

= c (k

^{2}+ k + 1) c (k^{2}– k + 1)= c

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

= c

^{2}(k^{4}+ k^{2}+ 1)RHS = a

^{2}+ b^{2}+ c^{2}We can write it as

= (ck

^{2})^{2}+ (ck)^{2}+ (c)^{2}So we get

= c

^{2}k^{4}+ c^{2}k^{2}+ c^{2}Taking out the common terms

= c

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