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 = (a2 + b2): (b2 + c2)
We can write it as
Therefore, LHS = RHS.
(iv) a2b2c2 (a-4 + b-4 + c-4) = b-2 (a4 + b4 + c4)
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.