Q) If x, y and z are in continued proportion, prove that:
π₯/π¦2. π§2 + π¦/π§2. π₯2 +π§/π₯2. π¦2 = 1/π₯3 + 1/π¦3 + 1/π§3
ICSE Specimen Question Paper – 2026
Step 1: We know that the a continued proportionΒ isΒ a sequence of numbers where the ratio between consecutive terms remains the same:
=> x/y = y/ z
=> x.z = y2 β¦β¦β¦β¦. (1)
Step 2: Next, let’s start solving from LHS:
LHS = π₯/π¦2. π§2 + π¦/π§2. π₯2 +π§/π₯2. π¦2
By taking LCM, we get:
LHS = (π₯(π₯2) + π¦(π¦2) + π§( π§2)) / (π₯2.π¦2.π§2)
= (π₯3 + π¦3 + π§3) / (π₯2.π¦2.π§2)
Step 3: By substuting value of y2 from equation (1) in denominator, we get:
LHS = (π₯3 + π¦3 + π§3)/(π₯2.(x.z).π§2)
= (π₯3 + π¦3 + π§3)/(π₯3.π§3)
= π₯3/(π₯3.π§3) + π¦3/(π₯3.π§3) + π§3/(π₯3.π§3)
= 1/π§3 + π¦3/(π₯3.π§3) + 1/π§3
Step 4: Again, by substuting value of y2 from equation (1) in nominator, we get:
LHS = 1/π§3 + π¦3/(y2)3 + 1/π§3
= 1/π§3 + π¦3/y6 + 1/π§3
= 1/π§3 + 1/π¦3 + 1/π§3
= RHS
Hence Proved !
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