Question

Prove that the product of the 2^nd and 3^rd terms of an arithmetic progression exceeds the product of the first and fourth by twice the square of the difference between the 1^st and 2^nd.

Answer 1

Let ‘a’ be the first term and ‘d’ be the common difference of A.P. 

Then, a₁ = a, a₂ = a + (2 – 1)d = a + d 

a₃ = a + (3 – 1)d = a + 2d, a₄ = a + (4 – 1)d = a + 3d 

We have to show that a₂.a₃ – a₁.a₄ = 2(a₂ – a₁)² 

LHS = a₂.a₃ – a₁.a₄ = (a + d)(a + 2d) – a(a + 3d) 

= a₂ + 3ad + 2d₂ – a₂ – 3ad = 2d₂ 

RHS = 2(a₂ – a₁)² = 2d₂ 

Since LHS = RHS. Hence proved.