Question

If the first and third terms of an AP are (a-b)^3 and (a+b)^3, what is the second term?​

Answer 1

Step-by-step explanation:

Given, 1st term =a=(a-b)^3=a^3 - b^3 –3ab(a-b)

3rd term=a3=(a+b)^3= a^3 + b^3 + 3ab(a+b)

We know, common difference is constant in AP

a2-a1=a3-a2

2a2=a1+a3=a^3 - b^3 –3ab(a-b)+a^3 + b^3 +3ab(a+b)=a^3 - b^3 – 3(a^2)b + 3a(b^2 )+ a^3 + b^3 + 3(a^2)b + 3a(b^2)=a^3+ a^3 + 3a(b^2 )+ 3a(b^2 )= 2a^3 + 6a(b^2)=2a( a^2 + 3(b^2))

Thus, a2= a( a^2 + 3(b^2))

Answer 2

Answer:-

Given:

first term of an AP - a = (a - b)³

Third term – a₃ = (a + b)³

We know that,

If a , b , c are in AP , then 2b = a + c.

(using formula of arithmetic mean)

So,

2 * a₂ = (a - b)³ + (a + b)³

using -

• (a - b)³ = a³ - b³ - 3a²b + 3ab²

• (a + b)³ = a³ + b³ + 3a²b + 3ab²

We get,

⟶ 2 * a₂ = a³ - b³ - 3a²b + 3ab² + a³ + b³ + 3a²b + 3ab²

⟶ 2 * a₂ = 2a³ + 6ab²

⟶ a₂ = 2(a³ + 3ab²)/2

⟶ a₂ = a³ + 3ab²

∴ The second term of the given AP is a³ + 3ab².

_________________________

Proof for,

• (a - b)³ = a³ - b³ - 3a²b + 3ab²

⟶ (a - b)(a - b)(a - b)

⟶ a(a - b) - b(a - b)(a - b)

⟶ a² - ab - ab + b² (a - b)

⟶ a(a² - ab - ab + b²) - b(a² - ab - ab + b²)

⟶ a³ - a²b - a²b + ab² - a²b + ab² + ab² - b³

⟶ a³ - b³ - 3a²b + 3ab²

_________________________

Proof for,

• (a + b)³ = a³ + b³ + 3a²b + 3ab²

⟶ (a + b)(a + b)(a + b)

⟶ a(a + b) + b(a + b)(a + b)

⟶ a² + ab + ab + b² (a + b)

⟶ a (a² + ab + ab + b²) + b(a² + ab + ab + b²)

⟶ a³ + a²b + a²b + ab² + a²b + ab² + ab² + b³

⟶ a³ + b³ + 3a²b + 3ab²