Question

If second, third, and fourth terms in the expansion of (x + a) ⁿ are 240, 720, and 1080 respectively, then find the value of n.

Answer 1

(x + a) ⁿ 

= ⁿC₀ xⁿ + ⁿC₁ x^{n – 1} a + ⁿC₂ x^{n – 2} a² + ⁿC₃ x^{n – 3} a³ + ........ + ⁿC_n aⁿ 

 ∴  T₂ = ⁿC₁ x^{n – 1} a = 240 ...(i) 

T₃ = ⁿC₂ x^{n – 2} a² = 720 ...(ii) 

T₄ = ⁿC₃ x^{n – 3} a³ = 1080 ...(iii)

On dividing (ii) by (i), we get

⇒ \(\frac{(n-1)a}{2x}\) = 3  ... (iv)

On dividing (iii) by (ii) we get

⇒ \(\frac{(n-2)a}{3x}\) = \(\frac{3}{2}\)  ... (v)

Now dividing (v) by (iv) we get 

 

= \(\frac{3/2}{3}\) 

⇒ \(\frac{2(n-2)}{3(n-1)} =\frac{1}{2}\) 

⇒ 4n – 8 = 3n – 3 

⇒ n = 5.