JEE Main 2005 — Binomial Theorem Question with Solution

Let r = 2

$$\therefore$$ 2nd, 3rd and 4th terms are in AP.

2nd term = T₂ = $${}^m{C}_1.y$$

Coefficient of T₂ = $${}^m{C}_1$$

3rd term = T₃ = $${}^m{C}_2.y^2$$

Coefficient of T₃ = $${}^m{C}_2$$

4th term = T₄ = $${}^m{C}_3.y^3$$

Coefficient of T₂ = $${}^m{C}_3$$

$$\therefore$$ 2.$${}^m{C}_2$$ = $${}^m{C}_1$$ + $${}^m{C}_3$$

$$\Rightarrow$$ $$2.\frac{m\left(m-1\right)}{1.2}$$ = $$\frac{m}{1}$$ + $$\frac{m\left(m-1\right)\left(m-2\right)}{\mathrm{1.2.3}}$$

$$\Rightarrow$$ 6m² - 6m = 6m +m(m² - 3m + 2)

$$\Rightarrow$$ 6m² - 6m = 6m + m³ - 3m² + 2m

$$\Rightarrow$$ 6m - 6 = 6 + m² - 3m + 2

$$\Rightarrow$$ m² - 9m + 14 = 0

Now put r = 2 at each option and find answer.

In option C, $$m^2-m(4r+1)+4r^2-2=0$$ putting r = 2 we get

m² - 9m + 14 = 0. So Option C is correct.