Choose the correct answer.

If the coefficients of 2^nd, 3rd and the 4^th terms in the expansion of (1 + x)ⁿ are in A.P., then value of n is:

[Hint: 2ⁿC₂ = ⁿC₁ + ⁿC₃ ⇒ n² - 9n + 14 = 0 ⇒ n = 2 or 7.]

Question

Choose the correct answer.

If the coefficients of 2^nd, 3rd and the 4^th terms in the expansion of (1 + x)ⁿ are in A.P., then value of n is:

[Hint: 2ⁿC₂ = ⁿC₁ + ⁿC₃ ⇒ n² - 9n + 14 = 0 ⇒ n = 2 or 7.]

Aicharya · Accepted Answer

Solution: $(1+\text{x})^\text{n}=\ ^\text{n}\text{C}_0+\ ^\text{n}\text{C}_1\text{x}+\ ^\text{n}\text{C}_2\text{x}^2+\ ^\text{n}\text{C}_3\text{x}^3+...+\ ^\text{n}\text{C}_\text{n}\text{x}^\text{n}$ So, coefficients of 2 nd , 3 rd and 4th terms are n C 1 , n C 2 and n C 3 , respectively. Given that, n C 1 , n C 2 and n C 3 , are in A.P. $\therefore2\ ^\text{n}\text{C}_2=\ ^\text{n}\text{C}_1+\ ^\text{n}\text{C}_3$ $\Rightarrow2\Big[\frac{\text{n}!}{(\text{n}-2)!2!}\Big]=\text{n}+\frac{\text{n}!}{3!(\text{n}-3)!}$ $\Rightarrow2\Big[\frac{\text{n}(\text{n}-1)}{2!}\Big]=\text{n}+\frac{\text{n}(\text{n}-1)(\text{n}-2)}{3!}\Rightarrow(\text{n}-1)=1+\frac{(\text{n}-1)(\text{n}-2)}{6}$ $\Rightarrow6\text{n}-6=6+\text{n}^2-3\text{n}+2\Rightarrow\text{n}^2-9\text{n}+14=0$ $\Rightarrow(\text{n}-7)(\text{n}-2)=0$ $\therefore\text{n}=2\ \text{or}\ \text{n}=7$ Since n = 2 is not possible, so n = 7.

Choose the correct answer. If the coefficients of 2 nd , 3rd and the 4 th terms in the expansion of (1 + x)…

Solution