Question

If coefficients of Tr Tr+1 Tr+2 terms of (1+x)^14 are in AP then r=?

• A. 5
• B. 9
• C. 5, 9
• D. No Solution

Answer 1

Answer: (C)

5, 9

Answer 2

The coefficients of the terms $T_r$, $T_{r+1}$, $T_{r+2}$ in the expansion of $(1+x)^{14}$ are $\binom{14}{r-1}$, $\binom{14}{r}$, and $\binom{14}{r+1}$ respectively. For these coefficients to be in an Arithmetic Progression (AP), the middle term's coefficient doubled must equal the sum of the first and third terms' coefficients:

$$2\binom{14}{r} = \binom{14}{r-1} + \binom{14}{r+1}$$

Dividing by $\binom{14}{r}$ and using the identity $\frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n-k+1}{k}$ and $\frac{\binom{n}{k+1}}{\binom{n}{k}} = \frac{n-k}{k+1}$, we get:

$$2 = \frac{r}{15-r} + \frac{14-r}{r+1}$$

Solving this equation leads to a quadratic equation $r^2 - 14r + 45 = 0$, which factors into $(r-5)(r-9)=0$. Thus, $r=5$ or $r=9$. Both values are valid as they satisfy the conditions for the existence of the binomial coefficients.