Q85 / 125 · BITSAT 2015 · Mathematics
If binomial coefficients of three consecutive terms of $(1+x)^{2}$ are in $H P$, then the maximum value of $\mathrm{n}$ is
- A1
- B2
- C0
- Correct answerDNone of these
Solution
Let the coefficients of rth, $(r+1)$ th, and (r $+$
2) th terms be in HP.
Then, $\frac{2}{{ }^{n} C_{r}}=\frac{1}{{ }^{n} C_{r-1}}+\frac{1}{{ }^{n} C_{r}+1}$
$$
\begin{array}{l}
\Rightarrow \quad 2=\frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}}=\frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}} \\
\Rightarrow \quad 2=\frac{\mathrm{n}-\mathrm{r}+1}{\mathrm{r}}+\frac{\mathrm{r}+1}{\mathrm{n}-\mathrm{r}} \\
\Rightarrow \mathrm{n}^{2}-4 \mathrm{nr}+4 \mathrm{r}^{2}+\mathrm{n}=0 \\
\Rightarrow(\mathrm{n}-2 \mathrm{r})^{2}+\mathrm{n}=0
\end{array}
$$
which is not possible for any value for $n$.
Hence, the correct answer is option: 'D'.