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If the coefficients of $r$ th and $(r+1)$ th terms in the expansion of $(3+7 x)^{29}$ are equal, then $r$ is equal to
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The correct answer is:
21
Coefficient of $r$ th term $=$ Coefficient of $(r+1)$ th term $\therefore{ }^{29} C_{r-1}(3)^{29-r+1}(7)^{r-1}={ }^{29} C_r(3)^{29-r}(7)^r$
$\begin{aligned} & \Rightarrow \quad{ }^{29} C_{r-1} \frac{3}{7}={ }^{29} C_r \\ & \Rightarrow \frac{29 !}{(r-1) !(29-r+1) !} \times \frac{3}{7}=\frac{29 !}{(29-r) ! r !} \\ & \Rightarrow \quad \frac{3}{7(29-r+1)}=\frac{1}{r} \\ & \Rightarrow \quad 3 r=203-7 r+7 \\ & \Rightarrow \quad 10 r=210 \\ & \Rightarrow \quad r=21 \\ & \end{aligned}$
$\begin{aligned} & \Rightarrow \quad{ }^{29} C_{r-1} \frac{3}{7}={ }^{29} C_r \\ & \Rightarrow \frac{29 !}{(r-1) !(29-r+1) !} \times \frac{3}{7}=\frac{29 !}{(29-r) ! r !} \\ & \Rightarrow \quad \frac{3}{7(29-r+1)}=\frac{1}{r} \\ & \Rightarrow \quad 3 r=203-7 r+7 \\ & \Rightarrow \quad 10 r=210 \\ & \Rightarrow \quad r=21 \\ & \end{aligned}$
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