If the coefficients of T_r, T_{r+1}, T_{r+2} | vedclass

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(D) The coefficients of $T_r, T_{r+1}, T_{r+2}$ in $(1+x)^{14}$ are $^{14}C_{r-1}, ^{14}C_r, ^{14}C_{r+1}$ respectively.Since they are in $A.P.$,we ha

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$9$

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(D) The coefficients of $T_r, T_{r+1}, T_{r+2}$ in $(1+x)^{14}$ are $^{14}C_{r-1}, ^{14}C_r, ^{14}C_{r+1}$ respectively.Since they are in $A.P.$,we have $2 \cdot ^{14}C_r = ^{14}C_{r-1} + ^{14}C_{r+1}$.Dividing by $^{14}C_r$,we get $2 = \frac{^{14}C_{r-1}}{^{14}C_r} + \frac{^{14}C_{r+1}}{^{14}C_r}$.Using the formula $\frac{^{n}C_k}{^{n}C_{k-1}} = \frac{n-k+1}{k}$,we have $\frac{^{14}C_{r-1}}{^{14}C_r} = \frac{r}{14-r+1} = \frac{r}{15-r}$ and $\frac{^{14}C_{r+1}}{^{14}C_r} = \frac{14-r}{r+1}$.So,$2 = \frac{r}{15-r} + \frac{14-r}{r+1}$.$2(15-r)(r+1) = r(r+1) + (14-r)(15-r)$.$2(15r + 15 - r^2 - r) = r^2 + r + 210 - 14r - 15r + r^2$.$2(-r^2 + 14r + 15) = 2r^2 - 28r + 210$.$-2r^2 + 28r + 30 = 2r^2 - 28r + 210$.$4r^2 - 56r + 180 = 0$.$r^2 - 14r + 45 = 0$.$(r-5)(r-9) = 0$.Thus,$r = 5$ or $r = 9$. Since $9$ is the only option provided,$r = 9$.

If the coefficients of $T_r, T_{r+1}, T_{r+2}$ terms of $(1+x)^{14}$ are in $A.P.$,then $r =$

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