In the binomial expansion of (p-q)^{14},if th | vedclass

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(B) The general term in the expansion of $(p-q)^{14}$ is given by $T_{r+1} = \binom{14}{r} p^{14-r} (-q)^r$. For the $7^{	ext{th}}$ term,$r=6$: $T_7

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$15$

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(B) The general term in the expansion of $(p-q)^{14}$ is given by $T_{r+1} = \binom{14}{r} p^{14-r} (-q)^r$. For the $7^{	ext{th}}$ term,$r=6$: $T_7 = \binom{14}{6} p^8 (-q)^6 = \binom{14}{6} p^8 q^6$. For the $8^{	ext{th}}$ term,$r=7$: $T_8 = \binom{14}{7} p^7 (-q)^7 = -\binom{14}{7} p^7 q^7$. Given $T_7 + T_8 = 0$,we have $\binom{14}{6} p^8 q^6 - \binom{14}{7} p^7 q^7 = 0$. This implies $\binom{14}{6} p^8 q^6 = \binom{14}{7} p^7 q^7$. Dividing both sides by $\binom{14}{6} p^7 q^6$,we get $p = \frac{\binom{14}{7}}{\binom{14}{6}} q$. Calculating the ratio of binomial coefficients: $\frac{\binom{14}{7}}{\binom{14}{6}} = \frac{14!}{7!7!} 	imes \frac{6!8!}{14!} = \frac{8}{7}$. So,$p = \frac{8}{7} q$,which means $7p = 8q$. We need to find $\frac{p+q}{p-q}$. Substituting $p = \frac{8}{7} q$: $\frac{\frac{8}{7}q + q}{\frac{8}{7}q - q} = \frac{\frac{15}{7}q}{\frac{1}{7}q} = 15$.

In the binomial expansion of $(p-q)^{14}$,if the sum of $7^{\text{th}}$ term and $8^{\text{th}}$ term is zero,then $\frac{p+q}{p-q}=$

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