If the 7th and 8th term of the binomial expansion (2a - 3b)n are equal, then (2a + 3b/2a - 3b) is equal to

Question

If the 7th and 8th term of the binomial expansion (2a - 3b)n are equal, then (2a + 3b/2a - 3b) is equal to (A) (13-n/n+1) (B) (n+1/13-n) (C) (6-n/13-n) (D) (n - 1/13-n)

Tardigrade Admin · Accepted Answer

We have, (2 a-3 b)n T7 =T8 ⇒ n C6(2 a)n-6(-3 b)6 = n C7(2 a)n-7(-3 b)7 ⇒ n C6(2 a) = n C7(-3 b) ⇒ (2 a/3 b) =-( n C7/ n C6) ⇒ (2 a/3 b)=-((n !/(n-7) ! 7 !)/(n !)(n-6) ! 6 !) ⇒ (2 a/3 b)=-(n-6/7) ⇒ (2 a/3 b)=(6-n/7) On applying componendo and dividendo, we get (2 a+3 b/2 a-3 b)=(6-n+7/6-n-7) =(13-n/-(n+1)) =-[(13-n/n+1)]

Q. If the $7^{t h}$ and $8^{t h}$ term of the binomial expansion $(2 a - 3 b)^{n}$ are equal, then $\frac{2 a + 3 b}{2 a - 3 b}$ is equal to

$\frac{13 - n}{n + 1}$

$\frac{n + 1}{13 - n}$

$\frac{6 - n}{13 - n}$

$\frac{n - 1}{13 - n}$

none of the above

Q. If the 7th and 8th term of the binomial expansion (2a−3b)n are equal, then 2a−3b2a+3b​ is equal to

n+113−n​

13−nn+1​

13−n6−n​