The correct option is B 10
The given expression may be written as (214+3−14)n.
The general term in this expansion is given by Tr+1= nCr×(214)(n−r)×(3−14)r
⇒Tr+1= nCr×2(n−r)4×(3−14)r ...(1)
Now, pth term from the end
=(n−p+2)th term from the beginning.
∴5th term from the end
=(n−5+2)th term from the beginning
=(n−3)th term from the beginning
Now, T5=T4+1
= nC4×2(n−4)4×3(−44)
= nC4×2(n−4)4×3−1 ...(2)
And, Tn−3=T(n−4)+1
= nCn−4×21×3−(n−4)4
(putting r=(n−4) in eqn (1))
= nC4×2×3−(n−4)4 ...(3)
[∵ nCn−r= nCr]
∴T5Tn−3=√61
⇒ nC4×2n−44×3−1 nC4×2×3−(n−4)4=√61
⇒2((n−4)4−1) ×3(n−44−1)=√6
⇒(2×3)(n−8)4=612
⇒(n−8)4=12
⇒n−8=2⇒n=10.
Hence, n=10.