Fifth term from the beginning is T4+1
i.e., T5 =

n

‌

C4(4√2)n−4(

1

4√3

)4 =

n

‌

C4(2)

n−4

4

(3)−

4

4

Fifth term from the end is Tn−5+2 i.e., Tn−3
∴ Tn−3 =

n

‌

Cn−4(4√2)4(

1

4√3

)n−4 =

n

‌

C4(2)

4

4

(3)

4−n

4

(Since

n

‌

Cn−4 =

n

‌

C4)
According to given condition, we have

n ‌ C4(2) n−4 4 (3)−1

n ‌ C4(2)(3) 4−n 4

=

√6

1

⇒ 2

n−4

4

−13−1−

4−n

4

= √6
⇒ 2

n−8

4

3

n−8

4

= √2√3 = (2)

1

2

(3)

1

2

On comparing, we get
⇒

n−8

4

=

1

2

⇒ 2n - 16 = 4 ⇒ n = 10