NCERT Solutions
Class 11 Maths
Binomial Theorem
Ex.Misc.Q.8
Find n, if the ratio of the fifth term from the beginning to the fifth term from the end
in the expansion of (4√2 + (1 ÷ 4√3)) n is √6: 1
In the expansion, (a + b) n = nC0 an + nC1 an-1 b + nC2 an-2 b2 + ……………... + nCn bn
Fifth term from the beginning = nC4 an-4 b4
Fifth term from the end = nCn-4 a4 bn-4
Therefore, it is evident that in the expansion of (4√2 + (1 ÷ 4√3)) n the fifth term
from the beginning is nC4 (4√2) n-4 (1 ÷ 4√3)4 and the fifth term from the end is nCn-4 (4√2)4 (1 ÷ 4√3) n-4
Now, nC4 (4√2) n-4 (1 ÷ 4√3)4 = nC4 {(4√2) n ÷ (4√2)4} (13
)
= nC4 {(4√2) n ÷ 2} ( )
= [n! ÷ {(6 × 4! × (n - 4)!}] × (4√2) n …………… (1)
and nCn-4 (4√2)4 (1 ÷ 4√3) n-4 = nCn-4 {2 × (4√3)4 ÷ (4√3)n}
= nCn-4 {2 × (3 ÷ (4√3) n)}
= [6n! ÷ {(4! × (n - 4)!}] × {1 ÷ (4√3) n} …………… (2)
It is given that the ratio of the fifth term from the beginning to the fifth term from the end is √6: 1.
Therefore, from equation (1) and (2), we get
[n! ÷ {(6 × 4! × (n - 4)!}] × (4√2) n: [6n! ÷ {(4! × (n - 4)!}] × {1 ÷ (4√3) n} = √6: 1
=> [(4√2) n ÷ 6] × [(4√3) n ÷ 6] = √6
=> (4√6) n ÷ 36 = √6
=> (4√6) n = 36√6
=> 
= 
=> 
= 
=> 
= 5
=> n = 10
Thus, the value of n is 10.