# Find n in the binomial (23+133)n if the ratio of 7th term from the beginning to the 7th term from the end is 16 - Mathematics

Sum

Find n in the binomial (root(3)(2) + 1/(root(3)(3)))^n if the ratio of 7th term from the beginning to the 7th term from the end is 1/6

#### Solution

The given expression is (root(3)(2) + 1/(root(3)(3)))^"n"

= (2^(1/3) + 1/3^(1/3))^"n"

General Term "T"(r + 1) = ""^n"C"_r x^(n - r) y^r

T7 = T6+1 = ""^n"C"_6 (2^(1/3))^(n - 6)  (1/(3^(1/3)))^6

= ""^n"C"_6 (2)^((n -6)/3) * (1/3^2)

= ""^n"C"_6 (2)^((n - 6)/3) * (3)^-2

7th term from the end = (n – 7 + 2)th term from the beginning

= (n – 5)th term from the beginning

So, "T"_(n - 6 + 1) = ""^n"C"_(n - 6) (2^(1/3))^(n - n + 6) (1/3^(1/3))^(n - 6)

= ""^n"C"_(n - 6) (2)^2 * (1/(3^((n - 6)/3)))

= ""^n"C"_(n - 6) (2)^2 (3)^((6 - n)/3)

We get (""^n"C"_6 ^((n - 6)/3) (3)^-2)/(""^n"C"_(n - 6) (2)^2 (3)^((6 - n)/3)) = 1/6

⇒ (""^n"C"_(n - 6) (2)^((n - 6)/3) (3)^-2)/(""^n"C"_(n - 6) (2)^2 (3)^((6 - n)/3)) = 1/6

⇒ (2)^((n - 6)/3 - 2) * (3)^(-2 (6 - n)/3) = 1/6

⇒ (2)^((n - 6 - 6)/3) * (3)^((-6 - 6 + n)/3) = 1/6

⇒ (2)^((n - 12)/3) * (3)^((n - 12)/3) = (6)-1

⇒ (6)^((n - 12)/3) = (6)^-1

⇒ (n - 12)/3 = – 1

⇒ n – 12 = – 3

⇒ n = 12 – 3 = 9

Hence, the required value of n is 9.

Concept: General and Middle Terms
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#### APPEARS IN

NCERT Mathematics Exemplar Class 11
Chapter 8 Binomial Theorem
Exercise | Q 14 | Page 143

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