Let P be the 7 text th term from the beginning and Q be the 7 text th term from the end in the expansion of (√[3]3+(1/√[3]4)) n where n ∈ N. If 12 P = Q, then find the value of n.

Question

Let P be the 7 text th term from the beginning and Q be the 7 text th term from the end in the expansion of (√[3]3+(1/√[3]4)) n where n ∈ N. If 12 P = Q, then find the value of n. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

P= n C6(3(1/3))n-6 ⋅(4(-1/3))6 Q = n C n -6(3(1/3))6 ⋅(4(-1/3)) n -6 ∴ ( Q / P )=12 ⇒(12)( n -6/3)=(12)1 ⇒ ( n -6/3)=1 ⇒ n =9

Q. Let $P$ be the $7^{th}$ term from the beginning and $Q$ be the $7^{th}$ term from the end in the expansion of $(33 + \frac{1}{3 4})^{n}$ where $n \in N$ . If $12 P = Q$ , then find the value of $n$ .

$P =^{n} C_{6} (3^{\frac{1}{3}})^{n - 6} \cdot (4^{\frac{- 1}{3}})^{6}$
$Q =^{n} C_{n - 6} (3^{\frac{1}{3}})^{6} \cdot (4^{\frac{- 1}{3}})^{n - 6}$
$∴ \frac{Q}{P} = 12 \Rightarrow (12)^{\frac{n - 6}{3}} = (12)^{1} \Rightarrow \frac{n - 6}{3} = 1 \Rightarrow n = 9$

Q. Let P be the 7th term from the beginning and Q be the 7th term from the end in the expansion of (33​+34​1​)n where n∈N. If 12P=Q, then find the value of n.

P=nC6​(331​)n−6⋅(43−1​)6 Q=nCn−6​(331​)6⋅(43−1​)n−6 ∴PQ​=12⇒(12)3n−6​=(12)1⇒3n−6​=1⇒n=9

Q. Let $P$ be the $7^{th}$ term from the beginning and $Q$ be the $7^{th}$ term from the end in the expansion of $(33 + \frac{1}{3 4})^{n}$ where $n \in N$ . If $12 P = Q$ , then find the value of $n$ .

$P =^{n} C_{6} (3^{\frac{1}{3}})^{n - 6} \cdot (4^{\frac{- 1}{3}})^{6}$
$Q =^{n} C_{n - 6} (3^{\frac{1}{3}})^{6} \cdot (4^{\frac{- 1}{3}})^{n - 6}$
$∴ \frac{Q}{P} = 12 \Rightarrow (12)^{\frac{n - 6}{3}} = (12)^{1} \Rightarrow \frac{n - 6}{3} = 1 \Rightarrow n = 9$