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 (Q/P)=12, then n is equal to

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 (Q/P)=12, then n is equal to  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

P= n C6 ⋅(3(1/3))n-6 ⋅(4-(1/3))6 Q= n Cn-6 ⋅(3(1/3))6 ⋅(4-(1/3))n-6 ∴ (Q/P)=12 P ⇒ (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 $\frac{Q}{P} = 12,$ then $n$ is equal to ____

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

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 PQ​=12, then n is equal to ____

P=nC6​⋅(331​)n−6⋅(4−31​)6 Q=nCn−6​⋅(331​)6⋅(4−31​)n−6 ∴PQ​=12P⇒(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 $\frac{Q}{P} = 12,$ then $n$ is equal to ____

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