Which term of G.P. 3, 3√3,9, …………….. equals to 243? A) 6�

1.	Which term of G.P. 3, 3√3,9, …………….. equals to 243? A) 6 B) 7 C) 8 D) 9
Answer» Correct option is (D) 9 Given G.P. is \(3, 3 \sqrt3,9,.......\) \(\therefore a_1=3,a_2=3\sqrt3\) \(\therefore\) Common ratio is r \(=\frac{a_2}{a_1}\) \(=\frac{3\sqrt3}3=\sqrt3\) Let \(a_n=243\) \(\therefore ar^{n-1}=243\) \((\because a_n=ar^{n-1})\) \(\Rightarrow3(\sqrt3)^{n-1}=243\) \(\Rightarrow3\frac{n-1}2=\frac{243}3\) \(=81=3^4\) \(\Rightarrow\frac{n-1}2=4\) \(\Rightarrow n-1=8\) \(\Rightarrow n=8+1=9\) Hence, \(9^{th}\) term of given G.P. equals to 243. Correct option is D) 9

Which term of G.P. 3, 3√3,9, …………….. equals to 243? A) 6 B) 7 C) 8 D) 9

Answer»

Correct option is (D) 9

Given G.P. is \(3, 3 \sqrt3,9,.......\)

\(\therefore a_1=3,a_2=3\sqrt3\)

\(\therefore\) Common ratio is r \(=\frac{a_2}{a_1}\)

\(=\frac{3\sqrt3}3=\sqrt3\)

Let \(a_n=243\)

\(\therefore ar^{n-1}=243\) \((\because a_n=ar^{n-1})\)

\(\Rightarrow3(\sqrt3)^{n-1}=243\)

\(\Rightarrow3\frac{n-1}2=\frac{243}3\)

\(=81=3^4\)

\(\Rightarrow\frac{n-1}2=4\)

\(\Rightarrow n-1=8\)

\(\Rightarrow n=8+1=9\)

Hence, \(9^{th}\) term of given G.P. equals to 243.

Correct option is D) 9