Question

which term of gp root 3,3,3root 3,..... is 729?

Answer 1

hello !

First term (a) = √3

Common ratio (r) = 3 / √3 = √3

Let nth term = 729

= a * r^n-1 = 729

= (√3) * (√3)^n-1 = 3^6

= (√3)n = (√3)^12

= n = 12

Answer 2

Given Sequence is root 3,3,3root3,...... is 729.

Let a be the first term and r be the common ratio.

nth term tn = 729.

first term a = root 3.

common ratio r = 3/root 3

  = root 3.


We know that nth term of GP tn = ar^n-1

                                                 729 = root 3 * (root3)^n-1

                                                 729 = (root 3)^n - 1 + 1

                                                 729 = (root 3)^n

                                                 729 = (3)n/2

                                                 3^6 = (3)n/2

                                                n/2 = 6

                                                n = 12.


Therefore the 12th term of the given sequence is 729.


Hope this helps!