Multiple choice

The $nth$ term of the series $3,\sqrt {3}, 1, ....$ is $\dfrac {1}{243}$, then $n$ is

  1. $12$
  2. $13$
  3. $14$
  4. $15$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

This is a geometric progression with first term a = 3 and common ratio r = 1/sqrt(3). The nth term is a * r^(n-1) = 3 * (1/sqrt(3))^(n-1) = 3^(1 - (n-1)/2) = 3^((3-n)/2). Setting 3^((3-n)/2) = 1/243 = 3^(-5), we get (3-n)/2 = -5, so 3-n = -10, n = 13.

AI explanation

Rewrite the terms of the series as powers of 3 to establish the pattern. The sequence becomes 3^1, 3^(1/2), 3^0, showing that the exponents form a decreasing arithmetic progression with the first term as 1 and a common difference of -1/2. The nth term is given by the formula 3^(1 - (n-1)/2). Setting this equal to 1/243, which is 3^(-5), we equate the exponents to get 1 - (n-1)/2 = -5. Solving this equation gives (n-1)/2 = 6, which means n is 13.

Keep practicing — related questions