Maths · Arithmetic and Geometric progressions
The term of the series is then is
The \( n t h \) term of the series \( 3, \sqrt{3}, 1, \ldots . \) is \( \frac{1}{243}, \) then \( n \) is
- A. 12
- B. 13
- C. 14
- D. 15
Step-by-step solution
The series 3, √3, 1, ... is a geometric progression with first term a=3 and common ratio r=1/√3. The nth term is given by a r^(n-1) = 3 * (1/√3)^(n-1) = 3^( (3-n)/2 ). Set equal to 1/243 = 3^{-5}, so (3-n)/2 = -5, giving 3-n = -10, hence n=13.
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