Sequences and Series 3 Question 15
15.
Does there exist a geometric progression containing 27,8 and 12 as three of its term? If it exists, then how many such progressions are possible?
(1982, 2M)
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Answer:
Correct Answer: 15. (Yes, infinite)
Solution:
- Let $27,8,12$ be three terms of a GP.
$ \begin{aligned} & \Rightarrow \quad t _m=27, t _n=8 \text { and } t _p=12 \\ & A R^{m-1}=27, A R^{n-1}=8 \\ & \text { and } A R^{p-1}=12 \\ & \therefore \quad R=\frac{27}{8}^{1 /(m-n)} \text { and } R=\frac{8}{12}^{1 /(n-p)} \\ & \Rightarrow \quad \frac{27}{8}^{1 /(m-n)}=\frac{2}{3}^{1 /(n-p)} \\ & \Rightarrow \quad 3^{3 /(m-n)} \cdot 3^{1 /(n-p)}=2^{1 /(n-p)} \cdot 2^{3 /(m-n)} \\ & \Rightarrow \quad \frac{3^{\frac{3}{m-n}+\frac{1}{n-p}}}{2^{\frac{1}{n-p}+\frac{3}{m-n}}}=1 \\ & \therefore \quad \frac{3}{m-n}+\frac{1}{n-p}=0 \text { and } \frac{1}{n-p}+\frac{3}{m-n}=0 \\ & \Rightarrow \quad 3(n-p)=n-m \text { and } \quad 2 n=3 p-m \end{aligned} $
Hence, there exists infinite GP for which 27,8 and 12 as three of its terms.
