Does there exist a G.P. containing 27,8 and 12 as three of its terms?n If it exists, how many such progressions are possible?

Let 27,8,12 be t_m, t_n, t_p of a G.P., whose first term is a and common ratio r.nThen t_m=a r^m-1=27, t_n=a r^n-1=8, t_p=a r^ p-1=12.nn n Therefore r=((27 / 8))^(1 / m-n)=((2 / 3))^(1 / n-p) n Therefore (3 / 3 m-n) * frac13^n-p=2^(1 / n-p) * (3 / 2 m-n) n Therefore fracfrac13^m-n+(1 / n-p)frac12^n-p+(3 / m-n)=1 => ((3 / 2))^(3 / m-n)+(1 / n-p)=1 n Therefore (3 / m-n)+(1 / n-p)=0 => 2 n=3 p-m n m=6 k, n=3 k, p=4 k, k in Nn nn∴ There exist infinite G.P's for which 27,8,12 are three of its terms.

Does there exist a G.P. containing 27,8 and 12 as three... | Sequence & Series (Mathematics)

Let

27,8,12

t_m, t_n, t_p

of a G.P., whose first term is

a

and common ratio

r

.\nThen

t_m=a r^{m-1}=27, t_n=a r^{n-1}=8, t_p=a r^{\mathrm{p}-1}=12

.\n

\n\begin{aligned}\n& \therefore r=\left(\frac{27}{8}\right)^{\frac{1}{m-n}}=\left(\frac{2}{3}\right)^{\frac{1}{n-p}} \\\n& \therefore \frac{3}{3 m-n} \cdot \frac{1}{3^{n-p}}=2^{\frac{1}{n-p}} \cdot \frac{3}{2 m-n} \\\n& \therefore \frac{\frac{1}{3^{m-n}+\frac{1}{n-p}}}{\frac{1}{2^{n-p}}+\frac{3}{m-n}}=1 \Rightarrow\left(\frac{3}{2}\right)^{\frac{3}{m-n}+\frac{1}{n-p}}=1 \\\n& \therefore \frac{3}{m-n}+\frac{1}{n-p}=0 \Rightarrow 2 n=3 p-m \\\n& m=6 k, n=3 k, p=4 k, k \in N\n\end{aligned}\n

\n∴ There exist infinite G.P's for which

27,8,12

are three of its terms.

Explanation

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