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Question: Does there exists a GP containing 8, 12 and 27 as three of its terms? If it exists how many such progressions are possible?

What I have tried: First of all I have made three equations as follows Here 27, 12, 8 are $p^{th},q^{th},r^{th} $ term of GP

$27=AR^{p-1} ...(1)$

$12=AR^{q-1} ....(2)$

$8=AR^{r-1} ....(3)$

Then I divided equation 1 by 2 and equation 2 by 3 and got the below equations.

$\frac{27}{12}=(\frac{3}{2})^2=R^{p-q} ....(4)$

$\frac{12}{8}=\frac{3}{2}=R^{q-r} ....(5)$

Then I got two more equations as follows

$p-q=2$

$q-r=1$

Then I have let $p=k$ and got $q=k-2$ and $r=k-3$

Now for each value of k we get different values of $q$ and $r$.

Now my question is that, is this enough to prove there can be innumerable progressions containing 27,8,12 as its terms?

wythagoras
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Kartik Watwani
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1 Answers1

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There are infinitely many such progressions. To see this, note that $12 = 1.5 \times 8$, and $27 = 1.5 \times 1.5 \times 8$.

Now have $a_n=8\cdot(1.5^{\frac{1}{q}})^n$ for some $q \in \mathbb N$. It will have 8 as the 0th therm, 12 as the $q$th term and 27 as the $3q$th term.

Your answer is also correct. It could be written as $a_n=(8\cdot 1.5^p) \times 1.5^n$ for some $p \in \mathbb N$.

Together with my answer you can even find all such progressions, they have the form $a_n=(8\cdot 1.5^{\frac{p}{q}}) \times (1.5^{\frac{1}{q}})^n$, for some $p,q \in \mathbb N$.

wythagoras
  • 25,026