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The numbers $a_1, a_2, a_3, . . .$ form an arithmetic sequence with $a_1 \ne a_2$. The three numbers $a_1, a_2, a_6$ form a geometric sequence in that order. Determine all possible positive integers $k$ for which the three numbers $a_1, a_4, a_k$ also form a geometric sequence in that order.

This is from Euclid 2015:Problem 7B

I got that:

$a_n = a_1 + (n-1)d = r^{n-1} a_1$ where $d$ is difference, and $r$ the geom. ratio.

But this doesnt give many leads, HINTS ONLY PLEASE.

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HINT :

Use the fact that $$\text{$a,b,c$ form a geometric sequence}\Rightarrow b^2=ac$$

Thus, you have $$a_2^2=a_1a_6\quad\text{and}\quad a_4^2=a_1a_k$$

Express these by $a_1$ and $d$.

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  • $\begingroup$ thanks (+1). I got: $a_1^2 + 2a_1d + d^2 = a_1 (a_1 + 5d)$ and $a_1^2 + 6a_1 d + 9d^2 = a_1 a_k$. The first yields, $a_1^2 + 2a_1d + d^2 - a_1^2 - 5a_1d = 0 \implies -3a_1 d + d^2 = 0 \implies d(d - 3a_1) = 0$ so $3a_1 = d$. Plugging to the second, $a_1^2 + 18a_1^2 + 81a_1^2 = a_1 a_k$ so $ a_k = 100a_1$. So will $k=99$? I got the value for $a_k$, but not $k$ itself. $\endgroup$ – Amad27 Nov 15 '15 at 17:49
  • $\begingroup$ @Amad27: $d=3a_1$ is correct. Then, you have $(a_1+3d)^2=a_1(a_1+(k-1)d)$, i.e. $(a_1+9a_1)^2=a_1(a_1+(k-1)\cdot 3a_1)$. You'll get $k=34$. Write $a_k$ as $a_1+(k-1)d$. $\endgroup$ – mathlove Nov 15 '15 at 17:54
  • $\begingroup$ Mmmm.. should have seen that. Thanks for helping! $\endgroup$ – Amad27 Nov 15 '15 at 17:56
  • $\begingroup$ @Amad27: You are welcome. $\endgroup$ – mathlove Nov 15 '15 at 17:57

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