# Geometric summation

The sum $a_1+a_2+a_3+...+a_n$ is geometric. If $a_1+a_3+a_5=455$ and $a_2+a_4+a_6=1365$, then the ratio between each consecutive term is $2$, $3$,$4$ or an other number? Answer is supposed to be $3$.

My progress:

Since this is a geometric sum we can write it as $\sum_{k=1}^{6}q^{k}=1365+455=1820$

Now I can test each of the above value of $q$ and calculate $\sum_{k=1}^{6}q^{k}$. None gives the answer $1820$.

If we instead assume that there exist such a number then it is equivalent for solving the equation $q+q^2+q^3+q^4+q^5+q^6=1820$ which has no integer solution. So I have two arguments here, are they false?

Have I missed something?

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Shouldn't you assume $a_k = c q^k$, instead of $a_k = q^k$? – Arin Chaudhuri Jan 8 '13 at 20:02
I see, yes I should. Thanks – EricAm Jan 8 '13 at 20:13

You have $a_1+a_3+a_5=455$ and $a_2+a_4+a_6=1365$. You know that in any geometirc summation $a_n=t^{n-1}a_1$, so $$a_1+a_3+a_5=455\to a_1(1+t^2+t^4)=455\\\ a_2+a_4+a_6=1365\to a_1(t+t^3+t^5)=1365$$ Now see the second identity above. If we factor $t_1$, we get: $$a_2+a_4+a_6=1365\to a_1(t+t^3+t^5)=1365\to a_1t(1+t^2+t^4)=1365$$ But $$a_1(1+t^2+t^4)=455$$ so $$a_1t(1+t^2+t^4)=1365\to t=\frac{1365}{455}=3$$

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Well done! + 1 :-) – amWhy Feb 21 '13 at 0:12