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The progression $a_1+a_2+...+a_n$ is geometric. If $a_1+a_3+a_5=455,$ and $a_2+a_4+a_6=1365$, what is the quotient $q$ of the progression?

This means that $$a_1+a_2+a_3+a_4+a_5+a_6=\sum_{n=1}^{6}a_n=\sum_{n=1}^{6}q^n=\frac{q^{7}-q}{q-1}=455+1365=1820.$$

I can't solve this equation of 7th degree.

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As $a_{n+1} = qa_{n}$ we have $$ a_2 + a_4 + a_6 = q(a_1 + a_3 + a_5), $$ or $1365 = q\cdot 465$. So $q = 3$.

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  • $\begingroup$ How do you know that $a_{n+1}=a_nq$? $\endgroup$ – Parseval Aug 27 '17 at 17:35
  • $\begingroup$ @Parseval as $\{a_n\}$ is a geometric progression $\endgroup$ – Anton Grudkin Aug 27 '17 at 17:44
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$$\begin{cases} a_{ 1 }+a_{ 3 }+a_{ 5 }=455 \\ a_{ 2 }+a_{ 4 }+a_{ 6 }=1365 \end{cases}\Rightarrow \begin{cases} { a }_{ 1 }+{ a }_{ 1 }{ q }^{ 2 }+{ a }_{ 1 }{ q }^{ 4 }=455 \\ { a }_{ 1 }q+{ a }_{ 1 }{ q }^{ 3 }+{ a }_{ 1 }{ q }^{ 5 }=1355 \end{cases}\Rightarrow \begin{cases} { a }_{ 1 }\left( 1+{ q }^{ 2 }+{ q }^{ 4 } \right) =455 \\ { a }_{ 1 }q\left( 1+{ q }^{ 2 }+{ q }^{ 4 } \right) =1355 \end{cases}\\ \frac { 1 }{ q } =\frac { 455 }{ 1355 } \Rightarrow q=3$$

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  • $\begingroup$ How come $a_3=a_1q^2$? $\endgroup$ – Parseval Aug 27 '17 at 17:34
  • $\begingroup$ it comes from the definition of $nth$ term's formula ${ a }_{ n }={ a }_{ 1 }{ q }^{ n-1 }$ $\endgroup$ – haqnatural Aug 27 '17 at 17:36
  • $\begingroup$ According to the answer above it's $a_n=qa_{n-1}$. This is confusing. $\endgroup$ – Parseval Aug 27 '17 at 17:38
  • $\begingroup$ @Parseval, $${ a }_{ 2 }=q{ a }_{ 1 }\\ { a }_{ 3 }=q{ a }_{ 2 }={ q }^{ 2 }a_{ 1 }\\ { a }_{ 4 }=q{ a }_{ 3 }=q\underset { { a }_{ 3 } }{ \underbrace { { q }^{ 2 }a_{ 1 } } } ={ q }^{ 3 }{ a }_{ 1 }\\ ......\\ { a }_{ n }={ { q }^{ n-1 }a }_{ 1 }$$ $\endgroup$ – haqnatural Aug 27 '17 at 17:44

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