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In a geometric progression, $S_2 = 7$ and $S_6 = 91$. Evaluate $S_4$. Alternatives: 28, 32, 35, 49, 84.

Here's what I tried so far:

$$ S_2 = \frac{a_1(1-r^2)}{1-r} \implies 1-r = \frac{a_1(1-r^2)}{7} \\ S_6 = \frac{a_1(1-r^6)}{1-r} \implies 1-r = \frac{a_1(1-r^6)}{91} $$

Then: $$ \frac{1-r^2}{1} = \frac{1-r^6}{13} \\ r^6 - 13r^2 + 12 = 0 $$

Now i can't solve this equation, perhaps there's an easier way…

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    $\begingroup$ The formulas you're using are for the sum of the first ever-so-many terms of the progression... is that what the $S_n$ are, the partial sums of the progression? Or are they the terms in the progression themselves? $\endgroup$ – Eevee Trainer May 7 '19 at 4:39
  • $\begingroup$ $S_n$ is the sum of the first n terms of the progression. $\endgroup$ – rodorgas May 7 '19 at 4:41
  • $\begingroup$ Given $r\ne\pm1$, you could have simplified $(1-r^6)/(1-r^2)=1+r^2+r^4$ $\endgroup$ – J. W. Tanner May 7 '19 at 5:04
  • $\begingroup$ You absolutely should explain such key pieces in the question body. Those are easy to miss in comments. Also, a series has infinitely many terms. Your sums don't. Read the tag descriptions before using them. $\endgroup$ – Jyrki Lahtonen May 7 '19 at 5:58
  • $\begingroup$ What are $S_2$ and $S_6$, precisely ? $\endgroup$ – Yves Daoust May 7 '19 at 6:07
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Let $x=r^2$ then we see $$x^3-13x+12=0$$ so $$x^3-x-12x+12=0$$ so $$ x(x-1)(x+1)-12(x-1)=0$$ so $$ (x-1)(x^2+x-12)=0$$ so $$ (x-1)(x+4)(x-3)=0.$$

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  • $\begingroup$ Ooops, sorry, my bad. I will delete this comment. $\endgroup$ – Yves Daoust May 7 '19 at 6:29
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Here's how to do it with only a quadratic equation (of sorts).

Denote the geometric sequence $a_1, a_2=a_1r, a_3=a_1r^2, a_4=a_1r^3, a_5=a_1r^4, a_6=a_1r^5...$

Then $S_2=a_1(1+r), S_4=a_1(1+r+r^2+r^3)=a_1(1+r)(1+r^2)$,

and $S_6=a_1(1+r+r^2+r^3+r^4+r^5)=a_1(1+r)(1+r^2+r^4).$

$\dfrac {S_6}{S_2}=\dfrac{91}7=13=1+r^2+r^4$ so $(r^2)^2+(r^2)-12=(r^2-3)(r^2+4)=0$

so $r^2=3$ so $S_4=S_2(1+r^2)=7(1+3)=28.$

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$$\frac{S_6}{S_2}=\frac{r^6-1}{r^2-1}=r^4+r^2+1=13$$ and

$$\frac{S_4}{S_2}=\frac{r^4-1}{r^2-1}=r^2+1.$$

With $s:=r^2+1$, we have $$s(s-1)+1=13$$

giving the two solutions

$$S_4=4\cdot 7$$ and $$S_4=-3\cdot7.$$

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  • $\begingroup$ $-21$ was not one of the alternatives offered by OP; presumably the progression is in real numbers, not complex $\endgroup$ – J. W. Tanner May 7 '19 at 6:39
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    $\begingroup$ @J.W.Tanner: then $-21$ is rejected. $\endgroup$ – Yves Daoust May 7 '19 at 6:40
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$$\begin{cases} S_2=a_1+a_1r=a_1(1+r)=7\\ S_6=a_1+a_1r+\cdots+a_1r^5=91\end{cases}\\ S_6-S_2=a_1r^2(1+r+r^2+r^3)=a_1r^2\cdot \frac{1-r^4}{1-r}=84\\ \frac{S_6-S_2}{S_2}=\frac{a_1r^2(1-r^4)}{a_1(1+r)(1-r)}=r^2(1+r^2)=12 \Rightarrow r^2=3$$ Hence: $$\begin{align}S_4&=a_1(1+r+r^2+r^3)=\\ &=a_1(1+r+r^2(1+r))=\\ &=a_1(1+r)(1+r^2)=\\ &=S_2(1+r^2)=\\ &=7(1+3)=\\ &=28.\end{align}$$

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