ratio of partial sums of the same geometric sequence
My kid was given this question:
In a geometric sequence, the proportion of (the sum of the first $12$ terms) to (the sum of the first $8$ terms) is $\frac{819}{51}$. Find the common ratio of the sequence.
The only formula thus far covered for the partial sum of a geometric sequence $(a_1q_i)_{i\ge0}$ is$$\frac{a_1(q^n-1)}{q-1}.$$ How does one solve this?
The solution I found is as follows, but it seems too roundabout for the context, so I wonder what more direct way there is:
The given information means$$\frac{q^{12}-1}{q^8-1}=\frac{819}{51}.$$ Let $u=q^4$. Then$$\frac{819}{51}=\frac{u^3-1}{u^2-1}=\frac{u^2+u+1}{u+1}$$ $$\frac{819-51}{51}=\frac{u^2}{u+1}$$ $$51u^2-768u-768=0$$so $u\in\lbrace-16/17,16\rbrace$ and $q=\pm2$.
1 answer
I think your approach is the expected one, but a shortcut if rigour is not required would be to note that the absolute value of the ratio must be greater than 1, or the proportion couldn't exceed $\frac{12}8$; but then the largest term dominates, so $$q^4 \approx \frac{819}{51} \approx 16.05$$ and then trial and error shows that $q^4 = 16$ works.
(Of course, this doesn't exclude other solutions: see "if rigour is not required" above).
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