Question
for a geometric progression, $S _4=10 S_2$. Find the common ratio.
✓
Answer
$S_4=10 S_2$
$S_n=\frac{a\left[r^n-1\right]}{r-1}$
$\therefore S_4=\frac{a\left[r^4-1\right]}{r-1} \text { and } S_2=\frac{a\left\lfloor r^2-1\right]}{r-1}$
Now, $S_4=10 S_2$
$\frac{a\left[r^4-1\right]}{r-1}=10 \times \frac{a\left[r^2-1\right]}{r-1}$
$\therefore r^4-1=10\left(r^2-1\right)$
$\therefore\left(r^2-1\right)\left(r^2+1\right)=10\left(r^2-1\right)$
$\therefore\left(r^2+1\right)=10$
$\therefore r^2=10-1=9$
$\therefore r= \pm 3$
$S_n=\frac{a\left[r^n-1\right]}{r-1}$
$\therefore S_4=\frac{a\left[r^4-1\right]}{r-1} \text { and } S_2=\frac{a\left\lfloor r^2-1\right]}{r-1}$
Now, $S_4=10 S_2$
$\frac{a\left[r^4-1\right]}{r-1}=10 \times \frac{a\left[r^2-1\right]}{r-1}$
$\therefore r^4-1=10\left(r^2-1\right)$
$\therefore\left(r^2-1\right)\left(r^2+1\right)=10\left(r^2-1\right)$
$\therefore\left(r^2+1\right)=10$
$\therefore r^2=10-1=9$
$\therefore r= \pm 3$
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