The sum of an infinite geometric series with positive terms is 3, and the sum of the cubes…

Question

The sum of an infinite geometric series with positive terms is $3$, and the sum of the cubes of its terms is $\dfrac{27}{19}$. Then the common ratio of this series is:

Accepted Answer

$\dfrac{a}{1-r}=3 \Rightarrow a=3(1-r)$. Sum of cubes: $\dfrac{a^3}{1-r^3} = \dfrac{27(1-r)^3}{(1-r)(1+r+r^2)} = \dfrac{27(1-r)^2}{1+r+r^2} = \dfrac{27}{19}.$

$$19(1-r)^2 = 1+r+r^2 \implies 18r^2-39r+18=0 \implies 6r^2-13r+6=0 \implies r=\frac{13\pm5}{12}.$$

Since $0<r<1$: $r=\dfrac{8}{12}=\dfrac23.$

Solutions · 1