Mathematics - Sequence & Series Question with Solution | TestHub

Let the G.P. be $a,ar,a{r}^2,\dots$. Since terms are positive, $a>0$ and $r>0$. For the infinite sum to converge, we must have $|r|<1$.

The sum of the G.P. is $S=\frac{a}{1-r}=4$. The series of cubes $a^3,(ar{)}^3,(a{r}^2{)}^3,\dots$ is also a G.P. with first term $a^3$ and common ratio $r^3$. Its sum is $S_3=\frac{a^3}{1-r^3}=24$. From the first equation, $a=4(1-r)$. Substitute this into the second equation: $$\frac{(4(1-r){)}^3}{1-r^3}=24$$ $$\frac{64(1-r{)}^3}{(1-r)(1+r+r^2)}=24$$ Since $r<1$, $1-r\ne 0$, so we can cancel $(1-r)$: $$\frac{64(1-r{)}^2}{1+r+r^2}=24$$

Divide by 8: $8(1-r{)}^2=3(1+r+r^2)⟹8(1-2r+r^2)=3+3r+3r^2⟹8-16r+8r^2=3+3r+3r^2$$5r^2-19r+5=0$$Usingthequadraticformula,$ r = $\frac{-(-19)\pm \sqrt{(-19{)}^2-4(5)(5)}}{2(5)}$ = $\frac{19\pm \sqrt{361-100}}{10}$ = $\frac{19\pm \sqrt{261}}{10}$$.Since$$\sqrt{261}$ = $\sqrt{9\times 29}$ = 3$\sqrt{29}$$.So$ r = $\frac{19\pm 3\sqrt{29}}{10}$$.Forthesumtoconverge,$ |r|<1$.$ 3$\sqrt{29}$ $\approx$ 3 $\times$ 5.385 = 16.155$.$ r_1 = $\frac{19+16.155}{10}$ $\approx$ 3.5155 > 1$(rejected).$ r_2 = $\frac{19-16.155}{10}$ $\approx$ 0.2845 < 1$(accepted).Thus,$ r = $\frac{19-3\sqrt{29}}{10}$ $.