Geometric series terms

The sum of the first two terms of the descending geometric sequence is five quarters, and the sum of the infinite geometric series formed from it is nine quarters. Write the first three terms of the geometric sequence.

Final Answer:

g₁ =  3/4
g₂ =  1/2
g₃ =  1/3 = 1:3
G₁ =  3.75
G₂ =  -5/2
G₃ =  5/3 = 5:3

Step-by-step explanation:

$$\begin{gathered} g_1+g_2=\frac{5}{4} \\ s=\frac{9}{4}=2.25 \\ {g}_1+q\cdot g_1=\frac{5}{4} \\ s=\frac{g_1}{1-q} \\ {g}_1+q\cdot g_1=5/4 \\ 9/4\cdot (1-q)=g_1 \\ {g}_1(1+q)=5/4 \\ 9/4\cdot (1-q)=g_1 \\ 9/4\cdot (1-q)\cdot (1+q)=5/4 \\ 9\cdot (1-x)\cdot (1+x)=5 \\ 9\cdot (1-x)\cdot (1+x)=5 \\ -9x^2+4=0 \\ 9x^2-4=0 \\ {x}_{1,2}=\pm \sqrt{4/9}=\pm 0.666666667 \\ {x}_1=0.666666667 \\ {x}_2=-0.666666667 \\ |q|<1 \\ q=x_1=0.6667=\frac{2}{3}\doteq 0.6667 \\ {g}_1=s\cdot (1-q)=2.25\cdot (1-0.6667)=\frac{3}{4}=0.75 \end{gathered}$$

Our quadratic equation calculator calculates it.

$g_2=q\cdot g_1=0.6667\cdot 0.75=\frac{1}{2}=0.5$

$g_3=q\cdot g_2=0.6667\cdot 0.5=\frac{1}{3}\approx 0.3333=1:3$

$$\begin{gathered} Q=x_2=(-0.6667)=-\frac{2}{3}\doteq -0.6667 \\ {G}_1=s\cdot (1-Q)=2.25\cdot (1-(-0.6667))=\frac{15}{4}=3\frac{3}{4}=3.75 \end{gathered}$$

$G_2=Q\cdot G_1=(-0.6667)\cdot 3.75=-\frac{5}{2}=-2\frac{1}{2}=-2.5$

$G_3=Q\cdot G_2=(-0.6667)\cdot (-2.5)=\frac{5}{3}=1\frac{2}{3}\approx 1.6667=5:3$

Try another example