Problems involving Geometric Progressions: Problems with Solutions

Solution:
The numbers form a geometric progression with first term 8 and common ratio [tex]\frac12[/tex].

Solution:
The common ratio is -2.

Solution:
[tex]r=\frac{a_{n+1}}{a_n}=\frac{a_2}{a_1}=\frac{15}{5}=3[/tex]

Solution:
By the definition [tex]r=\frac{a_2}{a_1}=\frac{5}{-1}=-5[/tex]

Solution:
Let denote with [tex]a_1, a_2, a_3 ...[/tex] the geometric progression terms.
[tex]\frac{20}{5}=\frac{a_3}{a_1}=\frac{a_1r^2}{a_1}=r^2[/tex], therefore [tex]r^2=4[/tex], which leads to two candidates for r: [tex]r=2[/tex] or [tex]r=-2[/tex]. Since [tex]{a_n}[/tex] is an increasing geometric progression, [tex]r>1>0[/tex] and [tex]r=2[/tex] remains the only answer.

Solution:
[tex]\frac{-40}{5}=\frac{a_4}{a_1}=\frac{a_1r^3}{a_1}=r^3=-8[/tex], so [tex]r=-2[/tex].

Solution:
Since the progression is alternating, we can conclude that [tex]r<0[/tex]. [tex]a_1r^2=a_3 \leftrightarrow r^2=\frac{a_3}{a_1}=\frac{5}{25}[/tex], so [tex]r=\frac{1}{5}[/tex] or [tex]r=-\frac{1}{5}[/tex]. The final answer is [tex]-\frac{1}{5}[/tex].

Solution:
[tex]a_1 \cdot r^3=2\cdot 3^3=2 \cdot 27=54[/tex]

Solution:
[tex]a_4=a_1\cdot r^3[/tex], therefore [tex]a_1=a_4\cdot r^{-3}=12\cdot (\frac{1}{3})^{-3}=12\cdot 3^3=12 \cdot 27=324[/tex]

Solution:
The common ratio of the geometric progression is [tex]r=\frac{a_2}{a_1}=\frac{10}{5}=2[/tex].
[tex]a_6=a_1 \cdot r^5=5 \cdot 2^5=5 \cdot 32=160[/tex]

Solution:
Since [tex]a_n[/tex] is increasing, [tex]a_n > a_1 > 0[/tex] for any [tex]n>1[/tex]. [tex]a_{\frac{5+1}{2}}=\sqrt{a_5\cdot a_1}[/tex], or [tex]a_3=\sqrt{2\cdot 162}=\sqrt{324}=18[/tex]

Solution:
The formula for the sum of the first n elements of a geometric progression is [tex]S_n=a_1\cdot\frac{1-r^n}{1-r}[/tex]. Substituting, we get [tex]S_5=2 \cdot \frac{1-3^5}{1-3}=2 \cdot \frac{1-243}{-2}=242[/tex]

Solution:
[tex]S_5=a_1 \cdot \frac{1-r^5}{1-r}=\frac{1-5^5}{1-5}=\frac{1-3125}{-4}=\frac{3124}{4}=781[/tex]

Solution:
We know, that if an infinite geometric series converges, the sum of its elements is given by the formula [tex]S=a_1.\frac{1}{1-r}[/tex]. In our case, [tex]S=1.\frac{1}{1-\frac{1}{2}}=\frac{1}{\frac{1}{2}}=2[/tex]