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 [tex]r[/tex]: [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 \times \frac{1}{1-r}[/tex]. In our case, [tex]S=1 \times \frac{1}{1-\frac{1}{2}}=\frac{1}{\frac{1}{2}}=2[/tex]

Solution:
The daily numbers form a geometric progression with ratio [tex]k[/tex]. From day [tex]6[/tex] to day [tex]11[/tex] there are five steps:
[tex]96k^{5}=3072\Rightarrow k^{5}=32\Rightarrow k=2[/tex]
From day [tex]3[/tex] to day [tex]6[/tex] there are three steps, so on day [tex]3[/tex] there were [tex]\dfrac{96}{2^{3}}=12[/tex] flies.

Solution:
The square of the middle term equals the product of its neighbours: [tex](x+y)^{2}=2x\times 2y=4xy[/tex].
[tex]x^{2}-2xy+y^{2}=0\iff (x-y)^{2}=0\iff x=y[/tex]
The terms are [tex]2x, 2x, 2x[/tex] with [tex]x\ne 0[/tex], so [tex]q=1[/tex].

Solution:
In a geometric progression [tex]a_{m}a_{n}=a_{k}a_{l}[/tex] whenever [tex]m+n=k+l[/tex] (both products equal [tex]a_{1}^{2}q^{m+n-2}[/tex]).
[tex]12+38=8+42=25+25\Rightarrow a_{12}a_{38}=a_{8}a_{42}=a_{25}^{2}[/tex]
[tex]a_{25}^{2}+9a_{25}^{2}=10\Rightarrow a_{25}^{2}=1\Rightarrow a_{8}a_{42}=1[/tex]

Solution:
The condition is [tex](x-3)^{2}=2x(x-7)[/tex]:
[tex]x^{2}-6x+9=2x^{2}-14x\iff x^{2}-8x-9=0\iff x=9[/tex] or [tex]x=-1[/tex]
For [tex]x=-1[/tex] the numbers are [tex]-8, -4, -2[/tex], which are negative.
For [tex]x=9[/tex] the numbers are [tex]2, 6, 18[/tex] ([tex]q=3[/tex]). So [tex]x=9[/tex].

Solution:
After each stroke [tex]\dfrac{9}{10}[/tex] of the air remains, so the remaining amounts form a geometric progression with ratio [tex]0. 9[/tex].
After [tex]10[/tex] strokes the remaining part is [tex](0. 9)^{10}\approx 0. 3487[/tex], so the pump removes
[tex]1-(0. 9)^{10}\approx 0. 6513\approx 65\%[/tex]

Solution:
The difference is the sum of all terms from [tex]a_{4}[/tex] on: [tex]S-S_{3}=a_{4}+a_{5}+\ldots=\dfrac{a_{4}}{1-q}[/tex].
[tex]a_{4}=\dfrac{a_{1}}{8}, S-S_{3}=\dfrac{a_{1}}{8}\times 2=\dfrac{a_{1}}{4}=2\Rightarrow a_{1}=8[/tex]

Solution:
[tex]a_{5}-a_{1}=a_{1}(q^{4}-1)=(q-1)\times a_{1}(1+q+q^{2}+q^{3})=(q-1)S_{4}[/tex]
[tex]480=240(q-1)\Rightarrow q=3[/tex]
Check: [tex]a_{1}=\dfrac{240}{1+3+9+27}=6, a_{5}-a_{1}=486-6=480[/tex].