Problems involving Geometric Progressions: Difficult Problems with Solutions

Solution:
[tex]\begin{array}{|l}a_1r^3-a_1r=18\\a_1r^4-a_1r^2=36\end{array}[/tex]
[tex]\begin{array}{|l}a_1r(r^2-1)=18\\a_1r^2(r^2-1)=36\end{array}[/tex]
Dividing them, we get [tex]\frac{a_1r^2(r^2-1)}{a_1r(r^2-1)}=\frac{36}{18}[/tex], or [tex]r=2[/tex]. Substituting into the first equation:
[tex]a_1\cdot 8-a_1\cdot 2=18[/tex]
[tex]a_1=3[/tex]
[tex]a_3=a_1r^2=3\cdot 4=12[/tex]

Solution:
[tex]a_6=a_1 \cdot r^5=15\cdot (-4)^5=-15\cdot 1024=-15360[/tex]

Solution:
[tex]S_{10}=a_1\frac{1-r^{10}}{1-r}=2\cdot\frac{1-(-2)^{10}}{1-(-2)}=-2.\frac{1023}{3}=-2\cdot 341=-682[/tex]

Solution:
We define a geometric progression [tex]{a_n}[/tex]: [tex]a_1=7[/tex] and [tex]r=7[/tex]. We need to find [tex]S_5=a_1.\frac{r^5-1}{r-1}=7.\frac{7^5-1}{7-1}=7.\frac{16806}{6}=7\cdot 2801=19607[/tex]

Solution:
[tex]a_4=-\sqrt{a_7.a_1}[/tex], since [tex]4=\frac{7+1}{2}[/tex] and the progression is alternating. Then [tex]a_4=-\sqrt{5.405}=-\sqrt{5.5.81}=-5.9=-45[/tex]

Solution:
We need to determine the value of the product [tex]a_1a_2a_3a_4a_5a_6a_7=(a_1a_7)(a_2a_6) (a_3a_5)a_4=a_4^2a_4^2a_4^2a_4=a_4^{7}=(a_1\cdot r^3)^7=(\frac{2}{11^3}.11^3)^7=2^7=128[/tex]

Solution:
The formula for the sum of an infinite geometric series is [tex]S=a_1.\frac{1}{1-r}[/tex]. Since [tex]a_n=(\frac{1}{3})^n, r=\frac{1}{3}[/tex] and [tex]a_1=6.\frac{1}{3}[/tex]. Then [tex]S=6.\frac{1}{3}.\frac{1}{1-\frac{1}{3}}=6.\frac{1}{3-1}=6.\frac{1}{2}=3[/tex]

Solution:
[tex]a_n=\frac{2^n}{3\cdot 3^n}=\frac{1}{3}.(\frac{2}{3})^n[/tex]. Then [tex]a_1=\frac{2}{9}[/tex] and [tex]r=\frac{2}{3}[/tex].
The infinite sum is [tex]S=a_1.\frac{1}{1-r}=\frac{2}{9}.\frac{1}{1-\frac{2}{3}}=\frac{2}{9}.\frac{1}{\frac{1}{3}}=\frac{2}{3}[/tex]

Solution:
[tex]a_1=\frac{2\cdot 3^1}{5}=\frac{6}{5}[/tex]. Obviously the common ratio [tex]r=3[/tex]. Then [tex]S_{4}=\frac{a_1}.\frac{1-r^{4}}{1-r}=\frac{6}{5}.\frac{1-3^4}{1-3}=\frac{6}{5}.\frac{80}{2}=48[/tex]

Solution:
Let denote the first term with [tex]a_1[/tex].
We know that [tex]S=a_1.\frac{1}{1-r}[/tex], so [tex]1-r=\frac{a_1}{S}[/tex], or [tex]r=1-\frac{a_1}{S}=1-\frac{4}{7}=\frac{3}{7}[/tex]

Solution:
Let [tex]a_1[/tex] be the first term.
From [tex]S=a_1.\frac{1}{1-r}[/tex], we have [tex]1-r=\frac{a_1}{S}[/tex], or [tex]r=1-\frac{a_1}{S}=1-\frac{9}{15}=\frac{6}{15}=\frac{2}{5}[/tex]

Solution:
[tex]S_4=a_1.\frac{r^4-1}{r-1}[/tex], therefore [tex]\frac{r^4-1}{r-1}=40[/tex], or [tex]r^4-1=40r-40[/tex]
[tex]r^4-40r+39=0[/tex]. But [tex]r^4-40r+39=r^4-r-39r+39=r(r^3-1)-39(r-1)=r(r-1)(r^2+r+1)-39(r-1)=(r-1)(r^3+r^2+r-39)[/tex].
[tex]r=1[/tex] is obviously not a solution (in this case, [tex]a_n=1[/tex] and [tex]S_4=4 \ne 40[/tex]), so it must be a root of the equation [tex]r^3+r^2+r-39=0[/tex].
[tex]r^3-27+r^2-9+r-3=0[/tex]
[tex](r-3)(r^2+3r+9)+(r-3)(r+3)+1(r-3)=0[/tex]
[tex](r-3)(r^2+3r+3+r+3+1)=0[/tex]
[tex](r-3)(r^2+4r+7)=0[/tex]. But [tex]r^2+4r+7=(r+2)^2+3>0[/tex], so it yields no solutions. Then the only root left is [tex]r=3[/tex].

Solution:
Let us denote the progression as [tex]a,ar,ar^2,...[/tex]. Its sum is [tex]S_1=6=\frac{a}{1-r}[/tex]. The squares of the progression are [tex]a^2, a^2r^2, a^2r^4, ...[/tex], which is another geometric progression - with a first term [tex]a^2[/tex] and common ratio [tex]r^2[/tex]. Its sum is [tex]S_2=\frac{a^2}{1-r^2}=\frac{a^2}{(1-r)(1+r)}[/tex]. Then
[tex]\frac{S_2}{S_1}=\frac{a^2}{1-r^2}.\frac{1-r}{a}=\frac{a}{1+r}=3[/tex]. We have
[tex]\begin{array}{|l}a=3+3r\\a=6-6r\end{array}[/tex]
[tex]\begin{array}{|l}2a=6+6r\\a=6-6r\end{array}[/tex]. We add them to get
[tex]3a=6+6[/tex], or [tex]a=4[/tex]

Solution:
[tex]0.272727(27)=27\cdot 10^{-2}+27\cdot 10^{-4}+27\cdot 10^{-6}+...[/tex], which is the sum of an infinite geometric series ([tex]a_1=27\cdot 10^{-2}; r=10^{-2}[/tex]) and by the formula it is
[tex]\frac{27 \cdot 10^{-2}}{1-10^{-2}}=\frac{\frac{27}{100}}{1-\frac{1}{100}}=\frac{27}{99}=\frac{3}{11}[/tex]

Solution:
By using [tex]a_n=a_1\cdot r^{n-1}[/tex]:
[tex]\begin{array}{|l}a_1r+a_1r^4-a_1r^3=10\\a_1r^2+a_1r^5-a_1r^4=20\end{array}[/tex]

[tex]\begin{array}{|l}a_1r(1+r^3-r^2)=10\\a_1r^2(1+r^3-r^2)=20\end{array}[/tex]. We divide them to get

[tex]\frac{a_1r^2(1+r^3-r^2)}{a_1r(1+r^3-r^2)}=\frac{20}{10}=2[/tex] => [tex]r=2[/tex]. Substituting into the first equation, we get [tex]a_2+8a_2-4a_2=10[/tex], or [tex]5a_2=10[/tex]

[tex]a_2=2[/tex].

Solution:
Let's the first term is a. The second term is 2 = ar. So $r = \frac{2}{a}$ The formula for infinite geometric series sum is [tex]S=\frac{a}{1-r} = 8[/tex]
[tex]S=\frac{a}{1-\frac{2}{a}} = 8[/tex]
[tex]\frac{a}{1-\frac{2}{a}} = 8[/tex]
[tex]a = 8(1-\frac{2}{a})[/tex]
[tex]a = 8-\frac{16}{a}[/tex]
[tex]a^2 = 8a - 16[/tex]
[tex]a^2 - 2\cdot4a + 4^2=0[/tex]
[tex](a-4)^2=0[/tex]
$a=4$

Solution:
Let us denote [tex]x_2=x_1r[/tex], [tex]y_1=x_1r^2[/tex], [tex]y_2=x_1r^3[/tex]. From Vieta's formulas, we know that
[tex]\begin{array}{|l}x_1+x_2=x_1(1+r)=3\\y_1+y_2=x_1(r^2+r^3)=x_1r^2(1+rr)=12\end{array}[/tex]. We divide them to get [tex]r^2=4[/tex], or [tex]r=2[/tex] (since the progression is strictly increasing and it can't be such if r is negative). Since [tex]x_1(r+1)=3[/tex], we have [tex]x_1=1, x_2=2, y_1=4, y_2=8[/tex].
From Vieta's formulas, [tex]a=x_1x_2=2[/tex] and [tex]-b=y_1y_2=32[/tex]. Therefore [tex]ab=2\cdot (-32)=-64[/tex].