Problems involving Geometric Progressions: Very Difficult Problems with Solutions

Solution:
We know that
[tex]\frac{a_{n+1}}{a_n}=2^n[/tex]
[tex]\frac{a_n}{a_{n-1}}=2^{n-1}[/tex]

[tex]\frac{a_2}{a_1}=2^{1}[/tex] We multiply all of the above in order to reach
[tex]\frac{a_{n+1}}{a_n}\cdot\frac{a_n}{a_{n-1}} \dots \frac{a_2}{a_1}=2^{n+(n-1)+\dots +1}[/tex]
[tex]\frac{a_{n+1}}{a_1}=2^{\frac{n(n+1)}{2}}[/tex], since [tex]1+2+...+n-1+n=\frac{n(n+1)}{2}[/tex]. Then
[tex]a_{n+1}=2^{\frac{n(n+1)}{2}}\cdot a_1=2^{\frac{n(n+1)}{2}}[/tex]. Substituting [tex]n+1[/tex] for [tex]n[/tex], we get
[tex]a_n=2^{\frac{n(n-1)}{2}}[/tex] and [tex]log_2(a_n)=\frac{n(n-1)}{2}[/tex]. We need [tex]log_2(a_{100})=\frac{100\cdot (100-1)}{2}=50 \cdot 99= 4950[/tex]

Solution:
[tex]a_{n+1}-a_n=3^n[/tex]
[tex]a_{n}-a_{n-1}=3^{n-1}[/tex]
.
.
.
[tex]a_2-a_1=3[/tex] We add all of these in order to reach [tex]a_{n+1}-a_n+a_n-a_{n-1}+a_{n-1} \dots -a_1=3^n+3^{n-1}+\dots+3[/tex] [tex]a_{n+1}-a_1=3 \times (1+3+\dots + 3^{n-1})=3 \times \frac{3^n-1}{3-1}=\frac{3^{n+1}-3}{2}[/tex]. Substituting [tex]n+1[/tex] for [tex]n[/tex] and [tex]a_1=1[/tex], we get
[tex]a_n=\frac{3^n-3}{2}+1[/tex]. [tex]a_{10}=\frac{3^{10}-3}{2}+1=29524[/tex]

Solution:
If we add the first and second equation of the system, we get [tex]2x-y+2z+x+y+z=a+4+2a+2[/tex] or [tex]x+z=a+2[/tex]. Substituting back into the first equation, we have [tex]y+a+2=a+4[/tex], which yields [tex]y=2[/tex]. The system becomes
[tex]\begin{array}{|l}x+z=a+2\\2x+2z=2a+4\\3x+4-3z=1-2a \end{array} [/tex]. The first and second equations are linearly dependent, so we remove one of them. The system becomes
[tex]\begin{array}{|l}3x+3z=3a+6\\3x-3z=-3-2a \end{array}[/tex]. We add those two equations together and get [tex]6x=-3-2a+3a+6=a+3[/tex], or [tex]x=\frac{a+3}{6}[/tex].
By subtracting the same two equations, we get [tex]6z=3a+6+3+2a=5a+9[/tex], or [tex]z=\frac{5a+9}{6}[/tex]. Since [tex]x,2,z[/tex] form a geometric progression, we know that [tex]zx=2^2=4[/tex], or [tex]\frac{a+3}{6} \times \frac{5a+9}{6}=4[/tex]
[tex](a+3)(5a+9)=144[/tex]
[tex]5a^2+24a+27=144[/tex]
[tex]5a^2+24a-117=0[/tex] : [tex]D=12^2+117\cdot 5=729=27^2[/tex], [tex]a_{1,2}=\frac{-12 \pm 27}{5}[/tex]. We know that [tex]a>0[/tex], so the negative value is discarded and the answer is [tex]a=\frac{-12+27}{5}=\frac{15}{5}=3[/tex]

Solution:
[tex](a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=364+2(ab+bc+ca)=26^2=676[/tex], so [tex]ab+bc+ca=156[/tex]. Let us denote [tex]b=ar[/tex] and [tex]c=ar^2[/tex] (geometric progressions). We have
[tex]a+b+c=a+ar+ar^2=26[/tex] and [tex]ab+bc+ca=a^2r+a^2r^3+a^2r^2=156[/tex]
[tex]a(1+r+r^2)=26[/tex] and [tex]a^2r(1+r+r^2)=156=6\cdot 26[/tex]. We divide both equations and get
[tex]\frac{a^2r(1+r+r^2)}{a(1+r+r^2)}=6[/tex], or [tex]ar=b=6[/tex].

Solution:
We shall generalize the problem: find the infinite sum [tex]S=1+2x+3x^2+4x^3+...[/tex] for [tex]|x| < 1[/tex].
[tex]1+2x+3x^2+4x^3+...=(1+x+x^2+x^3+...)+x+2x^2+3x^3+...=\\=(1+x+x^2+x^3+...)+(x+x^2+x^3+...)+x^2+2x^3+3x^4+...=1 \times (1+x+x^2+x^3+...)+x(1+x+x^2+x^3+...)+x^2(1+x+x^2+x^3+...)+...=\\=(1+x+x^2+x^3+...)(1+x+x^2+x^3+...)=(1+x+x^2+x^3+...)^2[/tex]. We can take out [tex]x[/tex] behind the parentheses only because there are infinitely many terms.
But [tex]1+x+x^2+x^3+...[/tex] is the sum of an infinite geometric series with [tex]a_1=1[/tex] and [tex]r=x[/tex] and its sum is [tex]1+x+x^2+x^3+...=1 \times \frac{1}{1-x}=\frac{1}{1-x}[/tex], therefore [tex]S=(\frac{1}{1-x})^2[/tex]. Plugging back [tex]x=\frac{1}{7}[/tex], we get [tex]S=(\frac{1}{1-\frac{1}{7}})^2=(\frac{1}{\frac{6}{7}})^2=(\frac{7}{6})^2=\frac{49}{36}[/tex]

Solution:
Let [tex]G[/tex] be the centroid of a triangle [tex]ABC[/tex] with area [tex]S[/tex], and [tex]G'[/tex] the reflection of [tex]G[/tex] in the midpoint of [tex]BC[/tex]. Triangle [tex]BGG'[/tex] has sides [tex]\dfrac{2}{3}m_{a}, \dfrac{2}{3}m_{b}, \dfrac{2}{3}m_{c}[/tex], and its area equals the area of [tex]BGC[/tex], i.e. [tex]\dfrac{S}{3}[/tex]. So the triangle built from the medians has area
[tex]\left(\dfrac{3}{2}\right)^{2}\times \dfrac{S}{3}=\dfrac{3}{4}S[/tex]
The areas of the new triangles form a geometric progression with first term [tex]\dfrac{3}{4}Q[/tex] and ratio [tex]\dfrac{3}{4}[/tex]:
[tex]\dfrac{\frac{3}{4}Q}{1-\frac{3}{4}}=3Q[/tex]

Solution:
The centres lie on the bisector, which makes [tex]30^{\circ}[/tex] with each arm. The centre of a circle of radius [tex]r[/tex] tangent to both arms is at distance [tex]\dfrac{r}{\sin 30^{\circ}}=2r[/tex] from the vertex.
External tangency of [tex]k_{n}[/tex] and [tex]k_{n+1}[/tex] means that the distance between their centres is [tex]r_{n}+r_{n+1}[/tex]: [tex]2r_{n+1}-2r_{n}=r_{n}+r_{n+1}\iff r_{n+1}=3r_{n}[/tex].
[tex]r_{1}+r_{2}+\ldots+r_{6}=\dfrac{3^{6}-1}{3-1}=364[/tex]

Solution:
The condition is [tex]a_{n}=a_{n+1}-a_{n-1}\iff a_{1}q^{n-1}=a_{1}q^{n}-a_{1}q^{n-2}\iff q^{2}-q-1=0[/tex].
Since [tex]a_{1}=7-3\sqrt{5}\gt 0[/tex] (because [tex]49\gt 45[/tex]) and the progression is increasing, [tex]q\gt 1[/tex], so [tex]q=\dfrac{1+\sqrt{5}}{2}[/tex].
[tex]q^{2}=q+1=\dfrac{3+\sqrt{5}}{2}, q^{4}=\left(\dfrac{3+\sqrt{5}}{2}\right)^{2}=\dfrac{7+3\sqrt{5}}{2}[/tex]
[tex]a_{5}=a_{1}q^{4}=\dfrac{(7-3\sqrt{5})(7+3\sqrt{5})}{2}=\dfrac{49-45}{2}=2[/tex]

Solution:
The [tex]k[/tex]-th term is [tex]\underbrace{11\ldots1}_{k}=\dfrac{10^{k}-1}{9}[/tex], so the sum is
[tex]\dfrac{1}{9}\left[(10+10^{2}+\ldots+10^{n})-n\right]=\dfrac{1}{9}\left(\dfrac{10^{n+1}-10}{9}-n\right)=\dfrac{10^{n+1}-9n-10}{81}[/tex]
Check: [tex]n=2:\ \dfrac{1000-28}{81}=12=1+11[/tex].

Solution:
In a geometric progression [tex]a_{k}a_{2n+1-k}=a_{1}a_{2n}[/tex] for every [tex]k[/tex], so [tex]\dfrac{1}{a_{k}}=\dfrac{a_{2n+1-k}}{a_{1}a_{2n}}[/tex]. Adding over all [tex]k[/tex]:
[tex]\dfrac{1}{a_{1}}+\ldots+\dfrac{1}{a_{2n}}=\dfrac{a_{1}+\ldots+a_{2n}}{a_{1}a_{2n}}\Rightarrow 24=\dfrac{12}{a_{1}a_{2n}}\Rightarrow a_{1}a_{2n}=\dfrac{1}{2}[/tex]
The [tex]2n[/tex] factors split into [tex]n[/tex] pairs [tex]a_{k}a_{2n+1-k}[/tex], each equal to [tex]\dfrac{1}{2}[/tex], so the product is [tex]\dfrac{1}{2^{n}}[/tex].

Solution:
If [tex]q=1[/tex], then [tex]\dfrac{S_{3n}}{S_{n}}=3[/tex], so [tex]q\ne 1[/tex]. Let [tex]t=q^{n}[/tex]. The terms [tex]a_{n+1}, \ldots, a_{2n}[/tex] are [tex]t[/tex] times the first [tex]n[/tex] terms, and [tex]a_{2n+1}, \ldots, a_{3n}[/tex] are [tex]t^{2}[/tex] times them:
[tex]S_{2n}=S_{n}(1+t), S_{3n}=S_{n}(1+t+t^{2})[/tex]
[tex]1+t+t^{2}=2\iff t^{2}+t-1=0\iff t=\dfrac{-1\pm\sqrt{5}}{2}[/tex]
Then [tex]\dfrac{S_{2n}}{S_{n}}=1+t=\dfrac{1\pm\sqrt{5}}{2}[/tex], and the positive value is [tex]\dfrac{1+\sqrt{5}}{2}[/tex].

Solution:
Let [tex]T=1+q^{2}+q^{4}+\ldots+q^{2n-2}\ge 1[/tex]. Then
[tex]a_{2}+a_{4}+\ldots+a_{2n}=a_{1}qT, a_{1}+a_{3}+\ldots+a_{2n+1}=a_{1}(1+q^{2}T), q+q^{3}+\ldots+q^{2n-1}=qT[/tex]
The equation becomes [tex]\dfrac{1+q^{2}T}{qT}=2+\dfrac{1}{qT}\iff 1+q^{2}T=2qT+1\iff qT(q-2)=0[/tex].
Since [tex]q\ne 0[/tex] and [tex]T\ne 0[/tex], [tex]q=2[/tex].

Solution:
Since [tex]1-\cos 2x=2\sin ^{2}x[/tex], the product of the first and the third number is [tex]1[/tex], so
[tex]\left(\cos x-\dfrac{1}{2}\right)^{2}=1, |\cos x|\le 1\Rightarrow\cos x=-\dfrac{1}{2}[/tex]
Then [tex]\sin ^{2}x=\dfrac{3}{4}[/tex], the three terms are [tex]\dfrac{3}{2}, -1, \dfrac{2}{3}[/tex] and [tex]q=-\dfrac{2}{3}[/tex].
[tex]a_{15}=a_{k}q^{15-k}\Rightarrow\dfrac{3}{2}\left(-\dfrac{2}{3}\right)^{15-k}=\dfrac{27}{8}\Rightarrow\left(-\dfrac{2}{3}\right)^{15-k}=\dfrac{9}{4}=\left(-\dfrac{2}{3}\right)^{-2}[/tex]
[tex]15-k=-2\Rightarrow k=17[/tex]

Solution:
[tex]f'(x)=3x^{2}-12x+9=3(x-1)(x-3)[/tex], so the extrema are at [tex]x=1[/tex] and [tex]x=3[/tex].
Let [tex]L=\log_{3}p[/tex]. Then
[tex]f(3)=L, f(2)=L+2, f(1)=L+4[/tex]
The square of the middle term equals the product of the other two:
[tex](L+2)^{2}=L(L+4)\iff 4=0[/tex] – impossible;
[tex]L^{2}=(L+2)(L+4)\iff L=-\dfrac{4}{3}[/tex]
[tex](L+4)^{2}=L(L+2)\iff L=-\dfrac{8}{3}[/tex]
Since [tex]p=3^{L}[/tex] increases with [tex]L[/tex], the smallest value is [tex]p=3^{-\frac{8}{3}}[/tex].

Solution:
Let [tex]128=54q^{k}[/tex] for some integer [tex]k\ne 0[/tex]. Then [tex]|q|^{k}=\dfrac{64}{27}=\left(\dfrac{4}{3}\right)^{3}[/tex].
Every term has the form [tex]\pm 54\left(\dfrac{4}{3}\right)^{r}[/tex] with rational [tex]r[/tex]. Comparing the powers of [tex]3[/tex] shows that this number is rational only when [tex]r=m[/tex] is an integer.
The number [tex]54\left(\dfrac{4}{3}\right)^{m}=2^{2m+1}\times 3^{3-m}[/tex] is natural only for [tex]m=0, 1, 2, 3[/tex], which gives [tex]54, 72, 96, 128[/tex]; all four are terms of the progression with [tex]q=\dfrac{4}{3}[/tex].
[tex]54+72+96+128=350[/tex]

Solution:
By the AM–GM inequality for the [tex]2010[/tex] numbers [tex]q^{2010}, 1, \ldots, 1[/tex]: [tex]q^{2010}+2009\ge 2010\sqrt[2010]{q^{2010}}=2010|q|\ge 2010q[/tex].
Equality holds only if [tex]q^{2010}=1[/tex] and [tex]|q|=q[/tex], i.e. [tex]q=1[/tex]. So all terms equal [tex]1[/tex] and [tex]S_{n}=n[/tex]:
[tex]S_{1}+S_{2}+\ldots+S_{2009}=1+2+\ldots+2009=\dfrac{2009\times 2010}{2}=2019045[/tex]
Note: the formula [tex]S_{n}=\dfrac{q^{n}-1}{q-1}[/tex] cannot be used here, because [tex]q=1[/tex].

Solution:
Suppose they are terms of a geometric progression with ratio [tex]q[/tex]. Then [tex]11=10q^{m}, 12=11q^{n}[/tex] for some non-zero integers [tex]m[/tex] and [tex]n[/tex].
Raising the first equality to the power [tex]n[/tex] and the second to the power [tex]m[/tex]: [tex]\left(\dfrac{11}{10}\right)^{n}=q^{mn}=\left(\dfrac{12}{11}\right)^{m}[/tex]. Hence
[tex]11^{m+n}=10^{n}\times 12^{m}=2^{n+2m}\times 3^{m}\times 5^{n}[/tex]
By the uniqueness of prime factorization (also with negative exponents) the powers of [tex]5[/tex] on both sides must be equal, so [tex]n=0[/tex] – a contradiction.