Arithmetic Progressions: Very Difficult Problems with Solutions

Solution:
By the property of an arithmetic progression we know that for any positive integer [tex]m < k[/tex] the following holds true: [tex]\frac{a_{k+m}+a_{k-m}}{2}=a_{\frac{k+m+k-m}{2}}=a_k=0[/tex] . But [tex]S_{2k-1}=a_{k-(k-1)}+a_{k-(k-2)}+...+a_{k+(k-2)}+a_{k+(k-1)}=a_{k-(k-1)}+a_{k + (k-1)} +a_{k-(k-2)}+a_{k+(k-2)}+...+a_k=0+0+...+0=0[/tex]

Solution:
Let us denote the number count on the left side with [tex]n[/tex]. The left side's value is the sum of the arithmetic progression, defined as [tex]a_1=1[/tex], [tex]d=3[/tex]. We know that for our chosen [tex]n[/tex], [tex]a_n=x[/tex], or [tex]1+(n-1)\cdot 3=x[/tex] and [tex]n=\frac{x+2}{3}[/tex].
[tex]S_n=\frac{2a_1+(n-1)\cdot d}{2}\cdot n=\frac{2\cdot 1+(n-1)\cdot 3}{2}\cdot n[/tex]. Now we substitute [tex]n-1[/tex] and [tex]n[/tex] with the respective functions of x:
[tex]S_n=\frac{2+x-1}{2}\cdot \frac{x+2}{3}=\frac{(x+1)(x+2)}{6}[/tex]
[tex]\frac{(x+1)(x+2)}{6}=925[/tex]
[tex]x^2+3x+2=5550[/tex]
[tex]x^2+3x-5548=0[/tex]: [tex]D=9+4\cdot 5548=22201=149^2[/tex]
X is obviously positive, so we don't count the negative root of the equation and we have
[tex]x=\frac{-3+149}{2}=\frac{146}{2}=73[/tex]

Solution:
We know that for any [tex]k \in [0;99][/tex]: [tex]a_{k+1}=3k+5[/tex] and for any [tex]l[/tex] in the same interval, [tex]b_{l+1}=4l+3[/tex]. We seek the number of occurences [tex]a_{k+1}=b_{l+1}[/tex] for any [tex]k,l[/tex] in the given interval. This is equivalent to
[tex]3k+5=4l+3[/tex]
[tex]4l=3k+2[/tex]
[tex]l=\frac{3k+2}{4}[/tex]. We now have [tex]l[/tex] as a function of [tex]k[/tex]. For the given values, [tex]l=\frac{3k+2}{4} \le \frac{3.99+2}{4}<75[/tex], so we will not go out of bounds for [tex]l[/tex]. By definition, [tex]l[/tex] is an integer, and so should be [tex]\frac{3k+2}{4}[/tex]. So we are searching for the values of [tex]k[/tex] between 0 and 99, for which [tex]\frac{3k+2}{4}[/tex] is an integer. Then the following must hold true: [tex]3k+2 \equiv 0 (mod 4)[/tex]
[tex]3k+2-4k \equiv 0 (mod 4)[/tex]
[tex]2-k \equiv 0 (mod 4)[/tex]
[tex]k \equiv 2 (mod 4)[/tex].
These numbers are the sequence [tex]2,6,10,...,98[/tex] and their count is 25.

Solution:
Let us denote [tex]C=\frac{a_{2n}+a_{2n-1}+...+a_{n+1}}{a_n+a_{n-1}+...+a_1}[/tex]. By adding [tex]1[/tex] to both sides, we get
[tex]C+1=\frac{a_{2n}+a_{2n-1}+...+a_{n+1}}{a_n+a_{n-1}+...+a_1}+\frac{a_n+a_{n-1}+...+a_1}{a_n+a_{n-1}+...+a_1}=\frac{a_{2n}+a_{2n-1}+...+a_1}{a_n+a_{n-1}+...+a_1}=\frac{S_{2n}}{S_n}[/tex]. Since [tex]C[/tex] is constant, then [tex]C+1[/tex] is also constant. Therefore, [tex]\frac{S_{2n}}{S_n}[/tex] must not depend on [tex]n[/tex]. But
[tex]\frac{S_{2n}}{S_n}=\frac{\frac{2a_1+(2n-1) \times d}{2} \times 2n}{\frac{2a_1+(n-1) \times d}{2}n}=2\frac{2+(2n-1)d}{2+(n-1)d}[/tex].
Let us denote [tex]\frac{C+1}{2}=R[/tex], which is also a constant. We have
[tex]\frac{2+(2n-1)d}{2+(n-1)d}=R[/tex]
[tex]2R+(n-1) \times d \times R=2+(2n-1) \times d[/tex]
[tex]2R+n \times d \times R-d \times R=2+2n \times d-d[/tex]
[tex]n \times d(R-2)=d \times R-2R-d+2=R(d-2)-(d-2)=(R-1)(d-2)[/tex]. This must hold true for any [tex]n[/tex]. Let us assume that both sides are non-zero. The left side changes with [tex]n[/tex], while the right side does not, which means they will be unequal at some point. Therefore both sides must be zero. [tex]n[/tex] and [tex]d[/tex] are non-zero (since [tex]\{a_n\}[/tex] is a non-constant progression), so for the left side to be zero, [tex]R[/tex] must be 2. Which leads to
[tex]0=1\times (d-2)[/tex], which means that [tex]d[/tex] must also be 2. The arithmetic progression is now defined, with [tex]a_1=1[/tex], [tex]d=2[/tex] and [tex]a_{15}=a_1+14\times d=1+14\times2=29[/tex]

Solution:
Let there be [tex]n[/tex] pumps. They start at equal intervals and stop together, so their working times [tex]t_{1}, t_{2}, \ldots, t_{n}=10[/tex] form an arithmetic progression.
Filling the tanker requires [tex]15n[/tex] pump-hours (all pumps together need [tex]15[/tex] hours), so
[tex]t_{1}+t_{2}+\ldots+t_{n}=\dfrac{n(t_{1}+10)}{2}=15n\Rightarrow t_{1}+10=30\Rightarrow t_{1}=20[/tex]

Solution:
In an arithmetic progression with [tex]2k-1[/tex] terms the middle term is [tex]a_{k}[/tex], so [tex]S_{2k-1}=(2k-1)a_{k}[/tex].
For [tex]k=17[/tex]: [tex]S_{33}=33a_{17}, T_{33}=33b_{17}[/tex]. Hence
[tex]\dfrac{a_{17}}{b_{17}}=\dfrac{S_{33}}{T_{33}}=\dfrac{3\times 33+5}{4\times 33-1}=\dfrac{104}{131}[/tex]

Solution:
The second progression gives [tex]S_{n}=2a_{n}-a_{1}[/tex]. Then the inequality becomes
[tex](2a_{n}-a_{1})^{2}+7a_{1}^{2}\lt 4a_{n}^{2}\iff 2a_{1}^{2}\lt a_{1}a_{n}[/tex]
So [tex]a_{1}\ne 0, \dfrac{a_{n}}{a_{1}}\gt 2[/tex].
From the sum formula: [tex]\dfrac{n(a_{1}+a_{n})}{2}=2a_{n}-a_{1}\iff a_{n}(4-n)=a_{1}(n+2)[/tex].
For [tex]n=4[/tex] we would get [tex]a_{1}=0[/tex], which is impossible. Otherwise
[tex]\dfrac{a_{n}}{a_{1}}=\dfrac{n+2}{4-n}\gt 2\iff 2\lt n\lt 4[/tex]
So [tex]n=3[/tex] (for example [tex]1, 3, 5[/tex]).

Solution:
[tex]S_{5}-S_{6}=-a_{6}\ge 6\Rightarrow a_{6}\le -6[/tex]
[tex]S_{5}-S_{4}=a_{5}\ge 6[/tex]
Hence [tex]d=a_{6}-a_{5}\le -12[/tex].
Since [tex]a_{k+6}=a_{k}+6d[/tex], the sum of the next six terms differs from [tex]S_{6}[/tex] by [tex]36|d|[/tex]: [tex]36|d|\le 450\Rightarrow |d|\le 12. 5[/tex].
So [tex]d=-12, a_{5}=6, a_{6}=-6[/tex], and
[tex]a_{1}=a_{5}-4d=6+48=54[/tex]

Solution:
Let [tex]a=a_{1}[/tex]. Then [tex]a+7d=a^{3}\Rightarrow 7d=a(a-1)(a+1)[/tex]; since [tex]d\ne 0[/tex], [tex]a\notin\{-1, 0, 1\}[/tex].
Since [tex]a^{2}[/tex] is a term, [tex]a^{2}=a+kd[/tex] for some integer [tex]k\ge 1[/tex]. Then
[tex]a(a-1)=\dfrac{k\,a(a-1)(a+1)}{7}\Rightarrow k(a+1)=7[/tex]
Since [tex]a[/tex] is an integer and [tex]k\ge 1[/tex], [tex]a+1\in\{1, 7\}[/tex]; as [tex]a\ne 0[/tex], [tex]a=6[/tex].
[tex]d=\dfrac{6\times 5\times 7}{7}=30[/tex]
Check: [tex]a^{2}=36=a_{2}, a^{4}=1296=6+43\times 30=a_{44}[/tex].

Solution:
Let the sides be [tex]a-1, a, a+1[/tex]. The triangle inequality [tex](a-1)+a\gt a+1[/tex] gives [tex]a\gt 2[/tex].
By the law of cosines [tex]\cos\gamma=\dfrac{(a-1)^{2}+a^{2}-(a+1)^{2}}{2a(a-1)}=\dfrac{a-4}{2(a-1)}=\dfrac{1}{2}-\dfrac{3}{2(a-1)}[/tex].
For [tex]a\gt 2[/tex] this is a continuous increasing function, which tends to [tex]-1[/tex] as [tex]a\to 2[/tex] and to [tex]\dfrac{1}{2}[/tex] as [tex]a\to\infty[/tex], without reaching these values.
[tex]-1\lt\cos\gamma\lt\dfrac{1}{2}\iff 60^{\circ}\lt\gamma\lt 180^{\circ}[/tex]

Solution:
Write the elements in increasing order: [tex]a, a+d, \ldots, a+7d[/tex], where [tex]a\ge 1, d\ge 1, a+7d\le 2008[/tex].
For a fixed [tex]d[/tex] there are [tex]2008-7d[/tex] choices of [tex]a[/tex]; this number is positive for [tex]1\le d\le 286[/tex].
[tex]\sum_{d=1}^{286}(2008-7d)=286\times 2008-7\times \dfrac{286\times 287}{2}=574288-287287=287001[/tex]

Solution:
[tex]S_{2026}=1013(a_{1}+a_{2026})=2026\Rightarrow a_{1}+a_{2026}=2[/tex]
If [tex]d=0[/tex], all terms equal [tex]1[/tex] and the second sum is positive, so [tex]d\ne 0[/tex].
Since [tex]\dfrac{1}{a_{i}a_{i+1}}=\dfrac{1}{d}\left(\dfrac{1}{a_{i}}-\dfrac{1}{a_{i+1}}\right)[/tex], the sum telescopes:
[tex]\dfrac{1}{d}\left(\dfrac{1}{a_{1}}-\dfrac{1}{a_{2026}}\right)=\dfrac{a_{2026}-a_{1}}{d\,a_{1}a_{2026}}=\dfrac{2025}{a_{1}a_{2026}}=-2025[/tex]
So [tex]a_{1}a_{2026}=-1[/tex], and [tex]a_{1}[/tex], [tex]a_{2026}[/tex] are the roots of [tex]t^{2}-2t-1=0[/tex]:
[tex]a_{1}=1\pm\sqrt{2}[/tex]

Solution:
Let [tex]t=4^{2x-x^{2}-2}[/tex]; then [tex]16^{2x-x^{2}-2}=t^{2}[/tex].
Since [tex]2x-x^{2}-2=-(x-1)^{2}-1\le -1[/tex], [tex]t[/tex] takes exactly the values in [tex]\left(0, \dfrac{1}{4}\right][/tex].
The condition is [tex]2at=3+t^{2}\iff a=f(t)=\dfrac{1}{2}\left(t+\dfrac{3}{t}\right)[/tex].
The function [tex]f[/tex] is decreasing for [tex]0\lt t\le\sqrt{3}[/tex], so on [tex]\left(0, \dfrac{1}{4}\right][/tex] it takes all values [tex]a\ge f\left(\dfrac{1}{4}\right)=\dfrac{49}{8}[/tex].

Solution:
For any numbers [tex](x_{1}+\ldots+x_{m})^{2}-(x_{1}^{2}+\ldots+x_{m}^{2})=2\sum_{i\lt j}x_{i}x_{j}[/tex].
Deleting [tex]a_{n}[/tex] removes the products [tex]a_{n}a_{i}[/tex], so the difference decreases by [tex]2a_{n}S_{n-1}=1440\iff a_{n}S_{n-1}=720[/tex].
The terms are increasing positive integers, so [tex]a_{k}\ge k\Rightarrow n\times \dfrac{n(n-1)}{2}\le 720\Rightarrow n\le 11[/tex].
For [tex]n=11[/tex] we need [tex](a_{1}+10d)(2a_{1}+9d)=144[/tex], and for [tex]n=10[/tex] we need [tex](a_{1}+9d)(a_{1}+4d)=80[/tex]; checking the few possible [tex]a_{1}\ge 1, d\ge 1[/tex] gives no solutions.
For [tex]n=9[/tex] we need [tex](a_{1}+8d)(2a_{1}+7d)=180[/tex], which holds for [tex]a_{1}=4, d=1[/tex]: the progression [tex]4, 5, \ldots, 12[/tex]. So the maximum is [tex]9[/tex].

Solution:
Let [tex]t=x^{4}[/tex]: [tex]t^{2}+at+a^{4}=0[/tex].
Four real roots require two distinct positive roots [tex]0\lt t_{1}\lt t_{2}[/tex]; then the real roots are [tex]-q, -p, p, q[/tex], where [tex]p=\sqrt[4]{t_{1}}, q=\sqrt[4]{t_{2}}[/tex].
They form an arithmetic progression when [tex]q-p=2p\iff q=3p\iff t_{2}=81t_{1}[/tex].
By Vieta’s formulas [tex]t_{1}t_{2}=81t_{1}^{2}=a^{4}\Rightarrow t_{1}=\dfrac{a^{2}}{9}, t_{2}=9a^{2}[/tex].
Also [tex]t_{1}+t_{2}=\dfrac{82a^{2}}{9}=-a\Rightarrow a=-\dfrac{9}{82}[/tex] (since [tex]a\ne 0[/tex]).

Solution:
We need [tex]x\ge 0[/tex], and [tex]x=0[/tex] gives [tex]d=0[/tex], so [tex]x\gt 0[/tex].
Let [tex]x=t^{12}, t\gt 0[/tex]; the numbers are [tex]t^{6}, t^{4}, t^{3}[/tex].
The condition is [tex]2t^{4}=t^{6}+t^{3}\iff t^{3}-2t+1=0\iff (t-1)(t^{2}+t-1)=0[/tex].
The root [tex]t=1[/tex] gives [tex]d=0[/tex], so [tex]t=\dfrac{\sqrt{5}-1}{2}[/tex].
[tex]t^{2}=\dfrac{3-\sqrt{5}}{2}, t^{4}=\dfrac{7-3\sqrt{5}}{2}[/tex]
[tex]x=\left(t^{4}\right)^{3}=161-72\sqrt{5}[/tex]

Solution:
Divide by [tex]d^{2}[/tex] and let [tex]y=\dfrac{a_{1}}{d}[/tex]; then [tex]\dfrac{a_{k}}{d}=y+k-1[/tex], and the condition becomes the quadratic equation
[tex](n-2)y^{2}+(n-1)(n-4)y+\dfrac{(n-1)(2n^{2}-13n+12)}{6}=0[/tex]
Its discriminant is [tex]D=-\dfrac{n(n-1)(n^{2}-7n+4)}{3}[/tex], so a real [tex]y[/tex] exists only if [tex]n^{2}-7n+4\le 0[/tex], i.e. [tex]n\le\dfrac{7+\sqrt{33}}{2}\lt 7[/tex].
The free term is never [tex]0[/tex] for natural [tex]n[/tex], so [tex]y\ne 0[/tex] and [tex]a_{1}\ne 0[/tex]. Hence [tex]n\in\{3, 4, 5, 6\}[/tex]; for example, for [tex]n=3[/tex] the progression [tex]3, 4, 5[/tex] works.