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 n. 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 n, [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 k,l 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 l as a function of k. 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 l. By definition, l is an integer, and so should be [tex]\frac{3k+2}{4}[/tex]. So we are searching for the values of k 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 n. But
[tex]\frac{S_{2n}}{S_n}=\frac{\frac{2a_1+(2n-1).d}{2}.2n}{\frac{2a_1+(n-1).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).d.R=2+(2n-1).d[/tex]
[tex]2R+n.d.R-d.R=2+2n.d-d[/tex]
[tex]n.d(R-2)=d.R-2R-d+2=R(d-2)-(d-2)=(R-1)(d-2)[/tex]. This must hold true for any n. 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, R 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]