Problems involving Progressions: Very Difficult Problems with Solutions

Solution:
From what is given, we know that [tex]b^2=ac[/tex] and [tex]4b=3c+a[/tex]. If we square the last equality, we get [tex]16b^2=a^2+6ac+9c^2[/tex]. Substituting [tex]b^2[/tex] for [tex]ac[/tex], the last becomes
[tex]a^2-10ac+9c^2=0[/tex], which means that either [tex]a=c[/tex] or [tex]a=9c[/tex]. Let us assume that [tex]a=c[/tex]. Then the arithmetic progression becomes [tex]c,2b,3c[/tex], which yields [tex]4b=3c+c=4c[/tex], or [tex]b=c[/tex], which is not a solution (since the geometric progression is not constant).
Then [tex]a=9c[/tex]. The arithmetic progression becomes [tex]9c,2b,3c[/tex], which means that [tex]4b=9c+3c=12c[/tex], or [tex]b=3c[/tex]. The geometric progression becomes [tex]9c,3c,c[/tex], which obviously has a common ratio [tex]r=\frac{1}{3}[/tex].

Solution:
Since [tex]a,b,c[/tex] form an arithmetic progression, we can say that
[tex]a=b-d[/tex], [tex]c=b+d[/tex]. [tex]b,c,d[/tex] form a geometric progression, so we can say that [tex]c=b.q[/tex] and [tex]d=bq^2[/tex]. Writing down the sums, we get the system:
[tex]\begin{array}{|l}b-d+bq^2=37\\b+bq=36\\b+d=bq\end{array}[/tex]. By adding the last two, we get [tex]2b+d+bq=36+bq[/tex], or [tex]d=36-2b[/tex]. We substitute into the first and third equations and get
[tex]\begin{array}{|l}b+2b-36+bq^2=37\\b+36-2b=bq\end{array}[/tex]
[tex]\begin{array}{|l}(3+q^2)b=73\\b(q+1)=36\end{array}[/tex]. We divide these two in order to get
[tex]\frac{3+q^2}{q+1}=\frac{73}{36}[/tex]
[tex]108+36q^2=73+73q[/tex]
[tex]36q^2-73q+35=0[/tex]. The solutions are [tex]q_1=\frac{5}{4}[/tex] and [tex]q_2=\frac{7}{9}[/tex]. They lead to [tex]b=\frac{36}{q+1}[/tex]:
[tex]b_1=\frac{36}{1+\frac{5}{4}}=\frac{36.4}{9}=16[/tex], [tex]c_1=b_1q=16.\frac{5}{4}=20[/tex], [tex]a_1=b_1-d=b_1-(c_1-b_1)=16-4=12[/tex] and [tex]d_1=c_1q=20.\frac{5}{4}=25[/tex]
and also [tex]b_2=\frac{36}{1+\frac{7}{9}}=\frac{36.9}{16}=\frac{81}{4}[/tex] with [tex]c_2=b_2q=\frac{81}{4}.\frac{7}{9}=\frac{63}{4}[/tex], [tex]a_2=b_2-(c_2-b_2)=\frac{81}{4}+\frac{18}{4}=\frac{99}{4}[/tex] and [tex]d_2=c_2q_2=\frac{63}{4}.\frac{7}{9}=\frac{49}{4}[/tex]. We seek non-integers, but [tex]a_1,b_1,c_1,d_1[/tex] do no suit this condition. Therefore the only answer left is [tex]d_2=\frac{49}{4}[/tex]

Solution:
Let us denote [tex]b=aq[/tex], [tex]c=aq^2[/tex].
We get the arithmetic progression [tex]a,aq,aq^2-64[/tex] and the geometric progression [tex]a,aq-8,aq^2-64[/tex]. Using the progressions respective properties of the middle member, we get the system
[tex]\begin{array}{|l}2aq=a+aq^2-64\\(aq-8)^2=a(aq^2-64) \end{array}[/tex]
[tex]\begin{array}{|l}2aq=a+aq^2-64\\a^2q^2-16aq+64=a^2q^2-64a \end{array}[/tex]
[tex]\begin{array}{|l}2aq=a+aq^2-64\\aq=4+4a \end{array}[/tex]. Multiplying the last one by q and applying it recursively, we get [tex]aq^2=4q+4aq=4q+16+16a[/tex]. We substitute into the first equation:
[tex]2aq=a+aq^2-64[/tex], or [tex]8+8a=a+4q+16+16a-64[/tex]
[tex]9a+4q=56[/tex], or [tex]q=-\frac{9a}{4}+14[/tex]. Substituting into [tex]aq=4+4a[/tex], we get
[tex]a(-\frac{9a}{4}+14)=4+4a[/tex]
[tex]-\frac{9a^2}{4}+14a=4+4a[/tex]
[tex]-9a^2+40a-16=0[/tex]
[tex]9a^2-40a+16=0[/tex]: [tex]D=20^2-9.16=16.25-16.9=16(25-9)=16^2[/tex]
[tex]a_{1,2}=\frac{20 \pm 16}{9}[/tex]. Since a is an integer, the only solution left is
[tex]a=\frac{20+16}{9}=4[/tex]. [tex]q=14-\frac{9a}{4}=14-9=5[/tex] and [tex]c=aq^2=4.5^2=100[/tex].

Solution:
15(1+4+9+16.........,)
It becomes special series
15[n(n+1)(2n+1)/6]
15[20(20+1)(40+1) /6] =
5*20*21*41 / 2 =
5*10*21*41 = 50*861 = 43,050