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 \times 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 \times 4}{9}=16[/tex], [tex]c_1=b_1q=16 \times \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 \times \frac{5}{4}=25[/tex]
and also [tex]b_2=\frac{36}{1+\frac{7}{9}}=\frac{36 \times 9}{16}=\frac{81}{4}[/tex] with [tex]c_2=b_2q=\frac{81}{4} \times \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} \times \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 [tex]q[/tex] 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 \times 16=16 \times 25-16 \times 9=16(25-9)=16^2[/tex]
[tex]a_{1,2}=\frac{20 \pm 16}{9}[/tex]. Since [tex]a[/tex] 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 \times 5^2=100[/tex].

Solution:
The [tex]k[/tex]-th term is [tex]3k \times 5k = 15k^2[/tex], so the sum is
[tex]15\left(1^2+2^2+\dots+20^2\right)[/tex]
Using [tex]1^2+2^2+\dots+n^2 = \frac{n(n+1)(2n+1)}{6}[/tex]:
[tex]15 \times \frac{20 \times 21 \times 41}{6} = 15 \times 2870 = 43\,050[/tex]

Solution:
Let [tex]a_{1}=q, d=b_{1}=b[/tex]. Then
[tex]4q+6b=32, b(1+q)=12[/tex]
[tex]b=\dfrac{16-2q}{3}\Rightarrow (16-2q)(1+q)=36\iff q^{2}-7q+10=0\iff q=2[/tex] or [tex]q=5[/tex]
For [tex]q=5[/tex] we get [tex]b=2\lt a_{1}[/tex], so [tex]q=2[/tex] and [tex]b_{1}=4\gt a_{1}=2[/tex]. Then [tex]b_{n}=4\times 2^{n-1}=2^{n+1}[/tex].
[tex]2^{10}=1024\lt 1996\lt 2048=2^{11}\Rightarrow n+1\ge 11\Rightarrow n=10[/tex]

Solution:
Subtracting the last two conditions: [tex]a_{3}-a_{1}=2d=12\Rightarrow d=6[/tex].
Let [tex]a_{2}=b_{1}=x[/tex]; then [tex]a_{1}=x-6[/tex], [tex]a_{3}=x+6[/tex] and [tex]b_{3}=a_{1}+102=x+96[/tex]. The sum gives
[tex]3x+x+b_{2}+x+96=192\Rightarrow b_{2}=96-5x[/tex]
[tex]b_{2}^{2}=b_{1}b_{3}\iff (96-5x)^{2}=x(x+96)\iff x^{2}-44x+384=0\iff x=12[/tex] or [tex]x=32[/tex]
For [tex]x=32[/tex] the geometric progression is [tex]32, -64, 128[/tex], which is not increasing. For [tex]x=12[/tex] it is [tex]12, 36, 108[/tex] (the arithmetic progression is [tex]6, 12, 18[/tex]). So [tex]b_{3}=108[/tex].

Solution:
[tex]a_{5}+a_{6}=2a_{1}+9d=k^{2}-k[/tex]
[tex]a_{2}-S_{4}-a_{5}=(a_{1}+d)-(4a_{1}+6d)-(a_{1}+4d)=-4a_{1}-9d\Rightarrow 4a_{1}+9d=k^{2}-k+2[/tex]
Subtracting the first equation: [tex]2a_{1}=2\Rightarrow a_{1}=1[/tex] for every [tex]k[/tex].
The derivative [tex]f'(x)=\dfrac{1-x^{2}}{(1+x^{2})^{2}}[/tex] changes sign from [tex]+[/tex] to [tex]-[/tex] at [tex]x=1[/tex], so the local maximum is [tex]f(1)=\dfrac{1}{2}[/tex].
[tex]b_{1}=1, q=\dfrac{1}{2}, S_{10}=\dfrac{1-\left(\frac{1}{2}\right)^{10}}{1-\frac{1}{2}}=\dfrac{1023}{512}[/tex]

Solution:
A sequence that is both an arithmetic and a geometric progression is constant: from [tex]a_{1}+a_{3}=2a_{2}[/tex] and [tex]a_{1}a_{3}=a_{2}^{2}[/tex] we get [tex](a_{1}-a_{3})^{2}=0[/tex], so [tex]a_{1}=a_{2}=a_{3}[/tex].
Here [tex]y\gt 0, a_{1}=2^{3x}y^{3}, a_{2}=\dfrac{2^{x}}{y}[/tex].
[tex]a_{1}=a_{2}\iff 2^{2x}y^{4}=1\iff 2^{x}y^{2}=1, a_{2}=a_{3}\iff 2^{x}=5y^{2}[/tex]
[tex]5y^{4}=1\Rightarrow y=\dfrac{1}{\sqrt[4]{5}}, 2^{x}=\sqrt{5}\Rightarrow x=\dfrac{1}{2}\log_{2}5[/tex]

Solution:
Let the roots be [tex]m-3t, m-t, m+t, m+3t[/tex]. By Vieta’s formulas their sum is [tex]4m=4[/tex], so [tex]m=1[/tex].
The sum of their squares is [tex]4+20t^{2}[/tex], so the sum of their pairwise products is [tex]\dfrac{4^{2}-(4+20t^{2})}{2}=6-10t^{2}[/tex]. By Vieta it equals [tex]6[/tex], so [tex]t=0[/tex] and all roots equal [tex]1[/tex]:
[tex](x-1)^{4}=x^{4}-4x^{3}+6x^{2}-4x+1\Rightarrow a=-4, b=1[/tex]

Solution:
The angles are [tex]\beta-\delta, \beta, \beta+\delta[/tex], so [tex]3\beta=180^{\circ}[/tex] and [tex]\beta=60^{\circ}[/tex].
By the law of sines the side opposite [tex]\beta[/tex] is [tex]b=2R\sin 60^{\circ}=R\sqrt{3}[/tex].
The tangent segment from the vertex of [tex]\beta[/tex] to the incircle equals [tex]p-b[/tex], where [tex]p[/tex] is the semiperimeter, so
[tex]r=(p-b)\tan 30^{\circ}=\dfrac{p-b}{\sqrt{3}}\Rightarrow p=b+r\sqrt{3}=\sqrt{3}(R+r)[/tex]
[tex]S=pr=\sqrt{3}\,r(R+r)[/tex]

Solution:
Let [tex]x=1+\dfrac{1}{n}[/tex]; then [tex]x-1=\dfrac{1}{n}[/tex] and [tex]S_{n}=\sum_{k=1}^{n}kx^{k-1}[/tex]. Then
[tex]S_{n}-xS_{n}=1+x+x^{2}+\ldots+x^{n-1}-nx^{n}=\dfrac{x^{n}-1}{x-1}-nx^{n}=n(x^{n}-1)-nx^{n}=-n[/tex]
[tex](1-x)S_{n}=-n\iff -\dfrac{1}{n}S_{n}=-n\iff S_{n}=n^{2}[/tex]

Solution:
Let [tex]d\gt 0[/tex] be the common difference and [tex]x=\dfrac{d}{a_{1}}\gt 0[/tex]. Then [tex]q=\dfrac{a_{1}+d}{a_{1}}=1+x[/tex], so
[tex]b_{n}=a_{1}(1+x)^{n-1}, a_{n}=a_{1}+(n-1)d=a_{1}\left(1+(n-1)x\right)[/tex]
By the binomial theorem, for [tex]n-1\ge 2[/tex]: [tex](1+x)^{n-1}\ge 1+(n-1)x+\dfrac{(n-1)(n-2)}{2}x^{2}\gt 1+(n-1)x[/tex].
Multiplying by [tex]a_{1}\gt 0[/tex] gives [tex]b_{n}\gt a_{n}[/tex].

Solution:
Let [tex]d[/tex] be the common difference. Since [tex]1[/tex] and [tex]\sqrt{2}[/tex] are terms, [tex]\sqrt{2}-1=sd[/tex] for some non-zero integer [tex]s[/tex], so [tex]d[/tex] is irrational, and every term has the form [tex]1+jd[/tex] with an integer [tex]j[/tex].
Suppose the distinct terms [tex]1+id, 1+jd, 1+ld[/tex] form a geometric progression: [tex](1+jd)^{2}=(1+id)(1+ld)\iff (2j-i-l)+(j^{2}-il)d=0[/tex].
If [tex]j^{2}\ne il[/tex], then [tex]d=\dfrac{i+l-2j}{j^{2}-il}[/tex] is rational – a contradiction. So [tex]j^{2}=il[/tex] and [tex]2j=i+l[/tex], hence [tex](i-l)^{2}=(i+l)^{2}-4il=4j^{2}-4j^{2}=0[/tex] and [tex]i=l[/tex] – again a contradiction.

Solution:
[tex]3a_{2}=15, b_{2}^{3}=27\Rightarrow a_{2}=5, b_{2}=3, a_{2}+b_{2}=8[/tex]
Put [tex]a_{1}=5-d, a_{3}=5+d, b_{1}=\dfrac{3}{q}, b_{3}=3q, u=a_{1}+b_{1}, v=a_{3}+b_{3}[/tex]. Then
[tex]uv=64, u+v=10+3\left(q+\dfrac{1}{q}\right)[/tex]
Since [tex]u+v\ge 2\sqrt{uv}=16[/tex], we have [tex]q\gt 0[/tex], so [tex]a_{3}=v-3q\lt v[/tex]. If [tex]v\le 32[/tex], then [tex]a_{3}\lt 32[/tex].
If [tex]v=64[/tex] and [tex]u=1[/tex], then [tex]3q^{2}-55q+3=0[/tex], and the smaller root [tex]q=\dfrac{55-7\sqrt{61}}{6}[/tex] gives
[tex]a_{3}=64-3q=\dfrac{73+7\sqrt{61}}{2}\gt 32[/tex]
This is the largest possible value.