Problems involving Progressions: Problems with Solutions

Solution:
We know that [tex]2b=1+c[/tex] and [tex]b^2=1 \times (c+1)=1+c=2b[/tex]. Therefore either [tex]b=0[/tex] or [tex]b=2[/tex]. But 0 cannot be a term of a geometric progression, so the only option left is [tex]b=2[/tex]. Then the arithmetic progression is [tex]1,2,c[/tex] or [tex]1+c=2\cdot 2[/tex] => [tex]c=3[/tex].

Solution:
The sum is [tex]a-1+b+2+c+13=(a+b+c)-1+2+13=341-1+15=355[/tex].

Solution:
Since [tex]a,1,b[/tex] is an arithmetic progression, [tex]a+b=2[/tex], therefore [tex]b=2-a[/tex]. By the geometric progression's property, we get [tex]a^2=b=2-a[/tex]
[tex]a^2+a-2=0[/tex]: [tex]a_1=1[/tex] or [tex]a_2=-2[/tex]. But the arithmetic progression is non-constant, therefore [tex]a \ne 1[/tex], so [tex]a[/tex] must be [tex]-2[/tex].
[tex]b=2-a=2-(-2)=4[/tex].

Solution:
By the property of the geometric progression, we get
[tex]a \times b=9 \times 243=3^2 \times 3^5=3^7[/tex]. We also know that [tex]\frac{a}{9}=\frac{b}{a}[/tex], or [tex]\frac{a^2}{9}=b[/tex] Substituting into the previous equation, we get [tex]\frac{a^3}{9}=3^7[/tex], or [tex]a^3=3^9[/tex], which means that [tex]a=3^3=27[/tex]. Then [tex]b=\frac{a^2}{9}=81[/tex] and their arithmetic mean is [tex]\frac{27+81}{2}=\frac{108}{2}=54[/tex].

Solution:
By the mean property of the arithmetic progression, we have [tex]3+a=2.9[/tex], or [tex]a=15[/tex]. The geometric progression becomes [tex]5,15,b[/tex]. By its mean property, [tex]b \times 5=15^2=225[/tex], or [tex]b=\frac{225}{5}=45[/tex].

Solution:
By the mean property of the geometric progression, we have [tex]2x=4^2[/tex], or [tex]x=8[/tex]. By the mean property of the arithmetic progression, we have [tex]3+y=2x=2 \times 8=16[/tex], so [tex]y=16-3=13[/tex].

Solution:
In an arithmetic progression [tex]a_{m}+a_{n}=a_{k}+a_{l}[/tex] whenever [tex]m+n=k+l[/tex]. Here the indices pair up with sum [tex]10[/tex]:
[tex]a_{1}+a_{9}=a_{2}+a_{8}=a_{3}+a_{7}=a_{4}+a_{6}=21[/tex]
[tex]4\times 21=84[/tex]

Solution:
Let the numbers be [tex]a, b, c[/tex]. From the arithmetic progression [tex]2b=a+c[/tex], so
[tex]3b=9\Rightarrow b=3, a+c=6[/tex]
From the geometric progression [tex]ac=b^{2}=9[/tex], so [tex]a[/tex] and [tex]c[/tex] are the roots of [tex]t^{2}-6t+9=0[/tex], i.e. [tex]a=c=3[/tex].

Solution:
From the arithmetic progression [tex]a+c=2\times 4=8[/tex].
From the geometric progression [tex]4^{2}=a(a+c)=8a\Rightarrow a=2[/tex].
Check: [tex]2, 4, 6[/tex] is an arithmetic and [tex]2, 4, 8[/tex] a geometric progression.

Solution:
Let the numbers be [tex]a, b, c[/tex] with [tex]a+b+c=35, b^{2}=ac[/tex]. The arithmetic progression gives
[tex]2b=(a-1)+(c-4)=a+c-5=30-b\Rightarrow b=10[/tex]
Then [tex]a+c=25, ac=100[/tex], so the numbers are [tex]5, 10, 20[/tex] (or [tex]20, 10, 5[/tex]), and the required sum is [tex]5+20=25[/tex].

Solution:
From [tex]1990[/tex] to [tex]2006[/tex] there are [tex]16[/tex] years.
Model A is an arithmetic progression: the population grows [tex]8[/tex] times by [tex]100[/tex] and becomes [tex]1000+8\times 100=1800[/tex].
Model B is a geometric progression with ratio [tex]0. 95[/tex]: the population becomes [tex]1000\times 0. 95^{16}\approx 440. 13[/tex].
[tex]1800-440. 13=1359. 87\approx 1360[/tex]

Solution:
Let the numbers be [tex]a, a+d, a+2d, a+3d[/tex]. If [tex]d=0[/tex], then [tex]q=1\ne d[/tex], so [tex]d\ne 0[/tex].
The geometric progression gives [tex](a+d)^{2}=a(a+3d)\iff d^{2}=ad\iff a=d[/tex].
[tex]q=\dfrac{a+d}{a}=2=d\Rightarrow a=d=2[/tex]
The numbers are [tex]2, 4, 6, 8[/tex].

Solution:
Let [tex]CD=a, BC=a+d, AD=a+2d, AB=a+3d[/tex]; since [tex]AB\ne CD[/tex], [tex]d\ne 0[/tex]. The perimeter gives [tex]4a+6d=18[/tex].
The height is [tex]BC[/tex] and the projection of [tex]AD[/tex] on [tex]AB[/tex] is [tex]AB-CD=3d[/tex], so
[tex](a+2d)^{2}=(a+d)^{2}+9d^{2}\iff 2ad=6d^{2}\iff a=3d[/tex]
[tex]18d=18\Rightarrow d=1, a=3[/tex]
The sides are [tex]CD=3, BC=4, AD=5, AB=6[/tex].

Solution:
The fourth term of the sequence is [tex]S_{4}-S_{3}=15-7=8[/tex], so [tex]b=8[/tex].
One root is [tex]0[/tex]; the other two are the roots of [tex]x^{2}+ax+8=0[/tex]. The roots are non-negative and distinct, so they are [tex]0, r, 2r[/tex] with [tex]r\gt 0[/tex]. By Vieta’s formulas
[tex]r\times 2r=8\Rightarrow r=2, a=-(r+2r)=-6[/tex]