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Problems involving Progressions
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Problems involving Progressions: Difficult Problems with Solutions
Problem 1
We are given the positive numbers [tex]a,b,c[/tex], which form an arithmetic progresion in the given order. We know that [tex]a+b+c=9[/tex]. The numbers [tex]a+1, b+1, c+3[/tex] form a geometric progression in the given order. Find c.
Solution:
Since [tex]a,b,c[/tex] form an arithmetic progression, [tex]c=b+d=a+2d[/tex] and [tex]a+b+c=a+a+d+a+2d=3a+3d=9[/tex], so [tex]a+d=b=3[/tex] and [tex]a+c=6 => c=6-a[/tex]. Since [tex]a+1,b+1,c+3[/tex] form a geometric progression, we have [tex](b+1)^2=(a+1)(c+3)[/tex], or [tex]16=(a+1)(6-a+3)=(a+1)(9-a)[/tex]
[tex]16=9a+9-a^2-a[/tex]
[tex]a^2-8a+7=0[/tex], therefore [tex]a=7[/tex] or [tex]a=1[/tex]. But if [tex]a=7[/tex], [tex]c=6-a=6-7=-1<0[/tex] and by definition c is positive, so [tex]a=7[/tex] is not a solution. Therefore [tex]a=1[/tex] and [tex]c=6-a=6-1=5[/tex].
Problem 2
Between the number 3 and the number [tex]b>3[/tex] lies the number [tex]a[/tex], such that [tex]3,a,b[/tex] form an arithmetic progression. The numbers [tex]3,a-6,b[/tex] form a geometric progression. Find
a
.
Solution:
From the properties of arithmetic and geometric progressions we have [tex]2a=3+b[/tex] and [tex](a-6)^2=3b[/tex]. We substitute [tex]b=2a-3[/tex] into the last equation to get
[tex]a^2-12a+36=6a-9[/tex]
[tex]a^2-18a+45=0[/tex]
The solutions to this quadratic equation are [tex]a=3[/tex] and [tex]a=15[/tex]. But if [tex]a=3[/tex], the arithmetic progression becomes [tex]3,3,b[/tex], which means that [tex]b=3[/tex], which contradicts with the fact that [tex]b>3[/tex]. Therefore the only solution remains [tex]a=15[/tex].
Problem 3
An arithmetic progression [tex]\{a_n\}[/tex]has 9 elements. [tex]a_1=1[/tex] and [tex]S_a=369[/tex] (the sum of all elements is 369). A geometric progression [tex]\{b_n\}[/tex] also has 9 elements. [tex]a_1=b_1[/tex] and [tex]a_9=b_9[/tex]. Determine the value of [tex]b_7[/tex].
Solution:
[tex]\{a_n\}[/tex] has 9 members and by the formula for the sum of an arithmetic progression, we have [tex]S_a=\frac{a_1+a_9}{2}.9=369[/tex], therefore [tex]a_1+a_9=82[/tex], or [tex]a_9=81[/tex]. It means that [tex]b_9=81[/tex]. Since [tex]b_9=b_1.q^8[/tex], or [tex]81=3^4=q^8[/tex], we have [tex]q^2=3[/tex].
[tex]b_7=b_1.q^6=1.(q^2)^3=3^3=27[/tex]
Problem 4
The numbers [tex]a,b,c[/tex] are all different and form an arithmetic progression in said order. The numbers [tex]b,a,c[/tex] form a geometric progression. Find the common ratio r of the geometric progression (assume that [tex]|r| > 1[/tex]).
Solution:
We know that [tex]b=\frac{a+c}{2}[/tex] and [tex]a^2=bc=\frac{a+c}{2}.c[/tex]
[tex]2a^2=c(a+c)=ac+c^2[/tex]. We divide the equation by [tex]a^2[/tex]:
[tex]2=\frac{c}{a}+(\frac{c}{a})^2[/tex], or
[tex](\frac{c}{a})^2-\frac{c}{a}-2=0[/tex]. But [tex]\frac{c}{a}=r[/tex] (two consecutive members of the geometric progression). So
[tex]r^2-r-2=0[/tex] with roots [tex]r_1=-1[/tex] and [tex]r_2=-2[/tex]. But [tex]r=-1[/tex] implies [tex]b=r^2c=1\cdot c=c[/tex], which contradicts that [tex]b \ne c[/tex].
Therefore [tex]r=-2[/tex].
Problem 5
The positive numbers [tex]a,b,c[/tex] form an arithmetic progression, and [tex]a+b+c=21[/tex]. If the numbers [tex]a+2, b+3, c+9[/tex] form a geometric progression, find
c
.
Solution:
Let us denote [tex]b=a+d, c=a+2d[/tex]. Since [tex]a+b+c=a+a+d+a+2d=3a+3d=21[/tex], we can deduce [tex]b=a+d=7[/tex]. The geometric progression becomes [tex]a+2, 10, c+9[/tex] and the arithmetic becomes [tex]a,7,c[/tex]. Which means that [tex]a+c=14[/tex] and [tex](a+2)(c+9)=100[/tex]. Substituting [tex]a[/tex] for [tex]14-c[/tex], we get
[tex](16-c)(9+c)=100[/tex]
[tex]144-9c+16c-c^2=100[/tex]
[tex]c^2-7c+44=0[/tex], which yields [tex]c_1=-4[/tex] and [tex]c_2=11[/tex], but [tex]c>0[/tex] and the only answer left is [tex]c=11[/tex].
Problem 6
The numbers [tex]a,b,c,64[/tex] form a geometric progression. [tex]a,b,c[/tex] are also respectively the first, fourth and eighth members of a non-constant arithmetic progression. Determine the value of [tex]a+b-c[/tex].
Solution:
We can denote [tex]b=a+3d[/tex], [tex]c=a+7d[/tex], where [tex]d[/tex] is the difference of the arithmetic progression the numbers are in. By the mean property of the geometric progression, we have
[tex](a+3d)^2=a(a+7d)[/tex]
[tex]a^2+6ad+9d^2=a^2=7ad[/tex]
[tex]ad=9d^2[/tex]. Since the arithmetic progression is non-constant, [tex]d \ne 0[/tex]. We divide by it to reach
[tex]a=9d[/tex]. Then [tex]b=9d+3d=12d[/tex] and [tex]c=9d+7d=16d[/tex]. But since [tex]a,b,c,64[/tex] is a geometric progression,
[tex]\frac{64}{c}=\frac{b}{a}[/tex]
[tex]\frac{64}{16d}=\frac{12d}{9d}=\frac{4}{3}[/tex]
[tex]d=\frac{64.3}{16.4}=3[/tex], respectively [tex]a=27[/tex], [tex]b=36[/tex] and [tex]c=48[/tex] and the answer is [tex]a+b-c=36+27-48=15[/tex]
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