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Problems involving Progressions
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Problems involving Progressions: Problems with Solutions
Problem 1
The numbers [tex]1,b,c[/tex] form an arithmetic progression and the numbers [tex]1,b,c+1[/tex] form a geometric progression. Find
c
.
Solution:
We know that [tex]2b=1+c[/tex] and [tex]b^2=1.(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].
Problem 2
The numbers [tex]a,b,c[/tex] form an arithmetic progression with sum [tex]a+b+c=341[/tex], while [tex]a-1, b+2, c+13[/tex] form a geometric progression. Find the sum of the terms of the geometric progression.
Solution:
The sum is [tex]a-1+b+2+c+13=(a+b+c)-1+2+13=341-1+15=355[/tex].
Problem 3
The sequence [tex]a,1,b[/tex] is a non-constant arithmetic progression. The sequence [tex]1,a,b[/tex] is a geometric progression. Find
b
.
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
a
must be
-2
.
[tex]b=2-a=2-(-2)=4[/tex].
Problem 4
Between the numbers
9
and
243
are the numbers
a
and
b
, such that [tex]9,a,b,243[/tex] is a geometric progression. Find the arithmetic mean of
a
and
b
.
Solution:
By the property of the geometric progression, we get
[tex]a.b=9.243=3^2.3^5=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].
Problem 5
The numbers [tex]3,9,a[/tex] form an arithmetic progression. The sequence [tex]5,a,b[/tex] is a geometric progression. Find
b
.
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.5=15^2=225[/tex], or [tex]b=\frac{225}{5}=45[/tex].
Problem 6
The numbers [tex]2,4,x[/tex] form a geometric progression and the sequence [tex]3,x,y[/tex] is an arithmetic progression. Determine the value of
y
.
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=2.8=16[/tex], so [tex]y=16-3=13[/tex].
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