Arithmetic Progressions: Problems with Solutions

Solution:
The common difference is -5.

Solution:
The numbers do not form an arithmetic progression.
If they formed they would be 2, 6, 10, 14, 18.

Solution:
[tex]d=3. 5-1=2. 5[/tex]
[tex]a_{10}=a_{1}+9d=1+9\times 2. 5=23. 5[/tex]

Solution:
As we know, the natural numbers form an arithmetic progression with [tex]a_1=1[/tex] and [tex]d=1[/tex]. Therefore [tex]S_{10}=\frac{2a_1+(10-1) \times d}{2} \times 10=\frac{2+9}{2} \times 10=11\cdot 5=55[/tex]

Solution:
5 consecutive numbers form an arithmetic progression with difference 1.
n = 5,
S(5) = 100,
d = 1
Let the first number be [tex]a[/tex]
It is unknown.

[tex]S(n) = \frac{n}{2}(2a+d(n-1))[/tex]

[tex]100 = \frac{5}{2} (2a + 1\cdot 4)[/tex]
[tex]100 \cdot \frac{2}{5} = 2a +4[/tex]
[tex]40 = 2a + 4[/tex]
[tex]2a = 36[/tex]
[tex]a = 18[/tex]
The first number is 18, and the other numbers are 18, 19, 20, 21, 22.

Solution:
Since [tex]{a_n}[/tex] is an arithmetic progression, the difference between any two consecutive members is the same and it is [tex]d=a_2-a_1=7-4=3[/tex]. The formula for the nth member is [tex]a_n=a_1+(n-1)\cdot d[/tex], substituting n=11 we get [tex]a_{11}=4+(11-1) \cdot 3=34[/tex]

Solution:
[tex]a_n=a_1+(n-1)\cdot d[/tex] => [tex]a_1=a_n-(n-1)\cdot d[/tex]. We substitute [tex]n=3[/tex] and get [tex]a_1=43-2\cdot 12=43-24=19[/tex].

Solution:
[tex]a_{19}=a_1+(19-1)\cdot d=a_1+18\cdot d[/tex]. Since [tex]a_n[/tex] is an arithmetic progression, [tex]d=a_{n+1}-a_n=a_2-a_1=8-15=-7[/tex]. We substitute the value of [tex]d[/tex] and [tex]a_1[/tex] and we have [tex]a_{19}=15-18\cdot 7=-111[/tex]

Solution:
[tex]a_n=a_1+(n-1) \times d=1+(n-1) \times 1=n[/tex], so [tex]a_{1083}=1083[/tex]

Solution:
[tex]d=a_2-a_1=5-(-11)=16[/tex]

Solution:
[tex]a_5=a_1+4d[/tex], [tex]a_2=a_1+d[/tex]. By subtracting the latter from the first, we get [tex]a_5-a_2=a_1+4d-a_1-d=3d[/tex], so [tex]3d=18-9=9[/tex], which means that [tex]d=3[/tex]

Solution:
[tex]a_{\frac{3+11}{2}}=\frac{a_3+a_{11}}{2}[/tex], or [tex]a_7=\frac{13+25}{2}=\frac{38}{2}=19[/tex]

Solution:
There is a direct formula for the sum of the first n elements - it is [tex]S_n=\frac{2a_1+(n-1)d}{2} \times n[/tex]. For [tex]n=10[/tex], we have [tex]S_{10}=\frac{2\cdot 15+9\cdot 3}{2}\cdot 10=\frac{30+27}{2}\cdot 10=57\cdot 5=285[/tex]

Solution:
The common ratio is -2.

Solution:
The sum we are searching for is [tex]a_2+a_4+a_6+a_8+a_10+a_{12}=[/tex]
[tex]=(a_1+d)+(a_1+3d)+(a_1+5d)+(a_1+7d)+(a_1+9d)+(a_1+11d)=[/tex]
[tex]=6\cdot a_1+(1+3+5+7+9+11)d=6a_1+36d=[/tex]
[tex]=6\cdot 7+36\cdot 4=42+144=186[/tex]

Solution:
Every term is the difference of two consecutive sums: [tex]a_{n}=S_{n}-S_{n-1}[/tex].
[tex]S_{12}=3\times 144-24=408[/tex]
[tex]S_{11}=3\times 121-22=341[/tex]
[tex]a_{12}=408-341=67[/tex]

Solution:
The angles are [tex]30^{\circ}, 30^{\circ}+d, 30^{\circ}+2d, 30^{\circ}+3d[/tex] and their sum is [tex]360^{circ}[/tex]:
[tex]120^{\circ}+6d=360^{\circ}\Rightarrow d=40^{\circ}[/tex]
[tex]30^{\circ}+3\times 40^{\circ}=150^{\circ}[/tex]

Solution:
Each inserted number is the mean of its neighbours: [tex]2a_{k}=a_{k-1}+a_{k+1}[/tex]. So the nine numbers form an arithmetic progression with [tex]a_{1}=16[/tex] and [tex]a_{9}=4[/tex].
[tex]S_{9}=\dfrac{9(16+4)}{2}=90[/tex]

Solution:
The numbers are [tex]a, a+2, \ldots, a+14[/tex], an arithmetic progression with [tex]d=2[/tex].
The mean of an arithmetic progression equals the mean of its first and last terms: [tex]\dfrac{a+(a+14)}{2}=a+7=19\Rightarrow a=12[/tex].
The first three numbers are [tex]12, 14, 16[/tex] and their mean is [tex]14[/tex].

Solution:
The [tex]k[/tex]-th addend is [tex]k\times 4k[/tex], where [tex]k=1, 2, \ldots, 20[/tex].
After the change it becomes [tex]k(4k-1)=k\times 4k-k[/tex].
So the sum decreases by [tex]1+2+\ldots+20=\dfrac{20\times 21}{2}=210[/tex].

Solution:
[tex]S_{9}=\dfrac{9(2a_{1}+8d)}{2}=9(3+4d)=27+36d[/tex]
[tex]135\lt 27+36d\lt 207\iff 108\lt 36d\lt 180\iff 3\lt d\lt 5[/tex]
All terms are integers, so [tex]d=a_{2}-a_{1}[/tex] is an integer too. Hence [tex]d=4[/tex].

Solution:
Let the numbers be [tex]a, a+1, \ldots, a+9[/tex]; their sum is [tex]10a+45[/tex].
Let the erased number be [tex]a+k, 0\le k\le 9[/tex]. Then
[tex]10a+45-(a+k)=2009\Rightarrow 9a=1964+k[/tex]
The number [tex]1964[/tex] leaves remainder [tex]2[/tex] when divided by [tex]9[/tex], so [tex]k=7[/tex] and [tex]a=219[/tex]. The erased number is [tex]219+7=226[/tex].