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:
d = 3.5 - 1 = 6 - 3.5 = 2.5
n = 10
a is the first term
10th term = a +(n-1)d = 1 + (10-1)2.5 = 1 + 9 × 2.5 = 1 + 22.5 = 23.5

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).d}{2}.10=\frac{2+9}{2}.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.

$S(n) = \frac{n}{2}(2a+d(n-1))$

$100 = \frac{5}{2} (2a + 1\cdot 4)$
$100 \cdot \frac{2}{5} = 2a +4$
$40 = 2a + 4$
$2a = 36$
$a = 18$
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 n=3 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 d 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).d=1+(n-1).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}.n[/tex]. For n=10, 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]