Arithmetic Progressions: Difficult Problems with Solutions

Solution:
The formula for the nth member of an arithmetic progression is [tex]a_n=a_1+(n-1)d[/tex]. Substituting with n=8, we get [tex]a_8=1+(8-1)\cdot 4=1+7\cdot 4=29[/tex]

Solution:
[tex]a_3=a_1+2d[/tex], so [tex]d=\frac{a_3-a_1}{2}=\frac{15-5}{2}=5[/tex]

Solution:
[tex]a_{\frac{5+1}{2}}=\frac{a_5+a_1}{2}[/tex], or [tex]a_3=\frac{2+14}{2}=8[/tex]

Solution:
For any arithmetic progression, [tex]\frac{a_{m}+a_{n}}{2}=a_{\frac{m+n}{2}}[/tex]. For m=2 and n=4: [tex]\frac{a_2+a_4}{2}=a_3[/tex], so [tex]a_2+a_4=2.a_3[/tex]. Substituting into the given relation, we get [tex]3.a_3=54[/tex] or [tex]a_3=18[/tex]

Solution:
[tex]a_1+a_2+a_3=a_1+a_1+d+a_1+2d=3a_1+3d=102[/tex], so [tex]a_1+d=34[/tex]. [tex]d=34-a_1=34-15=19[/tex]. We know that [tex]a_{10}=a_1+9d=15+9.19=15+171=186[/tex]

Solution:
[tex](a_3+a_{23})+(a_8+a_{18})+(a_{10}+a_{16})=2a_{13}+2a_{13}+2a_{13}=6a_{13}[/tex], since [tex]3+23=8+18=10+16=26=13.2[/tex].
[tex]6a_{13}=126[/tex], so [tex]a_{13}=21[/tex].
[tex]S_{25}=\frac{a_1+a_{25}}{2}\cdot 25=25\cdot a_{13}=25\cdot 21=525[/tex]

Solution:
We know that [tex]S_5=\frac{2a_1+(5-1)d}{2}.5=\frac{2.9+4d}{2}.5=(9+2d).5=45+10d=15[/tex], therefore [tex]10d=-30[/tex] and [tex]d=-3[/tex]

Solution:
[tex]a_1^2+a_2^2=a_1^2+2a_1a_2+a_2^2-2a_1a_2=(a_1+a_2)^2-2a_1a_2=5^2-2a_1a_2=13[/tex], therefore [tex]a_1a_2=6[/tex].
On the other hand, [tex]13=a_1^2+a_2^2=a_1^2-2a_1a_2+a_2^2+2a_1a_2=(a_1-a_2)^2+12[/tex], therefore [tex](a_1-a_2)^2=1[/tex] and [tex]|a_1-a_2|=\sqrt{1}=1[/tex]

Solution:
Let the arithmetic progression is - a, a+d, a+2d...
6(a + 5d) = 9(a + 8d)
6a + 30d = 9a + 72d
6a - 9a = 72d - 30d
-3a = 42d
a = 42d/-3
a = -14d
T15 = a + 14d
-14d + 14d = 0

Solution:
The traveled distances for each hour form an arithmetic progression with [tex]a_1=15km[/tex] and [tex]d=-1 km[/tex]. We must note that we only allow values for n, for which [tex]a_n[/tex] is positive (since the biker is not going backwards), or [tex]n \le 15[/tex] (since [tex]a_{16}=0[/tex]
We seek n, for which the sum [tex]a_1+a_2+\dots + a_n=54km[/tex]. But [tex]a_1+a_2+\dots+a_n=S_n=\frac{2a_1+(n-1)d}{2}.n[/tex]. Substituting with [tex]a_1[/tex] and [tex]d[/tex], we get
[tex]\frac{2\cdot 15km+(n-1).(-1 km)}{2}.n=54km[/tex] / [tex]: km[/tex]
[tex]\frac{30-n+1}{2}.n=54[/tex]
[tex]31n-n^2=108[/tex]
[tex]n^2-31n+108=0[/tex]: [tex]D=31^2-4.108=529=23^2[/tex], we have two values for n: [tex]n_1=\frac{31+23}{2}=27[/tex] and [tex]n_2=\frac{31-23}{2}=4[/tex]. But we have shown that [tex]n \le 15[/tex], so [tex]n=4[/tex] remains the only solution.

Solution:
[tex]a_8-a_5=a_1+7d-a_1-4d=3d=6[/tex], so [tex]d=2[/tex]. From [tex]a_1+a_5=a_1+a_1+4d=2a_1+8=22[/tex] we get [tex]a_1=7[/tex]. [tex]S_3=\frac{2a_1+2.d}{2}.3=(a_1+d).3=(7+2).3=27[/tex]

Solution:
[tex]a_{10}=a_{5}+5\cdot d[/tex], or [tex]15=5+5\cdot d[/tex], therefore [tex]d=2[/tex].
[tex]a_{5}=a_1+4\cdot d[/tex] =>
[tex]5=a_1+4 \cdot 2[/tex], which means that [tex]a_1=-3[/tex].