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If the sequence {an} is an arithmetic sequence and the first term A1 > 0, a2005 + a2006 > 0, the maximum natural number n of a2005a20060 is
a2005a20060>a2006
Sn=(a1+an)*n/2>0
a1+an>0
a2005+a2006>0
So a1 + a (2006 + 2004) > 0
a1+a4010>0
So n max = 4010
On the number axis, there are 2004 integral points, A1, a2.a3,. A2004, which are arranged equidistant from left to right. The integers they represent are A1, A2, A3,. 2004, and A1, A2, A3,. A2004 are continuous integers. (1) calculate the distance from A1 to a2004; (2) if A15 = - 18, calculate A1 and a2004; (3) a2004 = 2005, calculate the value of a1 + A2 + a3 +... + a2004
(1) Find the distance from A1 to a2004;
2004－1＝2003
(2) If A15 = - 18, find A1 and a2004;
∵a1,a2,a3,… , a2004 is a continuous integer, A15 = - 18,
∴a1=a15-14= -32,a2004=a1+2003= -32+2003=1971；
(3) A2004 = 2005, find the value of a1 + A2 + a3 +... + a2004
A2004=2005,
Then a1 + A2 + a3 + +a2004
=2+3+4+… +2005
=(2+2005)×2004÷2
=2007×2004÷2
=2011014
a1,a2,a3…… A2004 is a continuous integer, a2004 = 2005, find A1, A2, A3 Value of a2004
a1,a2,a3…… A2004 is a continuous integer, a2004 = 2005
a2003=a2004-1=2005-1=2004=2003+1
a2002=a2003-1=2004-1=2003=2002+1
..
an=n+1
So:
a1=2
a2=3
a3=4
..
a2004=2005
If | A1-1 | + (A2-2) 2 + | A3-3 | + (a4-4) 4 +... + | a2003-2003 | + (a2004-2004) 2004 = 0, calculate A1, A2,
∵|a1-1|+(a2-2)2+|a3-3|+(a4-4)4+...+|a2003-2003|+(a2004-2004)2004=0
And every term on the left side of the equation is non negative
0 for each item
∴a1-1=0,a2-2=0
∴a1=1,a2=2,an=n
In the known arithmetic sequence an, A1 = - 23, A15 = 33, find the general term formula of the sequence and A2012
Solution
a1=-23
an=a1+[n-1]d
n=15
a15=-23+14d=33
14d=56
d=4
So the tolerance is 4
The leading formula of sequence is an = - 23 + 4 [n-1]
=-23+4n-4
=-27+4n
a2012=-27+4x2012
=-27+8048
=8021
Welcome to ask
In the arithmetic sequence {an}, A1 = 1, d = 0, then A2012=
a2012=a1+2011d=1
In the arithmetic sequence {an}, d = 3. A1 + a3 + A5 + A99 = 80, find the sum of the first 100 terms
S100=(a1+a3+a5+...+a99)+(a2+a4+a6+...+a100)
=(a1+a3+a5+...+a99)+(a1+d+a3+d+a5+d+.+a99+d)
=2(a1+a3+a5+...+a99)+50d
=2*80+50*3
=310
It is known that an is an arithmetic sequence, where a1 = 25, A4 = 16, find the value of a1 + a3 + A5... + A19
a4=a1+3d
d=(a4-a1)/3=-3
The tolerance of A1, A3, A5, -- - A19 is - 6
a1+a3+a5+…… +a19
=10a1+10*9*(-6)/2
=-20
In the arithmetic sequence an, a1 + A2 + a3 = 32, a11 + A12 + A13 = 118. What is A4 + A10
Because a1 + A2 + a3 = 32, a11 + A12 + A13 = 118;
So a1 + A2 + a3 + a11 + A12 + A13 = 150;
A1 + A1 = a1 + A1 = a1 + A1 = a1 + A1 = a1 + A1 = a1 + A1 = a1 + A1 = a1 + A1 = a1 + A1 = a1 + A1 = a1 + A1 + A1 = a1 + A1 = a1 + A1 + A1 = a1 + A1 + A1 = a1 + a1 + A1 = a1 + A1 + A1 = a1 + A1 +;
So A4 + A10 = 50, OK?
In the arithmetic sequence {an}, if a1 + A2 + a3 + A4 + A5 = 30, A6 + A7 + A8 + A9 + A10 = 80, then a11 + A12 + A13 + A14 + A15=
∵ in the arithmetic sequence {an}, a1 + A2 + a3 + A4 + A5 = 30, A6 + A7 + A8 + A9 + A10 = 80
∴a3=6,a8=16
∴a1=2,d=2
∴an=2n+2
∴a11=22,a12=24,a13=26,a14=28,a15=30
∴a11+a12+a13+a14+a15=22+24+26+28+30=130
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