In the arithmetic sequence an, a1 + A2 +. + A50 = 200, A51 + a52 +... + A100 = 2700, then the value of A1 is?

In the arithmetic sequence an, a1 + A2 +. + A50 = 200, A51 + a52 +... + A100 = 2700, then the value of A1 is?

Analysis: according to the conditions given by the two equations subtraction, get the tolerance of the sequence, and then according to the sum of the first 50 items is 200, substitute into the summation formula to make the first item, the subject gives such conditions, can solve a series of problems of arithmetic sequence
∵a1+a2+… +a50=200 ①
a51+a52+… +a100=2700 ②
② The results are as follows: 50 × 50D = 2500,
∴d=1,
∵a1+a2+… +a50=200,
∴na1+ n(n+1)d/2=200,
∴50a1+25×51=200,
∴a1=-20.5,

In the arithmetic sequence {an}, if a1 + A2 +. + A50 = 200, A51 + a52 +. + A100 = 2700, then A1 is
In the arithmetic sequence, a1 + A2 + a3. + A50 = 200, A51 + a52 +. A100 = 2700, then A1 is

a1+ a2+ a3.+a50=200,a51+a52+.a100=2700,
The two formulas subtract
（a51-a1)+(a52-a2)+.(a100-a50)=2700-200=2500
50*50d=2500
d=1
a1+ a2+ a3.+a50=50*a1+(50-1)*50/2d=200
a1=-20.5

In known arithmetic sequence {an}, a1 + A2 +. + A50 = 200, A51 + a52 +... + A100 = 2700, find A1 =?

a1+a2+.+a50=200,a51+a52+...+a100=2700=a1+50d+a2+50d+.a50+50d=2700(a1+.a50)+50×50d=27002500d=2500d=1a1+a2+.a50=200（a1+a50）×50/2=200a1+a1+49d=82a1=-41a1=-41/2

In the arithmetic sequence {an}, a1 + A2 + +a50=200，a51+a52+… +If A100 = 2700, then A1 equals ()
A. -1221B. -21.5C. -20.5D. -20

∵a1+a2+… +a50=200   ①a51+a52+… +A100 = 2700 & nbsp; & nbsp; & nbsp; ② - ①: 50 × 50D = 2500, ∵ d = 1, ∵ a1 + A2 + +A 50 = 200, Na 1 + 12n (n-1) d = 200, a 1 + 25 × 49 = 200, a 1 = - 20.5, so C