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If the sum of the first n terms of the arithmetic sequence {an} is Sn, A5 = 11, S12 = 186, then A8 = () A. 18B. 20C. 21D. 22
From the properties of sequence, we get that a1 + A12 = A5 + A8, because S12 = & nbsp; 122 × (a1 + A12) = 186, so a1 + A12 = A5 + A8 = 31, because A5 = 11, so A8 = 20, so we choose B
If A5 + A8 = 20, then S12 =?
Arithmetic sequence
a5+a8=a1+a12=20
Sn=n(a1+an)/2
s12=(a1+a12)x12/2=20x6=120
Let the sum of the first n terms of the arithmetic sequence {an} be Sn, if S3 = 9, S5 = 20, then A7 + A8 + A9 = ()
A. 63B. 45C. 27D. 36
According to the properties of arithmetic sequence, we can get S3 = 3a2 = 9, S5 = 5A3 = 20, A2 = 3, A3 = 4, d = 4-3 = 1, a8 = A2 + 6D = 3 + 6 = 9, a7 + A8 + A9 = 3A8 = 27, so we choose C
In the arithmetic sequence, Sn is the sum of the first n terms. If S4 = a9-3, S5 = A8 + A10 + 13, calculate the tolerance
From S4 = a9-3, S5 = A8 + A10 + 13 = 2a9 + 13, A5 = s5-s4 = A9 + 16, and A9 = A5 + 4D (where D is the tolerance), so A5 = A9 + 16 = A5 + 4D + 16, so d = - 4
Let the sum of the first n terms of the arithmetic sequence {an} be Sn, if S3 = 9, S6 = 36, then A7 + A8 + A9=______ ．
A4 + A5 + A6 = s6-s3 = 36-9 = 27, A4 + A5 + A6 = (a1 + 3D) + (A2 + 3D) + (A3 + 3D) = (a1 + A2 + a3) + 9D = S3 + 9D = 9 + 9D = 27, so d = 2, then A7 + A8 + A9 = (a1 + 6D) + (A2 + 6D) + (A3 + 6D) = S3 + 18D = 9 + 36 = 45
The sum of the first n terms of the arithmetic sequence {an} is Sn, and the tolerance is d
a8+a9+a10+a11=2a9+2a10=0
a9+a10=0
d0
a10
Let the sum of the first n terms of the arithmetic sequence {an} be Sn, if S3 = 9, S6 = 36, then A7 + A8 + A9 =?
For arithmetic sequence, S3, s6-s3 and s9-s6 are arithmetic sequence
That is 9,27, s9-36
So s9-36 = 27 × 2-9 = 45
S9=81
So A7 + A8 + A9 = s9-s6 = 45
Let the sum of the first n terms of the arithmetic sequence {an} be Sn, if S3 = 9, S6 = 36, then A7 + A8 + A9=______ ．
A4 + A5 + A6 = s6-s3 = 36-9 = 27, A4 + A5 + A6 = (a1 + 3D) + (A2 + 3D) + (A3 + 3D) = (a1 + A2 + a3) + 9D = S3 + 9D = 9 + 9D = 27, so d = 2, then A7 + A8 + A9 = (a1 + 6D) + (A2 + 6D) + (A3 + 6D) = S3 + 18D = 9 + 36 = 45
Let the sum of the first n terms of the arithmetic sequence {an} be Sn, if S3 = 9, S6 = 36, then A7 + A8 + A9=______ ．
A4 + A5 + A6 = s6-s3 = 36-9 = 27, A4 + A5 + A6 = (a1 + 3D) + (A2 + 3D) + (A3 + 3D) = (a1 + A2 + a3) + 9D = S3 + 9D = 9 + 9D = 27, so d = 2, then A7 + A8 + A9 = (a1 + 6D) + (A2 + 6D) + (A3 + 6D) = S3 + 18D = 9 + 36 = 45
Let the sum of the first n terms of the arithmetic sequence {an} be Sn, if S3 = 9, S6 = 36, then A7 + A8 + A9=______ ．
A4 + A5 + A6 = s6-s3 = 36-9 = 27, A4 + A5 + A6 = (a1 + 3D) + (A2 + 3D) + (A3 + 3D) = (a1 + A2 + a3) + 9D = S3 + 9D = 9 + 9D = 27, so d = 2, then A7 + A8 + A9 = (a1 + 6D) + (A2 + 6D) + (A3 + 6D) = S3 + 18D = 9 + 36 = 45
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