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It is known that the sum of the first n terms of the arithmetic sequence {an} is Sn, A3 + A8 = 13, and S7 = 35. Then A7 = () A. 11B. 10C. 9D. 8
From the properties of arithmetic sequence, we can get: S7 = 7 (a1 + A7) 2 = 7 × 2a42 = 35, the solution is A4 = 5, and A3 + A8 = A4 + A7 = 13, so A7 = 8, so D is chosen
In the arithmetic sequence an, a8 > 0, A9
The difference is negative, and A8 > 0, A9
It is known that the sum of the first n terms of the arithmetic sequence {an} is SN. If 2A6 = A8 + 6, then S7 is ()
A. 49B. 42C. 35D. 24
Let the tolerance of the arithmetic sequence {an} be D, ∵ 2A6 = A8 + 6, ∵ 2 (a1 + 5d) = a1 + 7d + 6, which is transformed into a1 + 3D = 6, that is, A4 = 6. From the properties of the arithmetic sequence, we can get: A1 + A7 = 2a4. ∵ S7 = 7 (a1 + A7) 2 = 7a4 = 7 × 6 = 42
It is known that the sum of the first n terms of the arithmetic sequence {an} is SN. If 2A6 = A8 + 6, then S7 is ()
A. 49B. 42C. 35D. 24
Let the tolerance of the arithmetic sequence {an} be D, ∵ 2A6 = A8 + 6, ∵ 2 (a1 + 5d) = a1 + 7d + 6, which is transformed into a1 + 3D = 6, that is, A4 = 6. From the properties of the arithmetic sequence, we can get: A1 + A7 = 2a4. ∵ S7 = 7 (a1 + A7) 2 = 7a4 = 7 × 6 = 42
Let the sum of the first n terms of the arithmetic sequence {an} be SN. If S4 is greater than or equal to 12 and S7 is less than or equal to 28, then the maximum value of A7 is?
S4=4a1+4×3/2d≥12,∴2a1+3d≥6
S7=7a1+7×6/2d≤28,∴a1+3d≤4
A7 = a1 + 6D, let A7 = m (2A1 + 3D) + n (a1 + 3D) = (2m + n) a1 + (3m + 3n) d
∴1=2m+n 6=3m+3n
The solution is m = - 1, n = 3
∴a7=-(2a1+3d)+3(a1+4d)≤-6+12=6
The maximum value is 6
Given that SN is the sum of the first n terms of the arithmetic sequence {an}, and a1 + A7 = 8, find S7=
Because the sequence {an} is an arithmetic sequence, a1 + A7 = A2 + A6 = A3 + A5 = 8, and A4 is the median of the arithmetic of A1 and A7 = 4, so S7 = a1 + A2 + a3 + A4 + A5 + A6 + A7 = 8 × 3 + 4 = 28
Let Sn be the sum of the first n terms of the arithmetic sequence {an}, S6 = 36, Sn = 324, sn-6 = 144, then n = ()
A. 15B. 16C. 17D. 18
∵Sn=324，Sn-6=144，∴Sn-Sn-6=an-5+an-4+… +An = 180 and S6 = a1 + A2 + +A6 = 36, a1 + an = A2 + an-1 = A6 + an-5, 6 (a1 + an) = 36 + 180 = 216, a1 + an = 36, and D is selected from Sn = (a1 + an) N2 = 18N = 324, n = 18
The sum of the first n terms of the arithmetic sequence an is SN. We know that A5 = 11. A8 = 5
a8-a5=3d=5-11=-6
d=-2
a5=a1+(5-1)d=a1-8=11
a1=19
So an = a1 + (n-1) d = 21-2n
Sn=(a1+an)*n/2=(19+21-2n)*n/2=-n^2+20n
In the arithmetic sequence an, if A4 + A8 = 16, then SN is
From the topic, because the arithmetic sequence needs two independent conditions, there is only one condition that cannot be solved at present,
But according to symmetry and equality, S6 may be obtained;
a4+a8=16
(a6-2d)+(a6+2d)=16
2a6=16
a6=8
If the sum of the first n terms of the arithmetic sequence an is Sn and A4 + A8 = 0, then
A.S4
a4+a8=2a6=0
So A6 = 0
s6=s5+a6
So S6 = S5
Choose D
s5=s4+a5
A5 is not equal to 0, so B is not selected
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