Given that the general term formula of sequence {an} is an = 2n-37, then the value of n is?

Given that the general term formula of sequence {an} is an = 2n-37, then the value of n is?

Sn=a1+a2+a3+...+an
Sn=2(1+2+3+...+n)-37n
Sn=2*(1+n)*n%2-37n
Sn=n*n-36n
So when n = - (- 36)% 2, Sn is the smallest
That is, when n = 18, Sn is the minimum

If the general term formula of sequence is an = 2n-37, then the minimum value of Sn is n = what is Sn = at this time

∵an=2n-37
∴a1=-35
∴sn=(a1+an)*n/2
=(-35+2n-37)*n/2
=(n-36)n
=(n-18)∧2-18²
When n = 18, Sn is the minimum
sn(min)=18²=324

Given that the sequence {an} satisfies the n-th power of A1 = 1, an + 1 * an = 2, then s2012 =?

Odd items and even items are equally proportional. The first item of odd items is 1, and the common ratio is 2. There are 1006 items in total. S = 2 ^ 1006-1, the first item of even items is 2, and the common ratio is 2. There are 1006 items in total (and are exactly twice the odd items), so s = 3x2 ^ 1006-3

In the arithmetic sequence (an), A1 > 0, S16 > 0, S17

Let the tolerance be d 0, 17a1 + 136d0, so a1 + 7d > 0 is A8 > 0
From the second formula, we can get a1 + 8D

Let Sn be the sum of the first n terms of the arithmetic sequence {an}, and S16 ＞ 0, S17 = 0. If the largest mean value of Sn is SK, then the value of K is ()
A. 8b. 9C. 8 or 9D. 7 or 8

From the summation formula and properties of arithmetic sequence, we can get: S16 = 16 (a1 + a16) 2 = 16 (A8 + A9) 2 ＞ 0, S17 = 17 (a1 + A17) 2 = 17 × 2a92 = 0 ﹥ A8 + A9 ＞ 0, and A9 = 0, the sum of the first 8 terms of ﹥ sequence, or the sum of the first 9 terms, is the largest, so we choose C

It is known that the sum of the first n terms of the arithmetic sequence {an} is Sn, and A5 = 15, S15 > 0, S16

An = A0 + nd15 = A0 + 5da0 = 15-5dsn = n (A0 + an) / 2 = n (A0 + A0 + nd) / 2 = n (2a0 + nd) / 2 = DN ^ 2 / 2 + na0n = - A0 / D = (5d-15) / D = 5-15 / D when n = 5-15 / D, Sn has the maximum value. D = - 1, - 3, - 5, - 15N = 20, 10, 8, 6a0 = 20, 30, 40, 90s15 = 187.5, 112.5, 37.5, - 337.5s16 = 19

If A1 ＜ 0, S15 ＞ 0, S16 ＜ 0, what is the maximum value of Sn
That's what he wrote

S15=(a1+a15)*15/2>0 a1+a15>0 2a1+14d>0 d>-2a1/14
S16=(a1+a16)*16/2

Let Sn be the sum of the first n terms of the arithmetic sequence {an}, A12 = - 8, S9 = - 9, and prove that the sequence {Sn / N} is the arithmetic sequence

a12=a1+11d=-8 s9=（a1+a9)*9/2=9a1+36d=-9
So A1 = 3, d = - 1
sn/n=a1+(n-1)d/2=3+（1-n）/2 =7/2-n*1/2
So it's an arithmetic sequence

Let Sn be the sum of the preceding terms of the arithmetic sequence {an}, A12 = - 8, S9 = - 9, then S16 =?

S9= -9.
9*(a1+a9)/2=-9
a1+a9=-2
2a5=a1+a9=-2
a5=-1
And A12 = - 8
a1+a16=a5+a12=1-8=-9
S16=16*(a1+a16)/2=8*(-9)=-72

Let Sn be the sum of the first n terms of the arithmetic sequence {an}, A12 = - 8, S9 = - 9, then S16 = ()
A. -72B. 72C. 36D. -36

S9 = 12 (a1 + A9) × 9 = - 9, and a1 + A9 = 2a5, A5 = - 1. From the properties of arithmetic sequence, a1 + a16 = A5 + A12, so S16 = 16 (a1 + a16) 2 = 16 (A5 + A12) 2 = - 72, so choose a