If A1 = 1, then S4 = () A. 7B. 8C. 15D. 16

If A1 = 1, then S4 = () A. 7B. 8C. 15D. 16

∵ 4A1, 2A2, A3 are arithmetic sequences ∵ 4A1 + A32 = & nbsp; 2A2, ∵ 4A1 + A1 · q22 = 2a1q, that is, 4 + q22 = 2 & nbsp; Q ∵ q = 2 ∵ S4 = A1 (1 − Q4) 1 − q = 1 × (1 − 24) 1 − 2 = 15, so select C

If it is known that the 4th, 7th and 16th items of an arithmetic sequence with nonzero tolerance are the 4th, 6th and 8th items of an arithmetic sequence, then the common ratio of the arithmetic sequence is ()
A. 3B. 2C. ±3D. ±2

Because the tolerance D of the arithmetic sequence {an} is not equal to 0, the 4th, 7th and 16th items of the arithmetic sequence are in equal proportion sequence, that is, A72 = A4 · a16, that is, (a1 + 6D) 2 = (a1 + 3D) (a1 + 15d) {A1 = - 32D, so A4 = a1 + 3D = 32D, a7 = a1 + 6D = 92d, so Q2 = a7a4 = 92d32d = 3

1. In known sequence an, A1 = 8, and (2 * an + 1) + an = 6, the sum of the first n terms is Sn, then the minimum positive integer n satisfying the inequality sn-2n-4 whose absolute value is less than 1 / 2008 is?
A.12 B.13 C.15 D.16
2. Sequence 1,1 + 2,1 + 2 + 4 ,1+2+2^2+…… +2 ^ n-1, and Sn is greater than 1020?
A.7 B.8 C.9 D.10
*I accidentally hit it. It's actually 2 times an.

I can't understand the first question,
Question 2 a2-a1 = 2
a3-a2=2^2
a4-a3=2^3
.
an-a(n-1)=2^n-1
So the final result is an = 2 ^ n-2 + 1 = 2 ^ n-1
Then the sum of the first n terms is 2 ^ (n + 1) - (n + 2)
Ah, I will only bring it in and count it as 10
Choose D
Then the first question can be obtained from the meaning of the question
（2a(n+1)）+an=6
Then 2 (a (n + 1) - 2) = - (An-2)
So the sequence (An-2) is an equal ratio sequence with the common ratio of - 0.5 and the first term of 6
an-2=6*(-0.5)^(n-1)
an=.+2
So Sn = 2n + 4 - (- 0.5) ^ (n-2)
So the absolute value of sn-2n-4 = - (- 0.5) ^ (n-2)
It's only a big budget, but it's not too much trouble
That is, 0.5 ^ (n-2) is less than 1 / 2008
So that's 13
I think this is the way
What's your teacher's method