Let the sum of the first n terms of the equal ratio sequence {an} be Sn, S4 = 1, S8 = 17, and find the general term formula an

Let the sum of the first n terms of the equal ratio sequence {an} be Sn, S4 = 1, S8 = 17, and find the general term formula an

Let the common ratio of {an} be q, Q ≠ 1 be known from S4 = 1, S8 = 17, then A1 (Q4 − 1) Q − 1 = 1, ① A1 (Q8 − 1) Q − 1 = 17, ② Q8 − 1q4 − 1 = 17 and Q4 = 16 be obtained from Formula 1 and formula 2, so q = 2 or q = - 2 substitute q = 2 into Formula 1 to get A1 = 115, and a = 2n − 115 substitute q = - 2 into Formula 1 to get A1 = − 15, and an

The problem of finding the number of terms (n) in the equal ratio sequence
For example, the common ratio q = 21, the first term is 1, and the number of terms is 40
If we know s (n) = x, how to find the number of terms?
If the reverse is troublesome, then if we don't reverse, we know s (n) = x, how to find the number of terms?

You can use the value to see. Use X-1 to remove 21 until the final answer is 22. Add 2 times to the number of 21, which is the number of terms. The two times of addition are the one of subtracting 1, and the other is the one of 22

Insert n numbers between 14 and 78 to form an equal ratio sequence. If the sum of all items is 778, then the number of items ()
A. 4B. 5C. 6D. 7

Let the number of terms of this sequence be n + 2 and the common ratio be q, then 78 = 14 × QN + 1, so QN + 1 = 116, Sn + 2 = A1 (1 − QN + 2) 1 − q = 14 (1 − Q16) 1 − q = 778, q = - 12, QN + 1 = 116 = (- 12) 4, so n + 1 = 4, n = 3, N + 2 = 5, so B

Insert n numbers between 14 and 78 to form an equal ratio sequence. If the sum of all items is 778, then the number of items ()
A. 4B. 5C. 6D. 7

Let the number of terms of this sequence be n + 2 and the common ratio be q, then 78 = 14 × QN + 1, so QN + 1 = 116, Sn + 2 = A1 (1 − QN + 2) 1 − q = 14 (1 − Q16) 1 − q = 778, q = - 12, QN + 1 = 116 = (- 12) 4, so n + 1 = 4, n = 3, N + 2 = 5, so B

The first term of an equal ratio sequence is 1, and the number of terms is even
The first term of an equal ratio sequence is 1, the number of terms is even, the sum of odd terms is 85, and the sum of even terms is 170

The number of items is 2n, the common ratio is Q, the odd common ratio is Q ^ 2, and S1 = (1-Q ^ 2n) / (1-Q ^ 2) = 85, the even common ratio is Q ^ 2, and S2 = Q * (1-Q ^ 2n) / (1-Q ^ 2) = 170, S2 / S1 = q = 2, so the common ratio is q = 2
If it is carried in, S1 = (1-2 ^ 2n) / (1-2 ^ 2) = 85, 2 ^ 2n = 256, n = 4, number of items 2n = 8

An even number of items in an equal ratio sequence, the sum of all items is even

When the sequence is a natural number sequence, the proposition holds, otherwise it does not hold

Given an even number of terms, the first term is 1

There can be many, as long as an even number of items are given, and each item has the same common ratio
For example: 1 39 27 81 243
1 4 16 64

If we know that the first term of an equal ratio sequence is 1, the number of terms is even, the sum of odd terms is 85, and the sum of even terms is 170, then the number of terms of this sequence is ()
A. 2B. 4C. 8D. 16

Let the common ratio be Q. from the meaning of the question, we can get a1 + a3 + +an-1=85，a2+a4+… +an=170，a1q+a2q+… +an-1q=170，∴（a1+a3+… +An-1) q = 170, q = 2, an = 2N-1, Sn = A1 (1 − QN) 1 − q = A1 − anq1 − Q, (Q ≠ 1) 170 + 85 = 2N-1, n = 8

If the sum of even items is twice the sum of odd items, the first item is 1, and the sum of the two items in the middle is 24, then the number of items in the sequence is ()
A. 12B. 10C. 8D. 6

According to the meaning of the title, an = qn-1a2 + A4 + A6 + ··· + a2na1 + a3 + ··· + A2N − 1 = 2  q = 2, so an = 2N-1, we know that an + an + 1 = 24 ﹣ 2N-1 + 2n = 24 ﹣ n-1 = 3 ﹣ n = 4, 2n = 8, so the sequence has 8 terms

If the nth term of the equal ratio sequence with the first term of 3 is 48, then the 2N-1 term is 48

an=3q*n-1=48
a2n-1=3q*2（n-1）=3*16*16=768