Judge whether 9 is an item in the sequence [3 (2n-7)]. If so, please indicate which item it is

Judge whether 9 is an item in the sequence [3 (2n-7)]. If so, please indicate which item it is

Let 3 (2n-7) = 9
Then 2n-7 = 3
We get n = 5
N is an integer, so 9 is the fifth item in [3 (2n - 7)]
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Judge whether 9 is an item in the sequence {3 (2n-7)}. If so, please indicate the item

Yes, when n = 5, 3 (2n - 7) = 9, is the fifth term

Given the sequence 2, 3, 4, 6, 6, 8, 9, 12 and so on, find the sum of the first 101 items and judge which item is 2010?

Odd terms 2,4,6,8
Even terms 3,6,9
The first 101 terms are the sum of 51 odd and 50 even terms
The odd term (2 + 102) × 51 / 2 = 2652
Even term (3 + 150) × 50 / 2 = 3825
The sum of the first 101 items is 6477
2010 is 2009 in odd and 670 in even

In the sequence {an}, A1 = 2, the sum of the first n terms is SN. If the sequence {Sn / N} is an arithmetic sequence with tolerance of 2, then A3?

∵ sequence {Sn / N} is an arithmetic sequence with tolerance of 2, S1 / 1 = 2 ∵ S2 / 2 = 4, S3 / 3 = 6 ∵ S2 = 8, S3 = 18 ∵ A3 = s3-s2 = 10
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Given the tolerance d = 6 in the arithmetic sequence {an}, and an = 22, the first n terms and Sn = 28, find A1 and n

Because n = a1 + (n-1) d is 22 = a1 + (n-1) * 6. (1)
Because Sn = n * (a1 + an) / 2, that is 28 = n * (a1 + 22) / 2. (2)
The simultaneous solution (1) (2) is n = 7, A1 = - 14

Note that the sum of the first n terms of the arithmetic sequence {an} is Sn, if A1 = 12, S4 = 20, then S6=______ ．

The sum of the first n terms of ∵ arithmetic sequence {an} is Sn, A1 = 12, S4 = 20, ∵ A4 + A1 = 10, ∵ A4 = 192, ∵ d = 3, ∵ S6 = 6 × 12 + 6 × 52 × 3 = 48, so the answer is: 48

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

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

If S6 = 36, Sn = 324, sn-6 = 144 (n ＞ 6), then n is equal to ()

S(n-6)-S6
=a7+a8+.+a(n-6)
=[a7+a(n-6)]*(n-12)/2
=[a1+a(n)]*(n-12)/2=108 (1)
Sn=a1+a2+.+a(n)
=[a1+a(n)]*n/2=324 (2)
(1)÷(2)
(n-12)/n=108/324=1/3
n=3n-36
n=18

Let Sn be the sum of the first n terms of the arithmetic sequence {an}, given S6 = 36, Sn = 324, s (n-6) = 144, (n > 6), find the value of n

Sum of the first n terms of arithmetic sequence Sn = Na1 + n * (n-1) * D / 2
When n = 6
S6=6a1 +6*5*d/2
S6=6a1 +15d
36=6a1 +15d
a1=6-(5/2)d
Sn=na1 +n*(n-1)*d/2=324
Substituting A1 into
6n-5nd/2 +n*(n-1)*d/2=324
6n + n[n-6]d/2=324
d/2 = (324-6n)/[n(n-6)],
S(n-6)=[(n-6)a1 +(n-6)*(n-7)*d/2]=144
(6-5d/2)(n-6)+n(n-6)d/2 - 7(n-6)d/2=144
Substitute D / 2 obtained above into
Simplification
We get n = 18