Let the sum of the first n terms of the arithmetic sequence {an} be Sn, S4 = - 62, S6 = - 75 (1) find the sum of the general term an and the first n terms Sn (2) find the sum of the first n terms TN of the sequence {the absolute value of an}

Let the sum of the first n terms of the arithmetic sequence {an} be Sn, S4 = - 62, S6 = - 75 (1) find the sum of the general term an and the first n terms Sn (2) find the sum of the first n terms TN of the sequence {the absolute value of an}

(1) From Sn = Na1 + 1 / 2 * n * (n-1) * D
S4=-62
S6=-75
The solution is: D = 3
a1=-20
an=3n-24
Sn=-20+1/2*3n(n-1)
(2)
an≥0
n≥8
So TN=
When 1 ≥ n ≥ 8
Tn=Sn
When n > 8
Tn=-Sn+2S5

The sequence {an} is an arithmetic sequence, the sum of its first n terms is Sn, and S4 = - 62, S6 = - 75
(1) Find the general formula of {an} and the sum SN of the first n terms
(2) Find the value of a1 + A2 +... + A14

(1) S4 = 2 (a1 + A4), so a1 + A4 = - 31s6 = 3 (a1 + A6), so a1 + A6 = - 252 subtraction formula 2D = 6, so d = 3s4 = 4A1 + 6D = - 62, so A1 = - 20, so the general term an = a1 + (n-1) d = 3n-23sn = n (a1 + an) / 2 = n (3n-43) / 2 (2) A7 = - 2, a8 = 1, the first seven terms are negative, the last seven terms are positive, so the formula = - A1

In the arithmetic sequence an, A2 = 120, then A2 + A4 + A6 + A8=

Is there any mistake in your title
Is A5 = 120
∵ a2+a4=2a3, a6+a8=2a7
∴ a2 +a4+a6+a8=2（a3+a7）=4a5=480

If {an} is known to be an arithmetic sequence and A2 + A4 + A6 + A8 = 48, then S9 =?

Because 1 + 9 = 2 + 8 = 4 + 6
So in the arithmetic sequence
a1+a9=a2+a8=a4+a6
So a1 + A9 = 48 / 2 = 24
S9=(a1+a9)×9÷2=108

Given the function f (x) = 2 ^ x, the tolerance of arithmetic sequence {an} is 2. If f (A2 + A4 + A6 + A8 + A10) = 4,
Find the value of log2 [f (A1) f (A2) f () A3. F (A10)]

Log2 [f (A1) f (A2).. f (A10)] = log2 [f (A1) + log2f (A2) +.. log2f (A10) = log2 (2 ^ A1) + log2 * (2 ^ A2) +... Log2 (2 ^ A10) = a1 + A2 +... A10; the tolerance of arithmetic sequence {an} is 2; A2-2 = a1a4-2 = A3;.. a10-2 = A9; A2 + A4 + A6 + A8 + a10-10 = a1 + a3 + A5 + A7 + A9; a1 + A2 + a3 + a3

A3 + A8 > 0 S9 in arithmetic sequence an

Let {an} be an arithmetic sequence, and the sum of the first n terms is SN. If A6 = 2, S5 = 30, then S8 = ()
A. 31B. 32C. 33D. 34

Let the first term of the arithmetic sequence {an} be A1, and the tolerance be d. from A6 = 2, S5 = 30, we get a1 + 5D = 25a1 + 5 × 4D2 = 30, and the solution is A1 = 263d = − 43. S8 = 8A1 + 8 × 7d2 = 8 × 263 + 4 × 7 × (− 43) = 32

If A3 = 18-a6, then S8=______ ．

From the meaning of the question, we can get A3 + A6 = 18, from the nature of the arithmetic sequence, we can get a1 + A8 = 18, so S8 = 8 (a1 + A8) 2 = 4 × 18 = 72, so the answer is: 72

Let Sn be the sum of the first n terms of the arithmetic sequence {an}, S8 = 4A3, a7 = - 2, then A9 = ()
A. -6B. -4C. -2D. 2

∵ SN is the sum of the first n terms of the arithmetic sequence {an}, S8 = 4A3, a7 = - 2, i.e. 8A1 + 8 × 72d = 4 (a1 + 2D) a1 + 6D = - 2. The solution is & nbsp; A1 = 10, d = - 2, ∵ A9 = a1 + 8D = - 6, so a

Given the tolerance d > 0 of arithmetic sequence an, A4 + A6 = 10, A4 * A6 = 24
Let BN = 1 / (an * an + 1), the sum of the first n terms of BN is not TN. if TN > m is n * constant for any n, the maximum value of integer m is obtained

By solving the equation, a 4 = a 6 = 6D > 0 a 6 > a 4, the general term formula an = NbN = 1 / (an × a (n +)) = 1 / (n × (n + 1)) = ((n + 1) - n) / (n × (n + 1)) = 1 / n-1 / (n + 1) TN = B1 + B2 + B3 + +Bn=(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+…… +(1 / n-1 / (n + 1)) = 1-1 / (n + 1) in order to make TN > m hold, we can find t