In the arithmetic sequence {an}, a1 + A6 = 12, A4 = 7, then the value of A9 is __

In the arithmetic sequence {an}, a1 + A6 = 12, A4 = 7, then the value of A9 is __

From a1 + A6 = 12, A4 = 7,
have to
2a1+5d=12
a1+3d=7 ,
The solution shows that A1 = 1, tolerance d = 2,
∴a9=a1+8d=1+2 × 8=17.
So the answer is: 17

In the sequence {an}, A1 = 1, an + 1 = 2An / 2 + an (n is a positive integer), try to guess the general term formula of this sequence 1 / an + 1 = 1 / an + 1 / 2 1 / an = 1 + 1 / 2 (n-1) an = 2 / N + 1 I found the original question 1 / an + 1 = 1 / an + 1 / 2 1 / an = 1 + 1 / 2 (n-1). How did the second question come from? Can anyone help me I mean, 1 / an = 1 + 1 / 2 (n-1). How did this come from?

a(n+1)=2an/(2+an)
Reciprocal on both sides
1/a(n+1)=1/an+1/2
1/a1=1
{1 / an} is an equal difference sequence with 1 as the first term and 1 / 2 as the tolerance
1/an=1+(1/2)*(n-1)
1/an=(n+1)/2
an=2/(n+1)

It is known that {an} is an equal difference sequence, and the tolerance D ≠ 0, and A1, A3 and A9 form an equal ratio sequence in turn. Find the value of a1 + a3 + A9 / A2 + A4 + A6

Method 1 (basic quantity method):
Let the tolerance of an be d
∴a3=a1+2d,a9=a1+8d
a2=a1+d,a4=a1+3d,a10=a1+9d
∴a1+a3+a9=3a1+10d,a2+a4+a10=3a1+13d
∵ A1, A3 and A9 are in equal proportion sequence
∴a3/a1=a9/a3
∴a1^2+4d^2+4a1d=a1^2+8a1d
∴a1=d
∴(a1+a3+a9)/(a2+a4+a10)=(3a1+10d)/(3a1+13d)=13d/16d=13/16
Method 2 (simple method):
A1, A3 and A9 form an equal ratio sequence
A3 * A3 = A1 * A9, i.e
(a1+2d)^2=a1*(a1+8d)
Bracket open simplified 4D * d = 4A1 * D
With A1 = D
(a1+a3+a9)/(a2+a4+a10)=(3a1+10d)/(3a1+13d)=13/16

Given that the general term formula of the sequence {an} is an = n / N ^ 2 + 156 (n ∈ n +), what is the maximum term of the sequence

one hundred and fifty-six

In the arithmetic sequence {an}, we know that a1 + a6.a4 is equal to 7 and find A9

A1 + A6 = A3 + A4, so A3 = 0, so the tolerance is 7, so A9 is 42

In the known sequence {an}, an = n N2 + 156, then the maximum term of the sequence {an} is the second______ Item

∵ an = nn2 + 156 = 1n + 156n ≤ 1439 ∵ 1n + 156n ≤ 1439, if and only if n = 239, take the equal, and then from n ∈ n +, so the maximum term of the sequence {an} may be the 12th or 13th term, and ∵ when n = 12, A12 = 12122 + 156 = 125 and ∵ when n = 13, A13 = 13132 + 156 = 125, so the 12th or 13th term is the maximum