If the third term of an arithmetic sequence is 11 and the seventh term is 19, what is the 15th term?

If the third term of an arithmetic sequence is 11 and the seventh term is 19, what is the 15th term?

a3=11=a1+2d
a7=19=a1+6d
D = 2, A1 = 7
a15=a1+14d=7+28=35

Finding A1 S10 with known arithmetic sequence A8 = 4 A6 = 8

a1=18 s10=90

It is known that three numbers form an arithmetic sequence, the product of the first and last terms is 5 times of the middle term, and the sum of the last two terms is 8 times of the first term
Please calculate it and write down the calculation process

Let the arithmetic sequence be a, a + Q, a + 2q
According to the known conditions, the following equations can be obtained
a(a+2q)=5(a+q) （1）
(a+q)+(a+2q)=8a （2）
According to (2), q = 2A
Substituting 1 gives a = 3 and substituting 2 gives q = 6
That is, the arithmetic sequence is: 3,9,15

If the sixth term of an arithmetic sequence is 5 and the sum of the third term and the eighth term is 5, then the sum of the first nine terms of the arithmetic sequence is 5______ ．

Let the arithmetic sequence be {an}, ∵ A6 = 5, A3 + A8 = 5. From the properties of the arithmetic sequence, we can see that the sum of the first nine terms of the arithmetic sequence is S9 = 9A5 = 0, so the answer is: 0

If the first term of the arithmetic sequence is 7 and the ninth term is 1, then its fifth term is 1

∵ item 5 is the median of the first and the ninth,
The fifth term is (7 + 1) / 2 = 4