If the sum of the first six terms of the arithmetic sequence {an} is 8 and the sum of the first 12 terms is 20, then the sum of the first 18 terms of the arithmetic sequence {an} is 8

If the sum of the first six terms of the arithmetic sequence {an} is 8 and the sum of the first 12 terms is 20, then the sum of the first 18 terms of the arithmetic sequence {an} is 8

Let the first term A1, tolerance D, because A6 = a1 + 5D, A12 = a1 + 11d, so [A1 + (a1 + 5d)] × 6 △ 2 = 8 [A1 + (a1 + 11d)] × 12 △ 2 = 20, sort out 2A1 + 5D = 8 / 32a1 + 11d = 10 / 3, solve the equations, get A1 = 19 / 18; d = 1 / 9, so A18 = a1 + (18-1) × 1 / 9 = 19 / 18 + 17 × 1 / 9 = 53 / 18, so the sum of the first 18 terms = (a1 + a

If the sum of the first six terms of an arithmetic sequence is 10 and the first 12 terms is 30, then the sum of the first 18 terms of an arithmetic sequence is 10

60
S6 s12-s6 s18-s12 are also equal difference series (fill in the blanks)
S18-S12=30
S18=60
The big problem is still to be solved

There are 8 numbers in the arithmetic sequence. The first and last numbers are 24 and 25 respectively
The more detailed the answer, the better. Thank you!

Let the arithmetic sequence be an
Then A1 = 24, a8 = 25
So d = (a8-a1) / 7 = 1 / 7
So A2 = 169 / 7 A3 = 170 / 7 A4 = 171 / 7 A5 = 172 / 7 A6 = 173 / 7 A7 = 174 / 7

Finding the rule 1 5 9 13 17 how to deduce is a 4n-3 important process

The rule of the problem is: if the difference between two adjacent numbers is 4, then the tree of 4n-3 is 4n-4 + 4, that is 4N