Let an be an arithmetic sequence with positive items, odd items, nonzero tolerance, and the sum of items is 2004, then A2 =?

Let an be an arithmetic sequence with positive items, odd items, nonzero tolerance, and the sum of items is 2004, then A2 =?

The number of items can be calculated by subtracting the first item from the non item / tolerance plus 1

Arithmetic sequence A1, A2 If A2 + ak-1 = 18, then the number of terms is K=

Yi Zhi: a [1] + a [k] = a [2] + a [k-1] = 18
It is known that: (a [1] + a [k]) K / 2 = 81
So k = 9

Does 0 exist

It should be analyzed according to the combination of number and shape
But it's not like the way I wrote upstairs to solve it, because there is no limit on the order of arithmetic sequence, so it's not necessarily SiNx < cosx < TaNx < Cotx
In fact, the most possible solutions are SiNx ＜ cosx ＜ TaNx ＜ Cotx (0 ＜ x ＜ π / 4) and cosx ＜ SiNx ＜ Cotx ＜ TaNx (π / 4 ＜ x ＜ π / 2). However, in this range, there is no solution of arithmetic sequence
Take the former case as an example
If the arithmetic sequence is satisfied, then SiNx cosx = TaNx Cotx
sinx-cosx=sin²x-cos²x/sinxcosx
Because SiNx and cosx are obviously not equal, because if they are equal, it means that TaNx is also equal to Cotx and is equal to SiNx
SiNx + cosx = sinxcosx
Root sign 2Sin (x + π / 4) = 1 / 2sin2x, (0 < x < π / 4)
If you draw a graph, you can see that y = root sign 2Sin (x + π / 4) and y = 1 / 2sin2x have no intersection at 0 ＜ x ＜ π / 4, so there is no solution. Similarly, we analyze the case of π / 4 ＜ x ＜ π / 2
Limited space, said these. LZ own good experience