If {an} is an arithmetic sequence and D is a tolerance, then the new sequence formed by the sum of K items in this sequence will still be an arithmetic sequence, and the tolerance is DK ^ 2 Explain why the tolerance equals DK ^ 2

If {an} is an arithmetic sequence and D is a tolerance, then the new sequence formed by the sum of K items in this sequence will still be an arithmetic sequence, and the tolerance is DK ^ 2 Explain why the tolerance equals DK ^ 2

The tolerance of the new sequence formed by the sum of K items in this sequence:
D=(S2k-Sk)-Sk
=S2k-2Sk
=2ka1+2k(2k-1)d/2-2*[ka1+k(k-1)d/2]
=dk²

In the arithmetic sequence {an}, it is known that the sum of the first term to the K term is 310, and the sum of the K + 1 term to the 2K term is 910

In fact, the process is too complete for you. You can't exercise your thinking
Let me talk about the train of thought
The difference between item 1 and item K + 1 is KD, the difference between item 2 and item K + 2 is KD, and the difference between item K and item 2K is KD
So, 910-310 = k, Kd = K & # 178; D, similarly, how much more is the sum of the last K items than the sum of the middle K items? You can think about it carefully. It's very interesting

Given that the sequence an satisfies A1 = 2K, an. A (n-1) = 2kA (n-1) - K & # 178;, it is proved that the sequence (1 divided by an-k) is an arithmetic sequence
The general term formula of sequence an is obtained

1 / (an-k) - 1 / [a (n-1) - k] = [a (n-1) - an] / [(an-k) (a (n-1) - K)] = [a (n-1) - an] / [an * a (n-1) - K (an + a (n-1)) + K & # 178;] = [a (n-1) - an] / [2kA (n-1) - K (an + a (n-1))] = [a (n-1) - an] / [k * (a (n-1) - an)] = 1 / K is a constant, so that n = 1 ｛ 1 / (a1-k) = 1 / K ｛