The tolerance of arithmetic sequence an is 1, a1 + A2 + a3 + +A99 = 99, then A3 + A6 + A9 +A96 + A99 =? The answer is 66, RT

The tolerance of arithmetic sequence an is 1, a1 + A2 + a3 + +A99 = 99, then A3 + A6 + A9 +A96 + A99 =? The answer is 66, RT

Because a1 + A2 + a3 + +a99=99
So (a1 + A4 +... + A97) + (A2 + A5 +... + A98) + (A3 + A6 +... + A99)
=(a3-2d+a6-2d+...+a99-2d)+(a3-d+a6-d+...+a99-d)+(a3+a6+...+a99)
=3(a3+a6+...+a99)-2d*33-d*33
=3(a3+a6+...+a99)-99
=99
So A3 + A6 +... + A99 = (99 + 99) / 3 = 66

Given that the tolerance of the arithmetic sequence {an} is 1, and a1 + A2 +... + A98 + A99 = 99, then A3 + A6 + A9 +... + a96
A99 =? A1 + A2 +... + A97 + A98 + A99 = 99 launch a1 + A2 +... + A97 + A98 + A99 + 33d + 66d = 99 + 99 33d + 66d = 99 + 99 (how to come out)?

It's like this
structure
a1+a4+a7.a97=x
a2+a5+a8.a98=y
a3+a6+a9.+a96+a99=Z
and
y-x
=(a2-a1)+(a5-a4)..+(a98-a97)
=d+d+d+...d
=33d
So y = x + 33, the same as Z = y + 33
Z + z-33 + z-66 = 99
That is, 3Z = 99 + 99
z=66
As for the answer given by a1 + A2 +... + A97 + A98 + A99 + 33d + 66d = 99 + 99, the
It's 33d + 66d = 99, which is correct

It is known that A1 = 1 / 3 and the arithmetic mean of the first n terms is equal to 2N-1 times of the nth term. Find the first 5 terms and prove that an = 1 / (2n-1) (2n + 1) holds by mathematical induction

(a1+a2+a3+…… An) / N = (2n-1) * an, n respectively take 1 to 5 to get 5 terms, mathematical induction, the front is omitted, let's say the back Sn = (a1 + A2 + a3 +...) an)=n*(2n-1)*an 1*S(n+1)=(a1+a2+a3+…… A (n + 1)) = (n + 1) * (2n + 1) * a (n + 1) 2 * 2 * - 1 * a (n + 1) = (n + 1) * (2n + 1) * a (n + 1) - n * (...)