In the arithmetic sequence {an}, A3 + a7-a10 If in the arithmetic sequence {an}, A3 + a7-a10 = 8, a11-a4 = 4, then S13 =? Is there any simple method This is in addition to solving A1 and d by two equations, and then substituting them into the summation formula,

In the arithmetic sequence {an}, A3 + a7-a10 If in the arithmetic sequence {an}, A3 + a7-a10 = 8, a11-a4 = 4, then S13 =? Is there any simple method This is in addition to solving A1 and d by two equations, and then substituting them into the summation formula,

Using the properties of arithmetic sequence
a3+a7-a10=8 （1）
a11-a4=4 （2）
Because A3 + a11 = A7 + A7 = A10 + A4
（1）+（2）
a7=12
So S13 = (a1 + A13) * 13 / 2 = 2a7 * 13 / 2 = 13 * A7 = 156

The known arithmetic sequence an satisfies: A3 = 7, A5 + A7 = 26
Given BN = 1 / 4N ^ 2 + 4N, find the sum of the first n terms of BN. TN

1.
a3=a1+2d=7
a5+a7=2a1+10d=26 a1+5d=13
The equations are obtained
a1+2d=7
a1+5d=13
The solution is A1 = 3, d = 2
an=3+(n-1)*2=2n-1
two
Sn=(1/4)(1^2+2^2+...+n^2)+4(1+2+...+n)
=(1/4)n(n+1)(2n+1)/6+2n(n+1)
=n(n+1)(2n+49)/24

If {an} is an arithmetic sequence, and a7-2a4 = - 1, A3 = 0, then the tolerance d =? For a detailed explanation! I want to use this problem to understand this kind of problem, so please academic experts to solve it in detail, and list the steps and ideas, otherwise only for this problem, it will not be very useful for solving the problem in the future. Thank you

This problem only needs a general formula. An = am + (n-m) d
It is known that A3 = 0, A4 = A3 + (4-3) d = D;
a7=a3+(7-3)d=4d；
a7-2a4=-1 => 4d-2d=-1 => d=-1/2