If the tolerance of {an} is d = 1, A4 + A17 = 8, then A2 + A4 + A6 +... + A20 =? The online answer is like this a4+a17=8 So, 2a10 + D = 8 a10＝3．5 a11＝4．5 So, a2+a4+a6+...+a20 = 10 × a11 = 45

If the tolerance of {an} is d = 1, A4 + A17 = 8, then A2 + A4 + A6 +... + A20 =? The online answer is like this a4+a17=8 So, 2a10 + D = 8 a10＝3．5 a11＝4．5 So, a2+a4+a6+...+a20 = 10 × a11 = 45

a4+a17=a10+a11=a10+a10+d
∴a4+a17=2a10+1=8
a10=3.5
a11=a10+d=3.5+1=4.5
a2+a4+a6+...+a20
=(a2+a20)+(a4+a18)+.+(a10+a12)
=2A11 + 2A11 +. + 2A11 [5 2a10]
=10a11
=10×4.5
=45
I don't understand that step. Please continue to ask

In the arithmetic sequence {an}, the tolerance d = 1, A4 + A7 = 8, then the value of A2 + A4 + A6 +. + A20 is

The general term formula of arithmetic sequence an = a1 + (n-1) d;
a4+a7=a1+3d+a1+6d=2a1+9d=8;
Because d = 1, A1 = - 0.5
So A2 = 0.5
The sequence to be solved is an arithmetic sequence with a 2 = 0.5 as the first term and a tolerance of 2