An arithmetic sequence satisfies u3 + u5 - u6 = 6 and u8 + u4 = 52. Find S = u2 + u4 + u6 + ... + u2020.

Step-by-step solution

1. 1

   Rewrite every condition in terms of $u_1$ and $d$. Using $u_k=u_1+(k-1)d$: $u_3+u_5-u_6=(u_1+2d)+(u_1+4d)-(u_1+5d)=u_1+d=6$ $u_8+u_4=(u_1+7d)+(u_1+3d)=2u_1+10d=52\ \Longrightarrow\ u_1+5d=26$

2. 2

   Solve the two-by-two system. Subtracting the first from the second gives $4d=20$, so $d=5,\qquad u_1=6-5=1$

3. 3

   Describe the even-indexed sub-sequence. $u_{2k}=1+(2k-1)\cdot 5=10k-4$, which is itself an arithmetic sequence with first term $u_2=6$ and common difference $10$.

4. 4

   Count its terms. The indices run $2,4,\dots,2020$, that is $k=1$ to $k=1010$, so there are $1010$ terms ending at $u_{2020}=10(1010)-4=10096$.

5. 5

   Apply the arithmetic-series sum. $S=\frac{1010}{2}\left(u_2+u_{2020}\right)=505\,(6+10096)=505\cdot 10102=5101510$

6. 6

   Confirm by direct summation. Adding all $1010$ terms in Python returns $5101510$ as well, so the answer is B.