2022 AMC 10A Problem 20

Let the arithmetic sequence be $$a, a + d, a + 2d, a + 3d$$ and the geometric sequence be $$b, br, br^2, br^3.$$ Then $$a + b = 57 \tag*{(1)},$$ $$a + d + br = 60 \tag*{(2)},$$ and $$a + 2d + br^2 = 91 \tag*{(3)}.$$ Subtracting $(1)$ from $(2)$ and $(2)$ from $(3),$ we get $$d + b(r - 1) = 3$$ and $$d + br(r - 1) = 31.$$ Subtracting these, we get $$b(r - 1)^2 = 28.$$ Let $t=b(r-1)=br-b,$ which is an integer. Then $t^2=28b,$ so $t=14u$ and $b=7u^2$ for some nonzero integer $u.$ Because $a=57-b>0,$ we must have $u\in\{-2,-1,1,2\}.$ Now $r=1+\frac{t}{b}=1+\frac{2}{u}.$ The cases $u=-1$ and $u=-2$ give $r=-1$ and $r=0,$ respectively, so they cannot produce a positive geometric sequence. If $u=2,$ then $b=28, r=2, a=29,$ and $d=3-t=-25,$ making the third arithmetic term negative. Therefore $u=1,$ so $b = 7, r = 3, a = 50,$ and $d = -11.$ The arithmetic sequence is $$50,39,28,17,$$ and the geometric sequence is $$7,21,63,189.$$ The desired answer is $17 + 189 = 206.$ Thus, E is the correct answer.