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$$C_1, C_2, C_3, \ldots$$ are in A.P. $$C_1 = 8$$, $$C_2 = 10$$, $$C_3 = 12$$. If $$b_1 = 6$$ and $$b_n$$ be determined by $$b_n = b_{n-1} + C_n$$ then
$$C_n = 2n + 6$$, so $$C_2, \ldots, C_7 = 10, 12, 14, 16, 18, 20$$. Then $$b_2 = 16$$, $$b_3 = 28$$, $$b_4 = 42$$, $$b_5 = 58$$, $$b_6 = 76$$ and $$b_7 = 96$$. All four options a, b, c and d are correct.
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