Two arithmetic progressions, $$\(A\)$$ and $$\(B\)$$, each have 100 terms. Their terms are denoted by $$\(a_{1}, a_{2}, a_{3}, a_{4}, \ldots a_{100}\)$$ and $$\(b_{1}, b_{2}, b_{3}, b_{4}, \ldots b_{100}\)$$ respectively. It is given that $$\(a_{1}=b_{100}=1\)$$ and $$\(a_{100}=b_{1}=298\)$$. Find $$\(n\)$$ such that $$\(a_{n}-b_{n}=45\)$$.

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