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The fourth term of an arithmetic sequence is 141, and the seventh term is 132. The first term is _____.
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let the first term be x, and the with every term suppose that the seqeunce increases with r. so first term is x, second is x+r, third is x+2r, fourth is x+3r=141 and seventh is x+6r=132. solve for x.
you should find r to be -3 and x is 150
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