# Maths help. Arithmetic hard question

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The first four terms of an arithmetic sequence are 2, a – b, 2a +b + 7 and a – 3b, where a and b are constants. Find a and b. [5 marks]

Explain the steps you take.

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The general strategy (for any similar question as this) is to recognize that in an arithmetic sequence, u_(n)-u_(n-1) = u_(n+1) - u_(n). Suppose you are given u_1, u_2, u_3, u_4, you can plug in the given values to get a system of two equations, and you can solve them. Eg a-b-2 = 2a + b+7-(a-b) and 2a + b+7-(a-b) = a-3b - (2a + b+7)

An alternative strategy(one that can used for these particular terms), whether you recognize it or that you derive from the general strategy,  is that if something, a-b, something else, a-3b is part of an arithmetic sequence, then that "something else" must be a-2b, and that "something" is a. So the sequence is equivalent to a, a-b, a-2b, a-3b. Equating the two forms, a= 2. And a-2b = 2a + b + 7. Simplify, 3b = -a - 7 = -2-7=-9. b = -3

Verify the answers, 2, (2 + 3= 5), (4 - 3+7=8), (2+9=11) is indeed an arithmetic sequence.

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