Enterbot3 Posted October 22, 2018 Report Share Posted October 22, 2018 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. Reply Link to post Share on other sites More sharing options...
kw0573 Posted October 22, 2018 Report Share Posted October 22, 2018 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. Reply Link to post Share on other sites More sharing options...
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