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I'm looking for the sum of a 12-term sequence where the last term is 13 and common difference of -10. How do I find the first term so I can solve? What I have so far is ___= 12/2 (___+(11)-10).
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\[a+11*(-10)=13\] is the equation,solve for a.
a + 11(-10) = 13 a = 123 (12/2)(123+(11)-10) = 744
But the answer can only be one of the following: --> 605 --> 660 --> 748 --> 816
equation for sum has 2*a instead of a
u will get 816
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Wait, if a=123 2a= 246 The sum would come out too high.
(12/2)(2*123+(11)-10) =6(246-110)=816
last term(L)= a + (n-1)*d....
I think I may have entered it into my calculator wrong.. Thanks.
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