SOLUTION: the last three terms of an arithmetic sequence with 18 terms are 67, 72,77 . Find the first term and the sum of the series

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Question 999847: the last three terms of an arithmetic sequence with 18 terms are 67, 72,77 . Find the first term and the sum of the series
Found 2 solutions by josgarithmetic, AnlytcPhil:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
A%2B%28n-1%29%2A5=77 when n=18, so substitution of this index will give an equation with only A as the unknown, and is solvable.

A%2B%2818-1%29%2A5=77
A=77-17%2A5
A=77-85
highlight%28A=-8%29

Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!
The other tutor forgot the sum.

The hard way is to notice that 77-72 = 5 and 72-67 = 5, and then list
them from right to left like this, subtracting 5 each time 

-8 + -3 + 2 + 7 + 12 + 17 + 22 + 27 + 32 + 37 + 42 + 47 + 52 + 57 + 62 + 67 + 72 + 77

And get -8 for the first term

And then add them up, getting 621.

The easy way is to use the formula

a%5Bn%5D=a%5B1%5D%2B%28n-1%29d

where n=18, a%5B18%5D=77, d=5

a%5B18%5D=a%5B1%5D%2B%2818-1%295

77=a%5B1%5D%2B%2818-1%295

77=a%5B1%5D%2B%2817%295

77=a%5B1%5D%2B85

-8=a%5B1%5D

and the sum formula:

S%5Bn%5D=expr%28n%2F2%29%28a%5B1%5D%2Ba%5Bn%5D%29

S%5B18%5D=expr%2818%2F2%29%28-8%2B77%29

S%5B18%5D=9%2869%29

S%5B18%5D=621

Edwin