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ADDMATHS : PROGRESSION?
The sum of the first n terms of an arithmetic progression is given by Sn=n/2(5-3n)
Find
a) the sum of first 6 terms
b)the common difference
the first question was easy but im stuck at b)
please give me the full working
thanx in advance
1 Answer
- BrianLv 79 years agoFavorite Answer
a) S(6) = (6/2)*(5 - 3*6) = 3*(-13) = -39.
b) In general S(n) = (n/2)*(2*a(1) + (n - 1)*d).
So (n/2)*(2*a(1) + (n - 1)*d) = (n/2)*(5 - 3n) ----> 2*a(1) + (n - 1)*d = 5 - 3n.
Now S(1) = a(1), so S(1) = (1/2)*(5 - 3*1) = (1/2)*2 = 1, and thus a(1) = 1.
Thus 2*1 + (n - 1)*d = 5 - 3n ---> 2 + nd - d = 5 - 3n ---> 3 + d = n*(3 + d).
So for this to be true for all n, we must have d = -3.
So the first 6 terms are 1, -2, -5, -8, -11, -14, which do add to -39.
