Find the sum of the series 1+3.5+6+8.5+......+101.
Sn = (n/2)(a1 + an)
a1 = 1
But we need first to determine n.
To do that we need d:
d = 3.5-1 = 6-3.5 = 8.5-6 = 2.5
an = a1 + (n - 1)∙d
101 = a1 + (n - 1)∙2.5
Solve that and get n = 41
Sn = (n/2)(a1 + an)
S41 = (41/2)(1 + a41)
S41 = (41/2)(1 + 101)
S41 = (41/2)(102)
S41 = 2091
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Find the sum of the first 23 term of the AP 4-3-10.....
First find d:
d = -3 - 4 = -10 - (-3) = -7
Then use
Sn = (n/2)[2∙a1 + (n-1)∙d
with n = 23, a1 = 4, d = -7
An arithmetic series have 1st term as 4 and common difference as 1/2 find the first 20 terms.
Write the first term as 4 and keep adding 1/2 over and over until
you have written 20 terms. To get you started,
the 2nd term is 4 + 1/2 = 8/2 + 1/2 = 9/2,
the 3rd term is 9/2 + 1/2 = 10/2 = 5,
etc., etc.,
Edwin
