Question 1118070
Find the sum of the series 1+3.5+6+8.5+......+101.
<pre>
S<sub>n</sub> = (n/2)(a<sub>1</sub> + a<sub>n</sub>)

a<sub>1</sub> = 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

a<sub>n</sub> = a<sub>1</sub> + (n - 1)&#8729;d

101 = a<sub>1</sub> + (n - 1)&#8729;2.5

Solve that and get n = 41

S<sub>n</sub> = (n/2)(a<sub>1</sub> + a<sub>n</sub>)

S<sub>41</sub> = (41/2)(1 + a<sub>41</sub>)

S<sub>41</sub> = (41/2)(1 + 101)

S<sub>41</sub> = (41/2)(102)

S<sub>41</sub> = 2091

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</pre>
Find the sum of the first 23 term of the AP 4-3-10.....
<pre>
First find d:

d = -3 - 4 = -10 - (-3) = -7

Then use

S<sub>n</sub> = (n/2)[2&#8729;a<sub>1</sub> + (n-1)&#8729;d

with n = 23, a<sub>1</sub> = 4, d = -7 
</pre>
An arithmetic series have 1st term as 4 and common difference as 1/2 find the first 20 terms.
<pre>
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</pre>