Question 233191
4 + 8 + 12 + 16 + 20 + ...... + 100
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This is an arithmetic series because the difference between
any term (after the first) and the term that precedes it is
4.  So the common difference is given by {{{d = 4}}}.

The formulas for an arithmetic series are

1.  For the nth term:

{{{a[n] = a[1]+(n-1)d}}}

where {{{a[1]}}} is the first term

2.  For the sum of the terms through the nth term:

{{{S[n]=(n(a[1]+a[n]))/2}}}

In your problem, {{{a[1]=4}}}, {{{a[n]=100}}} and {{{d=4}}}, so

we use the first formula to find n:

{{{a[n] = a[1]+(n-1)*d}}}
{{{100 = 4+(n-1)*4}}}
{{{100 = 4+4(n-1)}}}
{{{100 = 4+4n-4}}}
{{{100 = 4n}}}
{{{25=n}}}

So {{{a[n]=a[25]=100}}}.

Then we substitute in the second formula:

{{{S[n]=(n(a[1]+a[n]))/2}}}
{{{S[25]=(25(4+a[25]))/2}}}
{{{S[25]=(25(4+100))/2}}}
{{{S[25]=(25(104))/2}}}
{{{S[25]=(25(cross(104)^52))/cross(2)}}}
{{{S[25]=25*52}}}
{{{S[25]=1300}}}

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Here is another way to do it without using formulas:

S = 4 + 8 + 12 + 16 + 20 + ...... + 100 

S = 4(1 + 2 + 3 + 4 + 5 + ...... + 25) 

so there are 25 terms:

Write the sum, S, forward, then backwards

S =   4 +  8 + 12 + 16 + 20 + ...... + 100
S = 100 + 96 + 92 + 88 + 84 + ...... +   4

Add the two equations term by terms:

 S =   4 +   8 +  12 +  16 +  20 + ...... + 100
 S = 100 +  96 +  92 +  88 +  84 + ...... +   4
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2S = 104 + 104 + 104 + 104 + 104 + ...... + 104

Since there are 25 terms, there are 25 104's above, so

2S = 25*104
2S = 2600
 S = 1300

Your teacher probably wants you to do it the first way, though.

Edwin</pre>