Question 324238
<pre><b>
If it's an arithmetic sequence like

3, 7, 11, 15, ...

Then the first term, 3, is called {{{a[1]}}}, the second term, 7, is called 
{{{a[2]}}}, etc.  The common difference is the amount that's added to each
number to get the next one. To get from 3 to 7 you add 4. To get from 7 to 11
you add 4, to get from 11 to 15 you add 4.  So 4 is the common difference d

The get the equation for the nth term, you use the formula

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

{{{a[n]=3+(n-1)*4}}}

{{{a[n]=3+4(n-1)}}}

{{{a[n]=3+4n-4}}}

{{{a[n]=4n-1}}}


and to get the sum of the first n terms {{{S[n]}}}:

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

{{{S[n]=(n/2)(3+(4n-1))}}}

{{{S[n]=(n/2)(2+4n)}}}

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

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



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f it's a geometric sequence like

4, 12, 36, 108, ...

Then the first term, 3, is called {{{a[1]}}}, the second term, 12, is called 
{{{a[2]}}}, etc.  The common ratio is the amount that each number is
multiplied by to get the next one. To get from 4 to 12 you multiply by 3. To 
get from 12 to 36 you multiply by 3, to get from 36 to 108 you multiply by 3.  
So 3 is the common ratio r.

The get the equation for the nth term, you use the formula

{{{a[n]=a[1]*r^(n-1)}}}

{{{a[n]=4*3^(n-1)}}}

and to get the sum of the first n terms {{{S[n]}}}:

{{{S[n]=(a[1](r^n-1))/(r-1)}}}

{{{S[n]=(4(3^n-1))/(3-1)}}}

{{{S[n]=(4(3^n-1))/2}}}

{{{S[n]=2(3^n-1)}}}

Edwin</pre>