Question 1200515
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Answer: <font color=red>5, 9, 13, 17, 21, ... </font>
Starting term = 5
Common difference = 4
The nth term formula is <font color=red>a(n) = 5+4*(n-1)</font> which fully simplifies to <font color=red>a(n) = 4n+1</font>



Explanation:


Sn = sum of the first n terms
Sn = 2n^2 + 3n
This was given to us in the instructions.


If n = 1, then Sn = S1 = the first term of the arithmetic sequence. 
Let's find that first term.
Sn = 2n^2 + 3n
S1 = 2*1^2 + 3*1
S1 = 2*1 + 3*1
S1 = 2 + 3
S1 = 5
The first term of the arithmetic sequence is 5.


Now let's find S2
Sn = 2n^2 + 3n
S2 = 2*2^2 + 3*2
S2 = 2*4 + 3*2
S2 = 8 + 6
S2 = 14
This represents the sum of the first two terms. We'll use it to determine what the second term would be:
S2 = term1+term2
14 = 5+term2
term2 = 14-5
term2 = 9


The arithmetic sequence so far is 5,9,...
This is sufficient info to determine the common difference
d = term2-term1
d = 9-5
d = 4


So we have:
a1 = first term = 5
d = common difference = 4
The nth term formula is:
a(n) = a1 + d*(n-1)
a(n) = 5 + 4*(n-1)
a(n) = 5 + 4n-4
a(n) = 4n + 1



Check:
<table border = "1" cellpadding = "5"><tr><td>n</td><td>{{{a[n]}}}</td><td>{{{S[n]}}} (method 1)</td><td>{{{S[n]}}} (method 2)</td></tr><tr><td>1</td><td>5</td><td>5</td><td>2n^2+3n = 2*1^2+3*1 = 5</td></tr><tr><td>2</td><td>9</td><td>5+9 = 14</td><td>2n^2+3n = 2*2^2+3*2 = 14</td></tr><tr><td>3</td><td>13</td><td>5+9+13 = 27</td><td>2n^2+3n = 2*3^2+3*3 = 27</td></tr><tr><td>4</td><td>17</td><td>5+9+13+17 = 44</td><td>2n^2+3n = 2*4^2+3*4 = 44</td></tr><tr><td>5</td><td>21</td><td>5+9+13+17+21 = 65</td><td>2n^2+3n = 2*5^2+3*5 = 65</td></tr></table>
The two {{{S[n]}}} columns have their results match up, so this helps confirm the answer.


Another way to check:
{{{S[n] = (n/2)*(a[1]+a[n])}}} One formula to quickly find the sum of the first n terms of an arithmetic sequence


{{{S[n] = (n/2)*(5+a[n])}}} Plug in the first term of 5


{{{S[n] = (n/2)*(5+4n+1)}}} Plug in the nth term formula of  a(n) = 4n+1


{{{S[n] = (n/2)*(4n+6)}}}


{{{S[n] = (n/2)*(4n)+(n/2)*(6)}}}


{{{S[n] = 2n^2+3n}}} Therefore, the answer is confirmed. 
Arguably this is better confirmation compared to the table of examples as shown above


Another possible arithmetic summation formula to use would be {{{S[n] = (n/2)*(2*a[1]+d*(n-1))}}}. 
Plug in {{{a[1] = 5}}} and {{{d = 4}}}. Simplifying should lead to {{{S[n] = 2n^2+3n}}}
I'll let you do this verification pathway.
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