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Question 1200515: Find the arithmetic sequence which has the sum of its n terms equal to 2n^2+3n
Thank you
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
Find the arithmetic sequence which has the sum of its n terms equal to 2n^2+3n
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To solve this problem, you need to know that
+----------------------------------------------------+
| the sum of any arithmetic progression is |
| a quadratic function of the index "n", |
+----------------------------------------------------+
and vice versa,
+----------------------------------------------------+
| any quadratic function of the integer index "n" |
| creates a unique arithmetic progression. |
+----------------------------------------------------+
Based on these facts, it is enough to find the first term of the progression
and the second term, and then calculate its common difference.
n= 1:  = 2*1^2+ 3*1 = 5.  = 5.
n= 2: = 2*2^2+ 3*2 = 14.  =  - = 14 - 5 = 9.
This arithmetic progression has the first term = 5 and the common difference d= - = 9 - 5 = 4.
The progression is 5, 9, 13, 17, . . .
Solved.
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On these properties of arithmetic progression, learn from the lessons
- Free fall and arithmetic progressions
- Uniformly accelerated motions and arithmetic progressions
- Increments of a quadratic function form an arithmetic progression
in this site.
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
Answer: 5, 9, 13, 17, 21, ...
Starting term = 5
Common difference = 4
The nth term formula is a(n) = 5+4*(n-1) which fully simplifies to a(n) = 4n+1
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:
| n |  |  (method 1) | (method 2) | | 1 | 5 | 5 | 2n^2+3n = 2*1^2+3*1 = 5 | | 2 | 9 | 5+9 = 14 | 2n^2+3n = 2*2^2+3*2 = 14 | | 3 | 13 | 5+9+13 = 27 | 2n^2+3n = 2*3^2+3*3 = 27 | | 4 | 17 | 5+9+13+17 = 44 | 2n^2+3n = 2*4^2+3*4 = 44 | | 5 | 21 | 5+9+13+17+21 = 65 | 2n^2+3n = 2*5^2+3*5 = 65 |
The two columns have their results match up, so this helps confirm the answer.
Another way to check:
 One formula to quickly find the sum of the first n terms of an arithmetic sequence
 Plug in the first term of 5
 Plug in the nth term formula of a(n) = 4n+1
 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  .
Plug in  and  . Simplifying should lead to
I'll let you do this verification pathway.
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