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The sum of the first $n$ terms of an arithmetic sequence is given by $3n+2n^2$. find the common difference of the sequence.

So the answer is 4, but I've got 2 - my workings are shown - is my method wrong or have i just made a mistake.

$$d=U_n-U_{n-1}$$ $(1)$ $$U_1+U_2+...+U_{n-1}+U_n=3n+2n^2$$ $(2)$ $$U_1+U_2+...+U_{n-1}=3(n-1)+2(n-1)^2$$ $(1)-(2)$ = $U_n=2n-2$

This would give me a sequence of 0,2,4 which has a common difference of 2 not 4 Any help would be appreciated.

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Getting to your calculation $2n^2 + 3n - 2(n-1)^2 - 3(n-1) = 2n^2 + 3n -2n^2 + 2*2n - 2 - 3n +3$ which gives $U_n=4n+1$. So it looks like you made an arithmetic error.

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Your subtraction is incorrect; the difference is $2(2n-1)+3=4n+1$, since $n^2-(n-1)^2=2n-1$.

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I assume that you are aware of the formula for the terms of an Arithmetic Progression.
If not, then the formula is given as, $$S_n = \dfrac{n}{2} \cdot \Bigl(2a + \left(n-1\right)d \Bigl)$$ The derivation is simple and you can see it on the internet or try deriving yourself, it is not that hard. Now, using this formula we have two equations that you can get by multiplying everything, namely, $$\dfrac{2an -nd}{2} = 3n$$ and $$\dfrac{n^2d}{2} = 2n^2$$ Clearly you can see that the answer for the common difference is $4$.
Edit: As pointed out in the other answers the formula you should get for the $n^{th}$ term $a_n= 4n +1$.

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You can also just use specific examples. The sum of the first term alone is $3\cdot 1+2\cdot1^2=5$. So the first term is $5$. The sum of the first two terms is $3\cdot 2+2\cdot2^2=14$, so the second term is $14-5=9$. The sum of the first three terms is $3\cdot 3+2\cdot3^2=27$ so the third term is $27-14=13$. The terms run $5,9,13\dots$ so the common difference is $4$.

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The above answers point out the mistake in your calculation, which is what I found too. Also, there's a simpler method to approach this question, if you're interested:

Given that-

$S_n$ = $3n + 2n^2 $ ...(1)

Let '$t_n$' denote the nth term of the sequence.

We could say that

$S_1$ = $t_1$ ;

So by substituting n=1 in (1), we get-

$S_1$ = 5

=> $t_1$ = 5

Also,

$S_2$ = $t_1 + t_2$ ...(2)

Again, by substituting n=2 in (1), we get-

$S_2$ = 14

Finally, substituting the values of $t_1$ and $S_2$ in (2), we get

$t_2$ = 9

Hence, the common difference of the given sequence is given by:

$d = t_2 - t_1 $

$ => d = 4$

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This is the formula to find the nth term of an AP

$$a_n=S_n + S_{n+1}$$

So you can get

$$a_1 = 5$$ $$a_2 = 9$$ $$a_3 = 13$$

Thus the common difference is 4

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  • $\begingroup$ Your answer adds nothing new to this one. And you should not use the word “above”, since you have no way of knowing by which order the answers will appear. $\endgroup$ – José Carlos Santos Dec 20 '18 at 14:13

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