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In a arithmetic progression sum of first four terms sum : $$a_1+a_2+a_3+a_4=124$$

and sum of last four terms : $$a_n+a_{n-1}+a_{n-2}+a_{n-3}=156$$ and sum of arithmetic progression is : $$S_n=350$$

$$n=?$$

How to find $n$? I tried using arithmetic progression sum formulas but getting negative or fractional numbers.

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One has $a_1+a_2+a_3+a_4+a_{n-3}+a_{n-2}+a_{n-1}+a_n=280$. The mean of those eight terms is $35$, so the mean of $a_1,\ldots,a_n$ is also $35$. The sum of those $n$ terms is $n$ times their mean, and is $350$. So now you can read off $n$.

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  • $\begingroup$ hohohoh man thank you a lot. God bless you $\endgroup$ – Serif Yaohim Dec 14 '18 at 16:25
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Use the formula $$a_n=a_1+(n-1)d$$ where $$a_{n+1}-a_{n}=d$$

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$$350=\dfrac{n(a_1+a_n)}2$$

Now $a_1+a_n=a_2+a_{n-1}=\cdots=\dfrac{124+156}4$

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Use standard formula of A.P. which is

$a_n = a + (n-1)d$

and a simple difference formula

$a_n - a_{n-1} = d$

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