Find number of terms in arithmetic progression

Question

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.

Answer 1

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 $n$.

Answer 2

Use the formula $$a_n=a_1+(n-1)d$$ where $$a_{n+1}-a_{n}=d$$

Answer 3

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

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

Answer 4

Use standard formula of A.P. which is

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

and a simple difference formula

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