# Find number of terms in arithmetic progression

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.

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 $$n$$.

• hohohoh man thank you a lot. God bless you – Serif Yaohim Dec 14 '18 at 16:25

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

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

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

Use standard formula of A.P. which is

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

and a simple difference formula

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