algebra-precalculusarithmetic-progressions
Find an arithmetic progression for which $$\begin{cases}S_n-a_1=48\\S_n-a_n=36\\S_n-a_1-a_2-a_{n-1}-a_n=21\end{cases}$$
I have tried to use the formula $$S_n=\dfrac{a_1+a_n}{2}.n$$ but it seems useless at the end. From the first equation $$a_1=S_n-48$$ and I tried to put it into the third, but it isn't very helpful. Thank you!
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 $n$.
Sure: Look at this figure and see if you can reason it out:
Best Answer
If we put the first equation into the third, we have
$$48-a_2-a_{n-1}-a_n=21$$
$$a_2+a_{n-1}+a_n = 27$$
$$3a_1+d+(n-2)d+(n-1)d=27$$
$$3a_1+(2n-2)d=27$$
$$3a_1+2(n-1)d=27\tag{A}$$
If we subtract the first two equations, we have
$$a_n-a_1=12$$
$$(n-1)d=12\tag{B}$$
From $(A)$ and $(B)$, we can solve for $a_1=1$.
Now substituting $a_1$ into the first equation, we can solve for $S_n=49$ and from the second equation$a_n = S_n-36=13$.
$$S_n = \frac{n}{2}(1+a_n)=49$$
$$n=\frac{98}{14}=7.$$
I will leave finding the common difference to you.