The question has changed over time, so we restate it, changing the notation a little. Then we generalize.
Problem: Let $(a_1,a_2,a_3,a_4)$ be a $4$-term arithmetic sequence such that
(i) $a_1^2+a_2^2+a_3^2+a_4^2=176$ and (ii) $a_1+a_2+a_3+a_4=24$.
Find the $a_i$.
The problem originally arose in a probabilistic context. We use probabilistic language in the solution.
Let $X$ be the random variable which takes on the value $a_i$ with probability $1/4$ ($i=1,2,3,4$).
Then (i) says that $E(X^2)=44$, and (ii) says that $E(X)=6$. So $X$ has variance $E(X^2)-(E(X))^2$, which is $8$.
Let $Y$ be a random variable which is uniformly distributed on the set $\{0,1,2,3\}$. Let $a=a_1$, and $d=a_2-a_1=a_3-a_2=a_4-a_3$. Then $X=a+ d\,Y.$ Note that $E(Y)=(0+1+2+3)/4=\frac{6}{4}$ and $E(Y^2)=(0^2+1^2+2^2+3^2)/4=14/4$. It follows that $Y$ has variance $(14/4-(6/4)^2)$, which is $5/4$.
Since $X=a+d\,Y$, $X$ has variance $d^2$ times the variance of $Y$. It follows that $8=\frac{5}{4}d^2,$ and therefore $d=\pm 4\sqrt{2}/\sqrt{5}$. We use the positive value of $d$. Work with the negative value is essentially the same.
Since $X=a+d\,Y$, we have $E(X)=a+d\,E(Y)$. Thus $6=a+\frac{6}{4}d.$ But we know $d$, and therefore we can compute $a$. Now that $a$ and $d$ are known, all the $a_i$ are known.
Generalization: Suppose that the sequence $(a_1,a_2, a_3,\dots, a_n)$ is an increasing $n$-term arithmetic sequence. Let $X$ be the random variable that takes on the value $a_i$ with probability $1/n$ ($i=1,2,\dots,n$). Suppose also that we know the mean $\mu$ and the variance $\sigma^2$ of $X$. We want to determine the $a_i$.
Let $a=a_1$ be the first term of our sequence, and $d$ the common difference. Let $Y$ be the random variable that takes on the values $0,1,2,\dots,n-1$, each with probability $1/n$.
Then $X=a+d\,Y$. It is a standard fact that $Y$ has variance $\frac{n^2-1}{12}$. (This can also be easily derived from the usual formula for $\sum_0^{n-1}i^2$.) But the variance of $X$ is $d^2$ times the variance of $Y$. It follows that $\sigma^2=\frac{n^2-1}{12}d^2,$ and now we know $d$.
Also, $E(Y)=\frac{n-1}{2}$. Since $X=a+d\,Y$, by taking expectations we obtain $\mu=a+\frac{n-1}{2}d,$ and now we know $a$, and therefore everything.