Sumception

Let $a$ and $d$ denote the first term and the common difference of this arithmetic sequence.

The sum of the first $n$ terms of this sequence is given by $S_n = \frac n2 (2a + (n- 1)d )$ .

We are given that $S_{100} = 16550$ , thus $\frac{100}2 (2a + 99d) = 16550 \Leftrightarrow {\color{#3D99F6}{2a + 99d = 331}}$ .

And we are also given that $S_1 + S_2 + \cdots + S_{100} = 585800$ . Thus, $\begin{aligned} \sum_{k=1}^{100} \left [ \frac k2 (2a + (k-1)d) \right] &=& 585800 \\ 2a \left( \sum_{k=1}^{100} k \right) + d \left( \sum_{k=1}^{100} (k^2-k) \right) &=& 585800\cdot2 \end{aligned}$

Using the algebraic identities, $\displaystyle \sum_{k=1}^{n} k = \frac {n(n+1)}2$ and $\displaystyle \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}6$ , the equation above simplifies to ${\color{#3D99F6}{101a + 3333d = 11716}}$ .

Solving these 2 linear equations simultaneously gives $a = \boxed{17}, d = 3$ .

Let a = first term and d = common difference. Let S k = the sum of the first k terms. Given that S 100 = 16550, we have that 16550 = (100/2) (a + a + 99d) = 100a + 4950d. Now, the sum from S_1 to S_100= 585800, where S_k = ka + [k(k - 1)/2] d. Forming the sum, 585800 = 5050a + 166650d. We have 2 simultaneous equations for a and d. Eliminating d, we have a = 17.

For T(n) = a+d(n-1), we were taught that sum formulation of S(n) = n(a+l)/2 = (n/2)[2a+d(n-1)], but sum of sum and so on are just an extension of the previous summations. Here, sum of S(n), T in this question, but let's call it R(n), equals R(n) = (n/2)((n+1)/3)[3a+d(n-1)], and sum of R(n) is Q(n) = (n/2)((n+1)/3)((n+2)/4)[4a+d(n-1)]. Just add another factor of n, plus an extra one both for nominator AND denominator, and change the coefficient of a to match the number of n factors (the same as the number of brackets in the formula).
T(100) = 100*101(3a+99d)/6.
and.
S(100) = 100(2a+99d)/2, we get a = 17.

Similar to Pi Han Goh's solution:

$\sum _{i=0}^{99} (a i+b)=16550\land \sum _{j=0}^{99} \left(\sum _{i=0}^j (a i+b)\right)=585800 \Rightarrow a\to 3,\,b\to 17$ . Since the first term is just $b$ , as $i=0$ , the answer is $17$ .