在不使用数组的情况下，我们可以通过先对原始序列排序，再根据等差数列的性质，逐一遍历所有可能的等差数列进行求和，并记录最长的等差数列的和作为结果返回。

代码示例：

def longest_arithmetic_sequence_sum(sequence):
    n = len(sequence)
    if n < 3: # 序列长度小于3时，任意两个数都是等差数列
        return sum(sequence)
    sequence.sort()
    max_length = 2 # 等差数列最小长度为2
    cur_length = 2 # 当前等差数列长度，默认为2
    diff = sequence[1] - sequence[0] # 等差数列差值
    cur_sum = sequence[0] + sequence[1] # 当前等差数列和
    max_sum = cur_sum # 最长等差数列和
    for i in range(2, n):
        if sequence[i] - sequence[i-1] == diff: # 如果当前数可以加入到当前等差数列
            cur_sum += sequence[i]
            cur_length += 1
            if cur_length > max_length: # 如果当前等差数列长度大于已知的最长等差数列长度
                max_length = cur_length
                max_sum = cur_sum
        else: # 如果当前数无法加入到当前等差数列
            diff = sequence[i] - sequence[i-1]
            cur_length = 2
            cur_sum = sequence[i-1] + sequence[i]
    return max_sum

使用示例：

>>> longest_arithmetic_sequence_sum([1,2,3,4,5])
15 # 最长等差数列为 [1,2,3,4,5]

>>> longest_arithmetic_sequence_sum([10,7,4,1,-2])
-4 # 最长等差数列为 [10,7,4,1,-2]

>>> longest