The arithmetic sum formula states that for an arithmetic sequence
a_1, a_2, ..., a_n we have the sum
a_1 + a_2 + ... + a_n = n((a_1 + a_n))/2 (derivation below)
In the given sequence, as the difference between terms is -6, the 31st term will be a_1 + 30(-6) = 174 - 180 = -6
Thus, as n = 31, the final sum will be
174 + 168 + 162 + ... + (-6) = 31*(174+(-6))/2 = 2604
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What follows is a derivation of the arithmetic sum formula, and is not needed for understanding the above solution.
An arithmetic sum with n terms, with initial term a_1, and difference d between terms is a sum of the form
sum_(k=0)^(n-1) (a_1 + kd) = a_1 + (a_1 + d) + (a_1 + 2d) + ... + (a_1 + (n-1)d)
Noting that there are n terms above, we can group the a_1's and factor out the d's from the remaining sum to obtain
sum_(k=0)^(n-1) (a_1 + kd) = na_1 + d(1+2+3+...+(n-1))
In general, the sum of successive integers from 1 to k is
1 + 2 + ... + k = (k(k+1))/2
Thus we have
1 + 2 + ... + (n-1) = ((n-1)(n-1+1))/2 = (n(n-1))/2
So, continuing,
na_1 + d(1+2+3+...+(n-1)) = (2na_1)/2+d(n(n-1))/2
Combining the terms with a common denominator of 2 and factoring out n gives us
(2na_1)/2+d(n(n-1))/2 = n((2a_1) + d(n-1))/2
2a_1 = a_1 + a_1 so some slight rearranging gives us
n((2a_1) + d(n-1))/2 = n((a_1) + (a_1+d(n-1)))/2
But a_1 + d(n-1)=a_n (the nth term in the sequence). Thus the final sum is
sum_(k=0)^(n-1) (a_1 + kd) = n(a_1 + a_n)/2
