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 #n#th term in the sequence). Thus the final sum is
#sum_(k=0)^(n-1) (a_1 + kd) = n(a_1 + a_n)/2#
