Question 1191495
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d = common difference
n = number of terms
{{{a[1]}}} = first term = 3
{{{a[n]}}} = nth term or last term = 58
{{{S[n]}}} = sum of the first n terms = 366



{{{S[n] = (n/2)*(a[1]+a[n])}}}


{{{366 = (n/2)*(3+58)}}}


{{{366 = n*(1/2)*(61)}}}


{{{366 = 30.5n}}}


{{{n = 366/(30.5)}}}


{{{n = 12}}}
There are 12 terms.
Phrased another way, the 12th term is 58, ie {{{a[12] = 58}}}


Use this to find the value of d below.
{{{a[n] = a[1]+d(n-1)}}}


{{{a[12] = 3+d(12-1)}}}


{{{58 = 3+d(12-1)}}}


{{{58 = 3+11d}}}


{{{11d+3 = 58}}}


{{{11d = 58-3}}}


{{{11d = 55}}}


{{{d = 55/11}}}


{{{d = 5}}}
The common difference is 5. 
This means we add 5 to each term to get the next term.
The sequence of twelve terms is 3, 8, 13, ..., 53, 58 and these terms add to 366.


Answer: 5
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