In an arithmetic sequence, a_4 = 12 and a_16 = 42. Find a_n, a_20, and the sum of the first 20 terms.

Answer 1

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Answer 2

a_n = a_1 + (n-1)d where d= common difference = 2 (42-12)/15 = 30/15 = 2

a_n = 12 + (n-1)(2)

sum of n terms is (n/2)(2a_1 + (n-1)d) = (n/2)(a_1 + a_n)

s_20 = (20/2)(24+38) = 10(62) = 620

sum of the first 20 terms = 620