Sum of an Arithmetic Progression Last Updated: October 11, 2005

Let a1 = first term of an AP Let an = last term of an AP And d = the common difference Hence, the A.P can be written as a1, a1 + d, a1 + 2d, …. an And the SUM OF A.P is Sn = a1 + (a1 + d) + (a1 + 2d) + …+ an OR Sn = an + (an - d) + (an - 2d) + …+ a1 Jeff Bivin -- LZHS

Summing it up Sn = a1 + (a1 + d) + (a1 + 2d) + …+ an Sn = an + (an - d) + (an - 2d) + …+ a1 Jeff Bivin -- LZHS

1 + 4 + 7 + 10 + 13 + 16 + 19 a1 = 1 an = 19 n = 7 Jeff Bivin -- LZHS

4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24 a1 = 4 an = 24 n = 11 Jeff Bivin -- LZHS

Find the sum of the integers from 1 to 100 a1 = 1 an = 100 n = 100 Jeff Bivin -- LZHS

Find the sum of the multiples of 3 between 9 and 1344 Sn = 9 + 12 + 15 + . . . + 1344 Jeff Bivin -- LZHS

Find the sum of the multiples of 7 between 25 and 989 Sn = 28 + 35 + 42 + . . . + 987 Jeff Bivin -- LZHS

Find the sum of the multiples of 11 that are 4 digits in length an = 9999 d = 11 Sn = 10 01+ 1012 + 1023 + ... + 9999 Jeff Bivin -- LZHS

Evaluate Sn = 16 + 19 + 22 + . . . + 82 a1 = 16 an = 82 d = 3 n = 23 Jeff Bivin -- LZHS

Review -- Arithmetic Sum of n terms nth term Jeff Bivin -- LZHS

Problem solving The sum of the first n terms of a progression is given by Sn = n2 + 3n. Find, in terms of n the nth term. Sn = n2 + 3n Sn-1 = (n-1)2 + 3(n-1) = n2 – 2n + 1 + 3n – 3 = n2 + n – 2 Tn = Sn – Sn – 1 = n2 + 3n – (n2 + n – 2) = 2n + 2 Jeff Bivin -- LZHS