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

2. 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

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

4.
a1 = 1
an = 19
n = 7
Jeff Bivin -- LZHS

5.
a1 = 4
an = 24
n = 11
Jeff Bivin -- LZHS

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

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

8. Find the sum of the multiples of 7 between 25 and 989
Sn =
Jeff Bivin -- LZHS

9. Find the sum of the multiples of 11 that are 4 digits in length
an = 9999
d = 11
Sn =
Jeff Bivin -- LZHS

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

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

12. 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 n – 3
= n2 + n – 2
Tn = Sn – Sn – 1
= n2 + 3n – (n2 + n – 2)
= 2n + 2
Jeff Bivin -- LZHS