(F) Sum of the First n Terms of an Arithmetic Progression

Posted on April 22, 2020 by user

1.2.3 Sum of the First nTerms of an Arithmetic Progression

(F) Sum of the First n terms of an Arithmetic Progressions

$\begin{array}{l} S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ S_{n} = \frac{n}{2} (a + l) \end{array}$

a = first term

d = common difference

n = the number of term

S_n = the sum of first n terms

Example:

Calculate the sum of each of the following arithmetic progressions.

(a) -11, -8, -5, ... up to the first 15 terms.

(b) 8, 10½, 13,... up to the first 13 terms.

(c) 5, 7, 9,....., 75 [Smart TIPS: The last term is given, you can find the number of term, n]

Solution:

(a)

$\begin{array}{l} - 11, - 8, - 5, ..... Find S_{15} \\ a = - 11, \\ d = - 8 - (- 11) = 3 \\ S_{15} = \frac{15}{2} [2 a + 14 d] \\ S_{15} = \frac{15}{2} [2 (- 11) + 14 (3)] = 150 \end{array}$

(b)

$\begin{array}{l} 8, 10 \frac{1}{2}, 13, ..... Find S_{13} \\ a = 8 \\ d = 10 \frac{1}{2} - 8 = \frac{5}{2} \\ S_{13} = \frac{13}{2} [2 a + 12 d] \\ S_{13} = \frac{13}{2} [2 (8) + 12 (\frac{5}{2})] = 299 \end{array}$

(c)

$\begin{array}{l} 5, 7, 9, ....., 75 \leftarrow (The last term l = 75) \\ a = 5 \\ d = 7 - 5 = 2 \\ S_{n} = \frac{n}{2} (a + l) \\ S_{36} = \frac{36}{2} (5 + 75) = 1440 \\ The last term l = 75 \\ T_{n} = 75 \\ a + (n - 1) d = 75 \\ 5 + (n - 1) (2) = 75 \\ (n - 1) (2) = 70 \\ n - 1 = 35 \\ n = 36 \end{array}$