 5.2.3 Sum of the First n Terms of an Arithmetic Progression

5.2.3 Sum of the First nTerms of an Arithmetic Progression

(F) Sum of the First n terms of an Arithmetic Progressions
$\overline{)\begin{array}{l}\text{}{S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\text{}\\ \text{}{S}_{n}=\frac{n}{2}\left(a+l\right)\end{array}}$
a = first term
d = common difference
n = the number of term
Sn = 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)
$\begin{array}{l}-11,-8,-5,\dots ..\text{Find}{S}_{15}\\ a=-11,\\ d=-8-\left(-11\right)=3\\ {S}_{15}=\frac{15}{2}\left[2a+14d\right]\\ {S}_{15}=\frac{15}{2}\left[2\left(-11\right)+14\left(3\right)\right]=150\end{array}$

(b)
$\begin{array}{l}8,10\frac{1}{2},13,\dots ..\text{Find}{S}_{13}\\ a=8\\ d=10\frac{1}{2}-8=\frac{5}{2}\\ {S}_{13}=\frac{13}{2}\left[2a+12d\right]\\ {S}_{13}=\frac{13}{2}\left[2\left(8\right)+12\left(\frac{5}{2}\right)\right]=299\end{array}$

(c)
$\begin{array}{l}5,7,9,\dots ..,75←\left(\text{The last term}l=75\right)\\ a=5\\ d=7-5=2\\ {S}_{n}=\frac{n}{2}\left(a+l\right)\\ {S}_{36}=\frac{36}{2}\left(5+75\right)=1440\\ \\ \text{The last term}l=75\\ {T}_{n}=75\\ a+\left(n-1\right)d=75\\ 5+\left(n-1\right)\left(2\right)=75\\ \left(n-1\right)\left(2\right)=70\\ n-1=35\\ n=36\end{array}$