Sum of First n Terms of an AP
Let AP series be $a_1,a_2,....,a_{n-1}, a_n $ where $a$ is the first term and $d$ be the common difference
Let $S_n$ be the sum of this series,so
$S_n$ = $a_1+a_2+....+a_{n-1}+ a_n $
Since $a_1=a,a_2=a+d,a_3=a+2d,.....,a_n=a+(n-1)d$
Hence, $S_n$=$a+(a+d)+(a+2d)+....+(a+(n-1)d)$ ..............(1)
Rewriting the terms in reverse order, we have
$S_n$=$(a+(n-1)d)+(a+(n-2)d)+....+(a+d)+a$ .................(2)
Adding (1) and (2), we get
$2S_n$=$[2a+(n-1)d]+[2a+(n-1)d]+.....+[2a+(n-1)d]+[2a+(n-1)d]$
$S_n$=$n/2$$[2a+(n-1)d]$
Sum of First $n$ terms of an AP in term of Last term,$l$
$S_n$=$n/2$$[2a+(n-1)d]$ =$S_n$=$n/2$$[a+a+(n-1)d]$
Since $a_n$ term is given by $a+(n-1)d$
Hence, $S_n$=$n/2$$(a+a_n)$
If $a_n$ is the last term, then $a_n=l$
So, $S_n$=$n/2$$(a+l)$
So, the sum of the first n terms of an AP is given by
$S_n$=$n/2$$[2a+(n-1)d]$
or
$S_n$=$n/2$$(a+l)$
$S_n$=$n/2$$[2a+(n-1)d]$
or
$S_n$=$n/2$$(a+l)$
The $n^{th}$ term of an AP is the difference of the sum to
first $n$ terms and the sum to first ${n – 1}$ terms of it, i.e., $a_n = S_n – S_{n – 1}$.
Example: Find the sum of the first 22 terms of the AP : 8, 3, –2, . . .Solution : Here, a = 8, d = 3 – 8 = –5, n = 22. We know that $S_n$=$n/2$$[2a+(n-1)d]$ $S_{22}$=$22/2$$[2(8)+(22-1)(-5)]$ = $-979$ So, the sum of the first 22 terms of the AP is – 979.Example: If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term. (Hint:Find the common difference $d$ from the first statement, and then find the 20th term of the AP)Solution : $S_n$=$n/2$$[2a+(n-1)d]$ 1050=$14/2$$[2(10)+(14-1)d]$ i.e $d=10$ nth term of an AP is given by, $a_n=a+(n-1)d$ $a_{20}=10+(20-1)10$ $a_{20}=200$ 20th term of the AP=200Homework: Based on the above question, solve the various variation in the above question. (1).If the sum of the first 14 terms of an AP is 1050 and its common difference is 10, find the 20th term. (2).If the sum of the first 14 terms of an AP is 1050 and its 14 term is 140, find the first term?. (3).If the sum of the first 14 terms of an AP is 1050 and its 14 term is 140, find the common difference $d$?.Example: How many terms of the AP : 24, 21, 18, . . . must be taken so that their sum is 78?Solution : Here, a = 24, d = 21 – 24 = –3, Sn = 78. We need to find n. $S_n$=$n/2$$[2a+(n-1)d]$ $78$=$n/2$$[2(24)+(n-1)(-3)]$ $3n^2 – 51n + 156 = 0 $ $n^2 – 17n + 52 = 0 $ $(n – 4)(n – 13) = 0$ $n = 4 or 13$ Both values of n are admissible. So, the number of terms is either 4 or 13.Remarks: 1. In this case, the sum of the first 4 terms = the sum of the first 13 terms = 78. 2. Two answers are possible because the sum of the terms from 5th to 13th will be zero. This is because a is positive and d is negative, so that some terms will be positive and some others negative, and will cancel out each other.