ARITHMETIC PROGRESSIONS 109
Here, a = 100, d = 50 andn = 21. Using the formula :
S = 2(1)✁
2
n
an d✂ ✄ ,we have S = ☎ ✆
21
2 100 (21 1) 50
2
✝ ✞ ✟ ✝ = ✠ ✡
21
200 1000
2
✞
=
21
1200
2
✝ = 12600
So, the amount of money collected on her 21st birthday is Rs 12600.
Hasn’t the use of the formula made it much easier to solve the problem?
We also use Sn in place of S to denote the sum of first n terms of the AP. We
write S 20 to denote the sum of the first 20 terms of an AP. The formula for the sum of
the first n terms involves four quantities S, a, d and n. If we know any three of them,
we can find the fourth.
Remark : The nth 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., an = Sn – Sn (^) – 1.
Let us consider some examples.
Example 11 : 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= ☛2(1)☞
2
n
an d✞ ✟
Therefore, S = ✌ ✍
22
16 21 ( 5)
2
✞ ✟ = 11(16 – 105) = 11(–89) = – 979
So, the sum of the first 22 terms of the AP is – 979.
Example 12 : If the sum of the first 14 terms of an AP is 1050 and its first term is 10,
find the 20th term.
Solution : Here, S 14 = 1050,n = 14,a = 10.
As Sn = ✎2(1)✏
2
✞ ✟
n
an d,so, 1050 = ✑ ✒
14
20 13
2
✞ d = 140 + 91d