1. Finding a Formula for Sum
of a Sequence
KAPIL VERMA
10 ‘A’
TH
21

2. Arithmetic Sequence
A sequence is arithmetic if
each term – the previous term = d
where d is a constant
e.g. For the sequence
2, 4, 6, 8, . . .
d = 2nd term – 1st term
= 3rd term – 2nd term . . . = 2
The 1st term of an arithmetic sequence is given
the letter a.

3. Arithmetic Sequence
An arithmetic sequence is of the form
a, a + d, a + 2d , a + 3d , . . .
Notice that the 4th term has 3d added so, for example,
the 20th term will be
a + 19d
The nth term of an Arithmetic Sequence is
t n = a + (n − 1)d

4. Arithmetic Series
When the terms of a sequence are added we get a
series
e.g. The sequence 2, 4, 6, 8, . . .
gives the series 2+ 4+ 6+ 8+ . . .
The Sum of an Arithmetic Series
We can derive a formula that can be used for
finding the sum of the terms of an arithmetic series

5. Arithmetic Series
e.g. Find the sum of the 1st 10 terms of the series
1+ 2+ 3+ 4+ . . .
Solution: Writing out all 10 terms we have
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10
a l
Adding the 1st and last terms gives 11.
Adding the 2nd and next to last terms gives 11.
The 10 terms give 5 pairs of size 11 ( = 55 ).
Writing this as a formula we have
n
(a + l ) where l is the last term
2

6. Arithmetic Series
With an odd number of terms, we can’t pair up all
the terms. e.g.
1+ 2+ 3+ 4+ 5+ 6+ 7
n
However, ( a + l ) still works since we can miss out the
2
middle term giving n = 6.
n 6
We get (a + l ) = (1 + 7)
2 2
Now we add the middle term

7. Arithmetic Series
With an odd number of terms, we can’t pair up all
the terms. e.g.
1+ 2+ 3+ 4+ 5+ 6+ 7
n
However, ( a + l ) still works since we can miss out the
2
middle term giving n = 6.
n 6
We get (a + l ) = (1 + 7)
2 2 1
Now we add the middle term which equals (1 + 7 )
2
7 n
Together we have (1 + 7 ) which is (a + l )
2 2

8. For any arithmetic series, the sum of n terms is
given by n
Sn = (a + l )
2
Since the last term is also the nth term,
l = a + ( n − 1)d
Substituting for l in the formula for the sum
gives an alternative form:
n
S n = ( 2a + ( n − 1)d )
2

9. SUMMARY
 An arithmetic sequence is of the form
a, a + d, a + 2d , a + 3d , . . .
 The nth term is t n = a + (n − 1)d
 The sum of n terms of an arithmetic series is
given by
n n
Sn = (a + l ) or Sn = ( 2a + ( n − 1)d )
2 2

10. e.g.1 Find the 20th term and the sum of 20
terms of the series:
2 + 5 + 8 + 11 + 14 + 17 + . . .
Solution: The series is arithmetic.
a = 2, d = 3 and n = 20
t n = a + (n − 1)d ⇒ u 20 = 2 + 19( 3) = 59
n
Either Sn = (a + l ) where l = u 20 = 59
2
20
⇒ S 20 = ( 2 + 59) = 610
2
n 20
or Sn = ( 2a + ( n − 1)d ) ⇒ S 20 = (4 + (19)3) = 610
2 2

11. e.g.2 The common difference of an arithmetic series
is -3 and the sum of the first 30 terms is 255.
Find the 1st term.
Solution: d = − 3, n = 30 and S30 = 255
n
Sn = ( 2a + ( n − 1) d )
2
30
⇒ 255 = ( 2a + 29( − 3))
2
⇒ 255 = 15( 2a − 87)
255
⇒ = 2a − 87
15
⇒ 17 + 87 = 2a ⇒ a = 52

12. Exercises
1. The 1st term of an A.P. is 20 and the sum of 16
terms is 280. Find the last term and the
common difference.
Solution: n
Sn = (a + l ) ⇒ 280 = 8( 20 + l ) ⇒ 15 = l
2
1
t n = l = a + (n − 1)d ⇒ 15 = 20 + 15d ⇒ d = −
3
10
2. Find the sum of the series given by
∑ 4n − 10
1
Substituting n = 1, 2 and 3, we get −6, −2, 2
We can see the series is arithmetic so,
n = 10 ⇒ u10 = 4(10) − 10 = 30 = ( l )
n
Sn = (a + l ) ⇒ S 10 = 5( − 6 + 30) = 120
2