1. Now to find the sum of the terms in an Arithmetic sequence.
The set of a, a + d, a + 2d, a + 3d…….., a + (n - 1)d
is an Arithmetic
sequence
but,…..
The sum of a + (a + d) + (a + 2d) +…….+ a + (n - 1)d
is an Arithmetic
series
A calculation to find the sum of the terms in series is needed!
First of all, let un = a + (n - 1)d = l
This is needed to create the calculation.

2.  2Sn = n(a + l) So, Sn = ½n(a + l)  Sn = ½n[a + a + (n - 1)d] i.e.
This is just written in reverse! 1
and so on
2nd last term
last term
Sn = a (a + d) + (a + 2d) (l - 2d) + (l - d) + l 1
Now,
Sn = l (l - d) + (l - 2d) +…+ (a + 2d) + (a + d) + a
Also, 2
So,
2Sn = (a + l) + (a + l) + (a + l) +…+ (a + l) + (a + l)
There are n of these 1 2 + 
2Sn = n(a + l)
For this sum you need to know the first & last terms of the series
So,
Sn = ½n(a + l)
since l = a + (n - 1)d
And, 
Sn = ½n[a + a + (n - 1)d]
For this sum you need to know the first term & the common difference
i.e.
Sn = ½n[2a + (n - 1)d]

3. The whole expression has been multiplied by
This is how to develop the formula to find the sum of the terms in a Geometric sequence:
If the set of terms a, ar, ar2, ar3, ar4, …, arn-1 ….
is a Geometric sequence, then
The sum of a + ar + ar2 + ar3 + ar4 + …+ arn-1 +…..
is a Geometric series.
So,
Sn = a + ar + ar2 + ar3 + ar4 + …+ arn-1 +….
The next step is tricky!
rSn = ar + ar2 + ar3 + ar4 + …+ arn-1 + …+ arn + …+
The whole expression has been multiplied by
Sn - rSn =
a arn
Because all the terms cancel each other out!
Sn (1 - r) =
a(1 - rn) r
By factorising!
On BOTH sides!
a(1 - rn)
So,
Sn =
Move (1 - r) to the right by dividing.
a(rn - 1)
Sn =
(r - 1)
(1 - r) OR
This one is used when r < 1
This one is used when r > 1

4. We can use this one because the 12th term will be the last one.
Now for some examples…….
First, some Arithmetic ones! 1.
The first term of an Arithmetic series is 5 and the 12th term is -14.
Find the sum of the first twelve terms.
Sn = ½n(a + l)
You have to remember this! 
We can use this one because the 12th term will be the last one.
S12 = ½  12 (5 - 14)
n = 12
l = -14
S12 = 6  -9
a = 5
S12 = -54 2.
Find the sum to 15 terms of …..
This is an Arithmetic series!
Sn = ½n[2a + (n - 1)d]
n = 15
d = 4
a = 3
S15 = ½  15[2  3 + (15 - 1)4]
S15 = ½  15  62
S15 = 465

5. Same answer but a lot easier the other way!3.
Find the sum to 7 terms of the geometric series ……...
a(1 - rn)
Sn =
(1 - r)
a(rn - 1)
Sn =
(r - 1) OR
It’s either
It’s this one,
because….
r > 1
a = 2
r = 3
n = 7
So,
If you used
S7 = 2186
Same answer but a lot easier the other way!
That was easy.
Now for a tough one!

6.   + + 5n + 5n Easy! & Now for the hard one! 4.
The sequence 8, 14, 32, ……. has its nth tern 3n + 5.
Find (i) the fourth term and (ii) the sum of the first n terms
(i) the fourth term
We are told that Un = 3n + 5
So U4 = 
Easy!
Now for the hard one!
U4 = 86
(ii) the sum of the first n terms
This is a tricky one!
Because the series is neither Arithmetic NOR Geometric
Now U1 =
& U2 =
Now add vertically 
&
U3 =
U4 =
It needs a SPECIAL method to solve!
And so on 
Sn = Sum of a
Geometric
series
Sum of n terms each of which is 5 +
+ 5n
So, Sn =
where + 5n
a = 3
So,
Exercise 9G has plenty more!
r = 3