Sum of 'n' terms of an arithmetico geometric sequence

Formula:

The sum of an arithmetico geometric sequence is generally calculated by applying the method discussed below. It's formula is as follows,

However, this formula is seldom used, since it is too complex to remember (if you do remember it, great!). On the other hand, its derivation is a sequential process, and thus is applied whenever you have to find the sum of an arithmetico geometric sequence. An example of application of this derivation is given below.

Derivation:
Consider an arithmetico geometric sequence of 'n' terms as follows:

a, (a + d)r, (a + 2d)r², .... , (a + (n - 1)d)r^{(n - 1)}

The sum of the above arithmetico geometric sequence is

Sn = a + (a + d)r + (a + 2d)r² + .... + (a + (n - 1)d)r^{(n - 1)}

Now multiply the above sum with constant ratio 'r',

r Sn = ar + (a + d)r² + (a + 2d)r³ + .... + (a + (n - 2)d)r^{(n - 1)} + (a + (n - 1)d)rⁿ 

 Now subtract the above two equations, so we obtain

Sn(1 - r) = a + dr + dr² + .... + dr^{(n - 1)} - (a + (n - 1)d)rⁿ

Notice that in the above equation, the terms from dr till dr^(n - 1) form a geometric progression with first term 'dr' and common ratio 'r'. Therefore we can rewrite the above equation as

Moving (1 - r) from the LHS to the RHS,

This is a formula for the sum of 'n' terms of an arithmetico geometric progression, but since it is too complex and long to remember, we apply the whole method explained above in problems on the sum of arithmetico geometric sequences. In short, this method involves three steps:

1. Write the sum of the arithmetico geometric sequence as a series
2. Multiply the above equation with 'r', the common ratio
3. Subtract the above two equations 

Now let us discuss an example of how the above method is applied in actual math problems to find the sum of arithmetico geometric sequences:

An example:

Question: Find the sum of the following arithmetico geometric sequence:

2, 5x, 8x², 11x³, ... upto 'n' terms

Solution:
The given arithmetico geometric sequence is formed from the arithmetic sequence 2, 5, 8, 11, .. and the geometric sequence 1, x, x², x³, ...

Therefore for the above arithmetico geometric sequence,

• First term of arithmetic sequence, a = 2
• Common difference of arithmetic sequence, d = 3
• Common ratio of geometric sequence, r = x

n^th term of the above arithmetico geometric sequence is: T_n = (a + (n - 1)d)r^{(n - 1)}

Substituting the values of a, d and r, we get: T_n = (2 + (n - 1)3)x^{(n - 1)} 
Simplifying, we get T_n = (3n - 1)x^{(n - 1)}

and (n - 1)^th term of the above sequence is T_{n - 1} = (a + (n - 2)d)r^{n - 2}

That is, T_{n - 1} = (2 + (n - 2)3)x^{n - 2} =  (3n - 4)x^{n - 2}
Now we can write the sum of the artihmetico geometric sequence as

S_n = 2 + 5x + 8x² + 11x³... + (3n - 1)x^{(n - 1)}

Multiplying the above sum by common ratio 'r',

x S_n =  2x + 5x² + 8x³ + ... + (3n - 4)x^{n - 2} + (3n - 1)xⁿ

Subtracting the above two equations,

S_n(1 - x) = 2 + 3x + 3x² + 3x³ + ... + (3n - 4)x^{n - 1} - (3n - 1)xⁿ

In the above equation, the terms ranging from 3x to (3n - 4)x^{n - 1} form a geometric series, therefore we can write the above equation as,

Dividing both sides by (1 - x) we obtain

This is the sum of the given arithmetico geometric sequence.