Arithmetic-Geometric Progression

The term-by-term multiplication of a geometric progression with the corresponding terms of an arithmetic progression produces an arithmetic–geometric sequence in mathematics. Simply put, the product of the nth term of an arithmetic sequence with the nth term of a geometric sequence is the nth term of an arithmetic–geometric series. Arithmetic–geometric sequences can be found in a variety of contexts, including the computation of expected values in probability theory. Consider the following sequence:

This is an arithmetic–geometric sequence. 

Arithmetic progressions 

Consider the following two common sequences: 1, 3, 5, 7, and 0, 10, 20, 30, 40, etc. It’s clear how these sequences are created. They all begin with a specific first term, and then we just add a predetermined value to the preceding term to produce subsequent terms. To get the following term in the first series, we add 2, and in the second sequence, we add 10. As a result, the difference between each sequence’s subsequent terms is a constant.

For example, our first sequence could be written as 1, 3, 5, 7, 9,… 1, 1 + 2, 1 + 2 2, 1 + 3 2, 1 + 4 2,…, and this could be written as a, a + d, a + 2d, a + 3d, a + 4d,…, where a = 1 is the first term and d = 2 is the common difference, and this could be written as a, a + d, a + 2d, a + 3 If we wanted to write down the nth term, we would have a + (n 1)d, because there must be (n 1) common differences between subsequent terms if there are n terms in the series, hence we must add (n 1)d to the starting value a. We also use the letter l to represent the last term of a finite series, as in this case, l = a + (n − 1)d .

The sum of an arithmetic series 

Sometimes we wish to add a sequence’s terms. If we wished to add the first n terms of an arithmetic progression, what would we get? S_n = a + (a + d) + (a + 2d) +… + (l – 2d) + (l – d) + l + (l + d) + l As we have added up the n terms of a sequence, we now have a series. We can find the sum of this arithmetic series by utilising a technique. Let’s rewrite the series in reverse order.

We get S_n = l + (l − d) + (l − 2d) + . . . + (a + 2d) + (a + d) + a 

We’re going to combine these two series now. We only get 2S_n on the left-hand side. However, on the right-hand side, we’ll combine the terms from both series so that each phrase from the first is added to the term in the second series vertically below it.

We get;

 2S_n  = (a + l) + (a + l) + (a + l) + . . . + (a + l) + (a + l) + (a + l), and on the right-hand side there are n copies of (a + l) so we get 2S_n  = n(a + l). But of course we want S_n  rather than 2S_n , and so we divide by 2 to get;  S_n  = ½ n(a + l). We have found the sum of an arithmetic progression in terms of its first and last terms, a and l, and the number of terms n. We can also find an expression for the sum in terms of the a, n and the common difference d. To do this, we just substitute our formula for l into our formula for S_n. 

From l = a + (n − 1)d , S_n  = ½ n(a + l)

 we obtain S_n  = ½ n(a + a + (n − 1)d)

 = ½ n(2a + (n − 1)d).

Geometric progressions 

Let’s use algebra to write down a general geometric progression. As with arithmetic progressions, we’ll consider a to be the initial term. However, there is no common ground here. Instead, because the ratio of successive terms is always constant, there is a common ratio. As a result, we’ll choose r as the common ratio.

With this notation, the general geometric progression can be expressed as a, ar, ar² , ar³ , . . . . So the nth can be calculated quite easily. It is ar^n-1 , where the power (n −1) is always one less than the position n of the term in the sequence. In our first example, we had a = 2 and r = 3, so we could write the first sequence as 2, 2 × 3, 2 × 3², 2 × 3³ , . . . .

In our second example, a = 1 and r = −2, so that we could write it as 1, 1 × (−2), 1 × -2² , 1 × -2³ ,  . . . .

The sum of a geometric series 

Let’s say we wish to find the sum of a geometric progression’s first n terms. The result is S_n = a + ar + ar² + ar³ +… + ar^(n-1), which is referred to as a geometric series. The idea is to multiply by r and then remove to get the sum:

S_n = a + ar + ar² + ar³+ . . . + ar^n-1

rS_n = ar + ar² + ar³+ . . . + ar^n-1 + arⁿ   (S_n − rS_n = a − arⁿ )

so that S_n(1 − r) = a(1 − rⁿ). 

Now divide by 1 − r (as long as r ≠1) to give 

S_n = a(1 – rⁿ) /(1 − r)

Conclusion

This unit covers sequences and series, as well as some basic examples of each. It also looks at arithmetic progressions (APs) and geometric progressions (GPs), as well as the series that go along with them. It’s critical to do plenty of practise exercises to master the concepts described here so that they become second nature.