Properties of a Geometric Progression

A geometric progression is a series in which any element after the primary is derived by multiplying the preceding element by a constant known as the common ratio, indicated by r. For example, the numbers 1, 2, 4, 8, 16, 32… maybe a geometric sequence with r = 2.

In this case, the next number in the series is double the previous number. In other words, multiplying 1 by 2 results in the number 2. When 2 is multiplied by 2, the result is 4. Similarly, when 4 is multiplied by 2, we get 8 and so on. So, what do you believe is going on? Is it possible to assert that the ratio of the two successive terms in the series is constant?

As a result, the ratio of the two subsequent words in this sequence may be a fixed integer. This type of sequence is known as progression. Furthermore, the progression is the sequence in which the main term is non-zero, and each subsequent term arises from multiplying the prior term by a fixed number.

Properties of a Geometric Progression

• When all the terms of a geometrical Progression are raised to an equivalent power, then the new series also forms a geometrical Progression.

Proof:

Let a1  a2  a3  a4  ……………… an be a geometrical progression with common ratio r. Then,

an + 1/an = r, for all n ∈ N ………………. (i)

Let k be a non-zero real. Consider the sequence a1k, a2k, a3k, …….., ank, ………..

We have, an +1k  /  ank = (an+1 / an)k = rk for all n ∈ N, [Using (i)]

Hence a1k    a2k  a3k …….., ank ……….. maybe a progression with common ratio rk.

• The product of the primary and, therefore, the last term is usually adequate to the merchandise of the terms equidistant from the start and the end of the finite progression.

Proof:

Let, a1  a2 a3  a4 ………………, an ………. be a geometrical progression with common ratio r. Then, Kth term from the start = ak = a1rk-1

Kth term from the top = (n – k + 1) the term from the last. 

 = an-k+1 = a1rn-k

Therefore, the kth term from the beginning) x (kth term from the end) = ak * an-k+1

= a1rk-1a1rn-k = a1 rn-1 = a1 * a1rn-1 = a1an for all k = 2, 3, ……, n – 1.

Hence, the merchandise of the terms is equidistant from the start and therefore, the end is usually the same and is adequate to the merchandise of the primary and the last term.

• Three non-zero quantities a, b, c are in progression if and as long as b2 = ac.

Proof:

A, b, c are in progression ⇔ b/a = c/b = common ratio ⇔ b2 = ac

Note: When a, b, c are in geometric progression, then b is understood because of the mean of a and c.

• When the terms of a geometrical Progression are selected at intervals, then the new series obtains a geometrical Progression.

• During a geometric progression of non-zero non-negative terms, the logarithm of every term is formed by an arithmetic progression and vice-versa.

i.e., If a1, a2, a3, a4, ………………, an, ………………… are non-zero non-negative terms of a geometrical progression, then log a1, log a2, log a3, log a4, …………………, log an, ……………………. forms a progression and vice-versa.

If a1, a2, a3, a4, ………………, an, ………………… may be a progression of non-zero non-negative terms with common ratio r. Then,

an = a1rn-1, for all n ∈ N

⇒ log an = log a1 + (n – 1) log r, for all n ∈ N

Let bn = log an = log a1 + (n – 1) log r, for all n ∈ N

Then, bn+1  – bn = [log a1 + n log r] – [log a1 + (n -1) log r] = log r, for all n ∈ N.

Clearly, bn+1  – bn = log r = constant for all n ∈ N. Hence, b1, b2, b3, b4, …………….., bn, ……. i.e., log a1, log a2, log a3, log a4, ………………., log an, ……….. be an arithmetic progression with common difference log r.

Conversely, log a1, log a2, log a3, log a4, ………………., log an, ……………… be an arithmetic progression with common difference d. Then,

log an+1 – log an = d, for all n ∈ N.

⇒ log (an +1/an) = d, for all n ∈ N.

⇒ an+1/an = ed, for all n ∈ N.

⇒  a1, a2, a3, a4, ………………, an, ………………….. may be a geometric progression with common ratio ed.

Conclusion

Take a piece of paper and fold it as many times as you can. So, how many times did you fold the paper? Maybe four to five times? Can you now locate the summit of the paper stack after it has been folded multiple times? Furthermore, how does one go about determining it? A progression might be the answer to the problem at hand. In the article, we learnt about various properties which make geometric progression different. We also proved some of the important properties.