SL 1.3 Geometric sequences and series Study Notes- IBDP Maths AA- SL & HL – New Curriculum 2021-2028

## IBDP Maths MAA Study notes -All topics

Topics: SL 1.3 Geometric sequences and series Study Notes- IBDP Maths AA- SL & HL based on New Curriculum 2021-2028

*GEOMETRIC SEQUENCE….*

**THE DEFINATION..**

I give you the first term of a sequence, say \(u_1\) =5 and I ask you to multiply by a fixed number, say r =2, in order to find the next term. The following sequence is generated:

**5, 10, 20, 40, 80, …**

Such a sequence is called **geometric.** That is, in a geometric sequence the ratio between any two consecutive terms is constant.

In other words , we can say , **geometric sequence **is a sequence where there is a common ratio between consecutive number.

*For determining geometric sequence , we only need *

- The
**first term** - The
**common ratio**

- Common ratio: The common ratio is the constant factor by which each term in a geometric sequence is multiplied to get the next term.

(b) \(u_1\) =5, r = 10 the sequence is 5, 50, 500, 5000, … (c) \(u_1\) =1, r = -2 the sequence is 1,-2, 4,-8, 16, … (d) \(u_1\) =1, r = \(\frac{1}{2}\) the sequence is 1, \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{16}\), \(\frac{1}{32}\), ….. (e) \(u_1\) =1, r = \(\frac{-1}{2}\) the sequence is 1, \(\frac{-1}{2}\), \(\frac{1}{4}\), \(\frac{1}{16}\), \(\frac{-1}{32}\), ….. |

**ATTENTION!!**

- The common ratio r may also be negative! In this case the signs alternate (+, -, +, -, …) [see (c) and (e) above].
- The common ratio r may be between -1 and 1, that is |r|<1. In such a sequence the terms approach 0 [see (d) and (e) above]

**THE \(N^{th}\) TERM FORMULA**

The nth term formula for a geometric sequence is **\(u_n =u_1r^{n-1}\)**

where,

- \(u_1\) is the first term
- n is the term number
- r is the common ratio

*Indeed, let us think: **In order to find \(u_5\) , we start from u1 and then multiply 4 times by the ratio r*

Hence, \(u_5 = u_1r^4\)

Similarly, \(u_{10} = u_1r^9\), \(u_{50} = u_1r^{49}\) and so on…..

(b) the 100th term
\(u_{100} = u_1r^{99} = 3.2^{99}\)
\(u_{10}=u_1r^9 = 196830 = 10 . r^9\) \(r^9= 19683\) \(r= \sqrt[9]{19683} = 3\) Therefore , \(u_3 = u_1r^2 = 10.3^2 = 90\) |

**REMEMBER!!**

Our first task in a G.S. is to find the basic elements, \(u_1\) and r , and then anything else.

**SUM OF N TERMS(\(S_n\)**

Given that r ≠ 1 , the result is given by

\(S_n = \frac{u_1(r^n-1)}{r-1}\) or \(S_n = \frac{u_1(1-r^n)}{1-r}\)

**Write notes here**