A numerical sequence is any application of ℕ or a part of ℕ to ℝ. The calculators provided here allow you to practice calculations on numerical sequences.

## Numerical sequences : games, quizzes and exercises

Quiz on numerical sequences (Numeric ...

## Numerical sequences : Reminder

A numerical sequence or a numerical progression is any application of ℕ or a part of ℕ to ℝ

### Direction of variation of a sequence: strictly increasing sequence, strictly decreasing sequence.

• To say that the sequence (u_(n)) is strictly increasing means that:
For any natural number n, u_(n+1)>u_(n)
• To say that the sequence (u_(n)) is strictly decreasing means that:
For any natural number n, u_(n)>u_(n+1).
To show that a sequence is increasing or decreasing:
• We can calculate the difference u_(n+1)-u_(n), if this difference is positive then the sequence is increasing, otherwise it is decreasing.
• We can also, if the sequence is positive and u_n!=0, calculate the ratio u_(n+1)/u_(n), if this ratio is greater than 1 the sequence is increasing, otherwise it is decreasing.

### Arithmetic sequences, geometric sequences

#### Arithmetic sequences (arithmetic progression)

To say that a sequence (u_(n)) is arithmetic means that there is a real r such that for any natural number n, u_(n+1)=u_(n)+r. The real r is called the common difference of the sequence (u_(n)).
If (u_(n)) is an arithmetic sequence of first term u_(0), and common difference r. Then for any natural number n, u_(n)=u_(0)+nr

##### Sum of consecutive terms of an arithmetic sequence

If S=a+...+k is the sum of p consecutive terms of an arithmetic sequence then S = p(a+k)/2. We deduce that 1+2+3+...+n=n(n+1)/2

#### Geometric sequences (geometric progression)

To say that a sequence (u_(n)) is geometric means that there is a real q such that for any natural n, u_(n+1)=qu_(n). The real q is called the common ratio for the sequence (u_(n)).
If (u_(n)) is a geometric sequence of first term u_(0), and common ratio q. Then for any natural number n, u_(n)=u_(0)*q^n

##### Sum of consecutive terms of a geometric sequence

If S=a+...+k is the sum of p consecutive terms of a geometric sequence of common ratio q (q != 1) then S = (a-k*q)/(1-q).
We deduce that 1+q+q^2+...+q^n=(1-q^(n+1))/(1-q)