## Algebra II/Trigonometry

## Means

**Arithmetic Mean:**

Use this when asked for (insert number here) arithmetic means of Term 1 (T1) and Term N (Tn) with difference (d).

Formula: Tn = T1 + (n-1) * d

**Geometric Mean:**

Use this when asked for (insert number here) geometric means of Term 1 (T1) and Term N (Tn) with ratio (r).

Formula: Tn = T1 * r ^ (n-1)

**Partial Sum - General:**

Use this when asked for partial sum of (insert number here) for (insert equation here).

k is the first term number, the top is the last term number and the right is the equation. it basically means to sum up (for the first problem) - 2(1), 2(2), and 2(3) (and the second problem) - 5(3) + 3, 5(4) + 3, ... 5(10) + 3

Example 1: the partial sum of 1-3 in the equation 2k:

3

**Σ 2k**

k=1

Example 2: the partial sum of 3-10 in the equation 5k + 3

10

**Σ 5k + 3**

k=3

**Partial Sum - Arithmetic:**

Use this when asked for partial sum (Sn) of (insert number here)(n) with the given 1st Term (T1) and difference (d) of an ARITHMETIC SEQUENCE.

Formula A: Sn = (n/2) * (2 * T1 + (n-1) * d)

OR

Use the following equation when not given the difference. for example, given T1 and T3 instead of T1 and d.

Formula B: Sn = (n/2) * (T1 + Tn)

**Partial Sum - Geometric:**

Use this when asked for partial sum (Sn) of (insert number here)(n) with the given 1st Term (T1) and ratio (r) of a GEOMETRIC SEQUENCE.

Formula: Sn = T1 * [(1 - r ^ n) / (1 - r)]

**Convergence:**

Use this where asked for when a certain geometric sequence will converge. The ratio (r) can be found by dividing the first term (T1) from the second term (T2) -> (T2/T1).

Make sure to use this formula ONLY when |r| < 1, if |r| > 1, then the sequence does not converge.

Sometimes this formula will be used to find the fraction of a repeating decimal. For example, 0.161616...You would then split up the repeating decimal by each 'repeating' term (for example: 16), getting a geometric sequence (for example 0.16, 0.0016, 0.000016, ...). at the end it would be the first term (T1) divided by one minus the ratio (r) -> (T1 / (1 - r)). If there is an extra term (for example 0.3161616...) then you would add the extra part in front at the end (for example the terms would be 0.016, 0.00016, 0.0000016, ...).at the end it would be (0.3 + (T1/ (1 - r))))

Formula: T1/(1-r)

**Combination w/o Calculator (or Find the Term from an Equation):**

This would be used when asked for the (insert number here) (N) term from an equation that has exponents ( (a ± b) ^ n). The Nth term's a exponent (na) and b exponent (nb) would both be used for factorials in the denominator of the overall ratio (na! * ba!). The original exponent would be in numerator as a factorial (n!). For example, "Find the 5th term of the equation (a + b) ^ 16. Well, the a exponent (na) would be 12 (16-5+1 -> original exponent - term # + 1) and the b exponent (nb) would be 4 (5-1 -> term # - 1), so you would get 12! * 4! on the bottom. (***TIP: The bottom terms should always add up to the original exponent). Finally, the original exponent is 16, so the numerator would be 16!. Your final equation would be 16! / (12! * 4!).

Formula: n! / (na! * ba!)