Week 1 – Arithmetic Sequences

Posted on by shannonf2015

 

This is what I have learned so far in the first week of Pre-Calc 11: 

 

My Arithmetic Sequence:

-27, -20, -13, -6, 1

d = 7

d = common difference (In my arithmetic sequence, 7 is being added each time constantly) 

Formula : t_n={t_1+(n-1)}d

Information that is given:

  1. t_{1} = -27  *t_{1} = first term in a sequence 
  2. n (n is the position of term) = 50
  3. d = 7
  4. t_{n} = t_{50}
  5. t_{50} = ?

Part 1: How to find : t_{50} (term 50 of the sequence) 

Step 1 :

Plug in information that’s given in the formula:

Formula: t_n={t_1+(n-1)}d

= t_{50}={-27+(50-1)(7)}

Step 2 : Solve

  • Caution : We have to use BEDMAS when doing the calculations of the formula.

So it would look like this:

= t_{50}={-27+(49)(7)}

= t_{50}={-27+343}

= t_{50}={316} 

 

How to determine general equation of t_{n} : 

Step 1: Place given information into formula.

Formula: t_n={t_1+(n-1)}d

= t_n={-27+(n-1)(-7)}

= t_n={-27+(-7n+7)}

= t_n={-20+(-7n)}

= t_n={-7n-20} 

 

Part 2: Determine the sum of these first 50 terms. 

Step 1 :

-27 + -20 + -13 + -6 + 1 } a series

Formula to find the sum : S_n=\frac{n}{2}{(t_1+t_n)}

Plug in given information:

  1. t_{1} = -27
  2. n (position of term)=50
  3. t_{50} = 316  <—— we got this answer from finding what is t_{50}
  4. S_{20} = ?

= S_{20}=\frac{50}{2}{(-27+316)}

Step 2: Solve (BEDMAS) 

= S_{20}={25}{(289)}

= S_{20}={7,225}