Tag: arithmetic sequence

Week 1- Precalc 11

yil2016 on

In the first week of Pre-calculus class I learned the arithmetic sequence.

It is worth mentioning that the difference between consecutive terms is constant and this constant value is called the common difference. We can use letter  “d”  to represent it and  “t” to reflect terms in a sequence.

For example: 2 , 4 , 6 , 8 , 10 , …. is a arithmetic sequence. Our goal is to find the common difference ” d ” of this arithmetic sequence. The simplest way is using t_2 mins t_1 , at the same time , using t_3minus t_ 2. If the two differences are the same, we can determine that the common difference is equal to this value. In this arithmetic sequence ” d ” is positive 2. We can figure out any term in this arithmetic sequence by looking at the first term and the common difference. t_n=t_1+d(n-1)

In addition, how to calculate the sum of a arithmetic sequence is just as important as t_n. When we compute the sum of the first n terms of the arithmetic progression, we need to use a formula S_n=\frac{n}{2}\cdot (t_1+t_n) .

 

EX: There have a arithmetic sequence: 3 , 7 , 11 , 15 ,  19 , …

a) Find t_{50} of this arithmetic sequence.

step 1: Find the common difference ” d ” and the fist term of this sequence “t_1“.

step 2: Substitute the value of ” d ” , ” t_1 ” & ” n ” into the formula.

d= 4 , t_1=3 , n= 50

t_{50}=t_1+(n-1)d=3+(49\cdot4)=199

 

b)Find s_{50} of this arithmetic sequence.

step 1: Find the ” d ” and ” t_{50} “.

step 2: Substitute the value of ” d ” , ” n ” , ” t_1 ” & ” t_ {50} ” into the formula.

d=4 , t_1=3 , t_{50}=199 , n=50

S_{50}=\frac{n}{2}\cdot (t_1+t_n) =\frac{50}{2}\cdot(3+199)=5050