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How many terms are in the arithmetic sequence 5, 1, −3, …, −99?

Hint: an = a1 + d(n − 1), where a1 is the first term and d is the common difference

 

A.) 27

B.) 28

C.) 29

D.) 30

Redsox123  May 11, 2017
 #1
avatar+20573 
+1

How many terms are in the arithmetic sequence 5, 1, −3, …, −99?
Hint: an = a1 + d(n − 1), where a1 is the first term and d is the common difference

 

\(\begin{array}{|rcll|} \hline 5, 1, -3, \ldots, -99 \\ a_1 &=& 5 \\ a_2 &=& a_1 + d \\ d &=& a_2 - a_1 \\ d &=& 1-5 \\ d &=& -4 \\\\ a_n &=& a_1 + (n-1)\cdot d \quad & | \quad a_1 = 5 \quad a_n=-99 \quad d = -4 \\ -99 &=& 5 + (n-1)\cdot (-4) \quad & | \quad -5 \\ -104 &=& (n-1)\cdot (-4) \quad & | \quad : (-4) \\ \frac{-104}{-4} &=& n-1 \\ \frac{104}{4} &=& n-1 \\ 26 &=& n-1 \quad & | \quad +1 \\ 27 &=& n \\ \hline \end{array}\)

 

 

In the arithmetic sequence 5, 1, -3, …, -99 are 27 terms.

 

laugh

heureka  May 11, 2017

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