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The first three terms of an arithmetic sequence are as follows.3, -1, -5

The first three terms of an arithmetic sequence are as follows.3, -1, -5-example-1
User Winand
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We will take a look at how we go about with arithmatic progressions.

Arithmmetic sequences are caetgorized by the following two parameters:


\begin{gathered} a\text{ = First term} \\ d\text{ = common difference} \end{gathered}

Where,


\begin{gathered} \text{The value of the first term is called ( a )} \\ \text{The common difference between each and every successive value in a sequence is called ( d )} \end{gathered}

We are given the following arithmetic sequence:


3\text{ , -1 , -5 , }\ldots

Now we will try to determine the values of the two parameters ( a and d ) from the given sequence as follows:


\begin{gathered} a\text{ = 3 }(\text{ first term value )} \\ d\text{ = (-1 ) - ( 3 ) = (-5 ) - ( -1 ) = -4 ( common difference )} \end{gathered}

Now to determine the value of any term number ( n ) in an arithmetic sequence we use the following formula:


a_n\text{ = a + ( n - 1 )}\cdot d

Where,


n\text{ is the term number}

So if we plug in the values of arithmetic sequence parameters into the general equation above we get:


\textcolor{#FF7968}{a_n}\text{\textcolor{#FF7968}{ = 3 + ( n - 1 ) }}\textcolor{#FF7968}{\cdot}\text{\textcolor{#FF7968}{ ( -4 )}}

Now we are to determine the values of term numbers ( n = 4 ) and ( n = 5 ). We will evaluate the ( an ) for each term number as follows:


\begin{gathered} \text{\textcolor{#FF7968}{For n = 4}} \\ a_4\text{ = 3 + ( 4 - 1 )}\cdot(-4\text{ )} \\ a_4\text{ = 3 - 12} \\ \textcolor{#FF7968}{a_4}\text{\textcolor{#FF7968}{ = -9}} \\ \\ \text{\textcolor{#FF7968}{For n = 5}} \\ a_5\text{ = 3 + ( 5 - 1 )}\cdot(-4\text{ )} \\ a_5\text{ = 3 - 1}6 \\ \textcolor{#FF7968}{a_5}\text{\textcolor{#FF7968}{ = -}}\textcolor{#FF7968}{13} \end{gathered}

Hence, the next two consecutive numbers in the arithmetic sequence would be:


3\text{ , -1 , -5 ,}\text{\textcolor{#FF7968}{ -9}}\text{ , }\text{\textcolor{#FF7968}{-13}}

User Gean Ribeiro
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