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The third term of an arithmetic sequence is 14 and the ninth term is -1. Find the first four...

Question:

The third term of an arithmetic sequence is {eq}14 {/eq} and the ninth term is {eq}-1 {/eq}. Find the first four terms of the sequence.

Arithmetic progression

An A.P. is a sequence of numbers in which the numbers are arranged in such a manner so the difference between the two successive terms is always constant and known as the common difference(d).

A general A.P. can ce represented as:

{eq}\displaystyle \color {blue} { a, a+d, a+2d, a+3d, .................., a+(n-1)d, ........... }{/eq}

Here {eq}\displaystyle a= first~term {/eq}

{eq}\displaystyle d = common ~ difference {/eq}

And {eq}\displaystyle n = number~of ~ terms {/eq}

nth term of an A.P.:

{eq}\displaystyle \color {blue} { A_n = a + (n-1)d }{/eq}

Answer and Explanation: 1

Given:

The third term of A.P. is {eq}\displaystyle 14 {/eq}

And we know that :

{eq}\displaystyle \color {blue} { A_n = a + (n-1)d }{/eq}

So

{eq}\displaystyle A_3 = a + (3-1)d = 14 {/eq}

{eq}\displaystyle a + 2d = 14 {/eq}

{eq}\displaystyle a = 14 - 2d {/eq} .......(1)

And

The ninth term is {eq}\displaystyle -1 {/eq}

So

{eq}\displaystyle A_9 = a+ (9-1)d = -1 {/eq}

Using the eq.(1)

{eq}\displaystyle 14-2d +8d = -1 {/eq}

{eq}\displaystyle 6d = -1 -14= -15 {/eq}

{eq}\displaystyle d = \frac{-15}{6} = \frac{-5}{2} {/eq}

Put the value in eq.(1)

{eq}\displaystyle a = 14 - 2\times (\frac{-5}{2}) = 14 + 5 = 19 {/eq}

Now

{eq}\displaystyle A_2 = a+d = 19 + (-\frac{-5}{2} ) = \frac{38 - 5 }{2} = \frac{33}{2} {/eq}

And

{eq}\displaystyle A_3 = 14 {/eq}

And

{eq}\displaystyle A_4 = a + 3d = 19 + 3\times (\frac{-5}{2} ) = 19 - \frac{15}{2} = \frac{38-15}{2} = \frac{23}{2} {/eq}

So the first four terms of the A.P. are:

{eq}\displaystyle 19,\frac{33}{2}, 14,\frac{23}{2},.......... {/eq}


Learn more about this topic:

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Arithmetic Series: Formula & Equation

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Chapter 26 / Lesson 8
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An arithmetic series is the sum of a sequence in which each term is computed from the previous one by adding (or subtracting) a constant. Discover the equations and formulas in an arithmetic series.


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