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

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}