Question 1154147
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Let x be the very central point for the given 4 terms of the AP.


In other words, let x be the average of the 4 terms of the AP


    x = {{{(a[1]+a[2]+a[3]+a[4])/4}}}.


Next, let d be the common difference of the AP.


Our goal is to find "x" and "d".


It is clear that 

    {{{a[1]}}} = x - 1.5d;

    {{{a[2]}}} = x - 0.5d;

    {{{a[3]}}} = x + 0.5d;

    {{{a[4]}}} = x + 1.5d.



Therefore,  {{{a[1]}}} + {{{a[4]}}} = 2x.


It implies  2x = 8, and hence  x = 4.



Also,  {{{a[2]*a[3]}}} = 15.


It implies  (x-0.5d)*(x+0.5d) = 15,  or

            x^2 - 0.25d^2 = 15.


Substitute here x= 4 to get

            0.25d^2 = 4^2 - 15 = 16 - 15 = 1.


Thus  d^2 = {{{1/0.25}}} = 4;  hence,  d = {{{sqrt(4)}}} = +/- 2.


The problem is just solved. We know the central point x = 4  and the common difference d = +/-2.


If d = 2, then  {{{a[1]}}} = 4-1.5*2 = 1;  {{{a[2]}}} = 1+2 = 3;  {{{a[3]}}} = 3+2 = 5,  and  {{{a[4]}}} = 5 + 2 = 7.


If d = -2,  then we have the same sequence in the REVERSED order:  {{{a[1]}}} = 7;  {{{a[2]}}} = 5;  {{{a[3]}}} = 3  and  {{{a[4]}}} = 1.
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Solved.