Four Function Arithmetic Sequence

An arithmetic progression is of the form $\ a, a+d , a+2d ,..., a+nd$ so the difference between consecutive terms is constant (given by d), which we can use to create several equations to solve. $\ (a+b) - (a-b) = 2b \ ..........(1)$ Now we can use (1) to create a 2nd equation: $\ (a-b) - ab = 2b \Rightarrow\ a(1-b) = 3b \ .....(2)$ Now using (1) and (2) together produces: $\ ab - \frac{a}{b} = \frac{a(b-1)(b+1)}{b} = \frac{-3b(b+1)}{b} = -3(b+1) = 2b$ $\therefore\ b= - \frac{3}{5}$ To obtain the value of a we can use equaiton (2) as follows: $\ a = \frac{3b}{1-b} = \frac{-9/5}{8/5} = - \frac{9}{8}$ Finally the next term in the sequence is given by: $\ a+3b = -9 (\frac{1}{8} + \frac{1}{5}) = -\frac{117}{40} = -2.925$

$a=-\frac { 9 }{ 8 } \\ b=-\frac { 3 }{ 5 } \\ Hence,\quad the\quad terms\quad are\\ \frac { 75 }{ 40 } ,\frac { 27 }{ 40 } ,-\frac { 21 }{ 40 } ,-\frac { 69 }{ 40 } ,-\frac { 117 }{ 40 }$

Let d be the common difference. (1) a + b - (a - b) = d, (2) b = d/2, (3) a - b - ab = d, a(1 - b) = b + d = 3d/2, a = (3d)/(2 - d), (3) ab - a/b = d, [(3d)/(2 - d)] (d/2) -[(3d)/(2 - d)] (2/d) = d, [(3d)/(2 - d)] (d/2 - 2/d) = d, [(3d)/(2- d)] (d^2 - 4)/2d = d, 3d(d - 2)(d + 2) = 2d^2(2 - d). Either d = 2 or 5d^2 = -6d, d = -1.2. If d = 2, a = infinity, so d = -1.2. Then b = -.6, a = -1.125, a + b = -1.725, and the next term would be -1.725 + (-1.2) = - 2.925.