128 x 4 = 512
In a 128x4 microchip, there are 128 memory cells, each capable of storing 4 bits of data. This configuration implies that there are 128 registers, with each register corresponding to one memory cell. Therefore, the microchip has 128 registers and 128 memory cells.
The value of the nth term of an Arithmetic Progression is given by a + (n - 1)d, where a is the first term and d is the common difference.t5 = a + (5 - 1)d = a + 4d = -1/2t9 = a + (9 - 1)d = a + 8d = -1/128Subtracting the first equation in bold from the second equation gives :-4d = -1/128 - (-1/2) = -1/128 -(-64/128) = 63/128 therefore d = 63/(128x4) = 63/512Substituting for d in the first equation a + (4x63)/512 = -1/2 : a = -1/2 - 252/512 = -508/512.t3 = -508/512 + (3 - 1)63/512 = -508/512 + 126/512 = - 382/512 = -191/256