Question

The numbers t(t2 + 1), -t 2 /2 and 6 are three consecutive terms of an A.P. if t be real then find the next two terms of the A.P.

Answer 1

Answer:

14 and 22

Step-by-step explanation:

as these are three consecutive terms of ap then

-t^2/2 - t(t^2+1) = 6 + t^2/2

multiplying by -1 both side we get

t^2/2 + t(t^2+1) = -t^2/2 - 6

2(t^2/2) = -t(t^2+1) - 6

t^2 = -t^3 - t - 6

t^3+t^2+t+6=0 --------(1)

it is clear from above cubic polynomial that

t = -2 is a root of above polynomial, then

t^3+t^2+t+6 = (t+2)(t^2-t+3)

as it is clear from eq. (1) that

(t+2)(t^2-t+3)=0

t^2-t+3=0

as discriminamt for this quadratic equation is less than zero then it results in imaginary roots

for t to be real

t= -2

the the given terms of ap are -10 , -2 , 6

common difference is 8

then next two terms of ap are 14 and 22