Question

given that (y+2) (y+3) and (2y^2+1) are consecutive terms of an arithmetic progression, find the possible value of y​

Answer 1

Answer:

-3/2 and 1

Step-by-step explanation:

Given the  arithmetic sequence  (y+2) (y+3) and (2y²+1), the common difference is gotten by taking the difference in their terms. For example if we have 3 terms T1, T2, T3... the common difference d = T2-T1 = T3-T2

From the  sequence given;

T1 = y+2, T2 = y+3 and T3 = 2y²+1

d = y+3-(y+2) = 2y²+1- (y+3)

open the parenthesis

y+3-y-2 = 2y²+1- y-3

1 = 2y²+1- y-3

1 = 2y²- y-2

2y²- y-2-1 = 0

2y²- y-3 =0

Factorize the resulting expression

2y²- y-3 =0

2y²- 2y+3y-3 =0

2y(y-1)+3(y-1) = 0

(2y+3)(y-1) = 0

2y+3 = 0 and y-1 = 0

2y = -3 and y =1

y = -3/2 and 1

Hence the possible values of y are -3/2 and 1