Answer:
<em>-3/2 and 1</em>
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
<em>Hence the possible values of y are -3/2 and 1</em>
Answer:
315
Step-by-step explanation:
I believe this should be right
