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

-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

the entire expression should be subtracted from 30. to subtract the quantity, distribute the negative sign, which changes the signs on the terms. so, the –2 changes to positive and gets added to 30 instead of subtracted from it.

hope this ! ^w^

1/3

step-by-step explanation:

to find this, you can divide 12 by 4 which is 3 or you can multiply 4 until you reach 12 (4x3=12). both should give you 3, meaning theres three times less bananas than there are oranges.