Question

Find 4 consecutive terms in an AP such that the sum of the middle number is 18 and the product of the end two terms is 45

Answer 1

let the four consecutive terms be a-3d,a-d,a+d,a+3d

now according to question, a-d+a+d=18

2a=18

a=9

(a-3d)(a+3d)=45

as (x-y)(x+y)=x²-y²

so it becomes a²-9d²=45

as a=9

81-9d²=45

81-45=9d²

36=9d²

d²=4

d=±2,a=9

so ap can be made when d=2 or d=-2

when d=2

9-3(2),9-2,9+2,9+3(2)

3,7,11,15

when d=-2 ap will be 15,11,7,3 ans

Answer 2

here is your answer. . .

pl......s mark my answer as a brainlist one.....