Question

find four consecutive terms in an ap such that the sum of the middle twk term is 18 and the product of two end is 45​

Answer 1

Answer:

therfore terms are 3,7,11,15 (when d positive) and 15,11,7,3(when d is negative)

Step-by-step explanation:

let  (a-3d),(a-d),(a+d) ,(a+3d) be four consecutive terms in an A.P.

(a-d)+(a+d)=18

2a=18

a=9

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

(9-3d)(9+3d)=45

9²-(3d)²=45

81-9d²=45

-9d²=45-81

-9d²=(-36)

9d²=36

d²=4

d=2

hence,

(a-3d)= 9-3×2=9-6=3

(a-d)=9-2=7

(a+d)=9+2=11

(a+3d)=9+3×2=9+6=15

therfore terms are 3,7,11,15 (when d positive) and 15,11,7,3(when d is negative)

Answer 2

Answer:

3,7,11,15 and 15,11,7,3