Question

find 4 number in Ap such that the sume of second and third term is 22 and product is firsr and 4th rerm is 85​

Answer 1

Answer:

AP= 5, 9, 13,17...

Step-by-step explanation:

let the numbers in the AP be a, a+d,a+2d, a+3d

sum of 2nd and 3rd term=22

therefore, a+d+a+2d=22

2a+3d=22

a=22-3d/2   -----eq1

product of 1st and 4th term=85

therefore, a(a+3d)=85

substituting the value of a from eq1, we get (22-3d/2)(22-3d/2+3d)=85

(22-3d/2)(22-3d+6d/2)=85

(22-3d/2)(22+3d/2)=85

(22/2)^2-(3d/2)^2=85

121-9d^2/4=85

d^2=36(4)/9

d^2=16

d=4

from eq1 a=22-3d/2

a=22-3(4)/2

a=22-12/2

a=5

therefore, AP=5,9,13,17....