Question

the sum of 4 numbers in an A.P is 40 and the product of the first and last os to the product of middle two as 2:3 find them (assume that the four consecutive terms is A.P are a-3d,a-d,a+d,a+3d)​

Answer 1

Answer:

see the attachment

Step-by-step explanation:

hope it helps you

Answer 2

Answer:

Thus no are 4,8,12,16

Step-by-step explanation:

let the four consecutive terms is A.P are a-3d,a-d,a+d,a+3d

sum =a-3d+a-d+a+d+a+3d=40

4a=40

a=10

Also I *IV / II*III =2/3

(a-3d)(a+3d) /)a-d)*(a+d)=2/3

a²-9d² /a²-d²=2/3

3(a²-9d²)=2*(a²-d²)

3*100-27d²=2*100-2d²

25d²=100

d²=100/25=4

d=+-2

If d =2

the no are 10-6,10-2,10+2,10+6

Or 4,8,12,16

If d=-2

Then  no are 10+6,10+2,10-2,10-6

16,12,8,4

Thus no are 4,8,12,16