Question

15. The sum of three numbers of an AP is 3 and the product of the 1st and 3rd numbers is
-35. Find the three numbers​

Answer 1

Answer:

Ok for example

Step-by-step explanation:

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

So, 3a =18

a=6

Thus,

(a-d) (a+d) =5d

(6-d)(6+d) = 5d

d^2 + 5d - 36 = 0

d^2 +9d - 4d - 36 = 0

d(d+9) - 4(d+9) = 0

(d+9)(d-4) = 0

d = - 9 or 4

So, the three nos are 2,6,10 or 15,6,-3

You can solve by this method

Hope it helps you

Answer 2

Step-by-step explanation:

Let a−d,a,a+d are in A.P

(a−d)+a+(a+d)=−33a=−3a=−1 

Product=8

(a−d)(a)(a+d)=8(d+1)(d−1)=8d2−1=8d=±3

Numbers in A.P are −4,−1,2