Question

the sum of three numbers in AP is 21 and their product is -35 find the numbers

Answer 1

Let  a-d , a , a+d are the required three numbers of AP,

where a = any term of AP

           d = common difference

a-d = previous term of a

a+d = next term of a

According to given condition

Their sum is

a-d +a+ a+d = 21

  3a = 21

  a = 7

Also their product is

(a-d)(a)(a+d) = -35

putting a = 7 , we get

(7-d)(7)(7+d) = -35

(7-d)(7+d) = -35 ÷ 7

49 - d^2 = - 5

  49 + 5 = d ^ 2

 54 = d ^2

d = 7.35    , d = -7.35

Case 1 : when d = 7. 35

The numbers are

a - d ,  a , a + d

7- 7.35 , 7 , 7+ 7. 35

-0.35 , 7 , 14.35

Case 2 : when d = - 7. 35

The numbers are

  a - d ,     a ,   a + d

7- (-7.35) , 7 ,  7+ (-7. 35)

7 + 7.35 , 7 , 7 - 7.35

 14.35 ,  7 , -0.35

Answer 2

Let a - d , a , a + d are three consecutive terms in AP.
A/C to question,
sum of three consecutive terms in AP = 21
a - d + a + a + d = 21
3a = 21
a = 7 ........(1)

product of three consecutive terms = -35
(a - d)a(a + d) = -35
a³ - ad² = -35
from equation (1),
343 - 7d² = -35
343 + 35 = 7d²
378 = 7d²
54 = d²
d = ±√54

so, numbers are 7 + √54, 7, 7 - √54
or 7 - √54, 7 , 7 + √54