Question

Find the four no in ap whose sum is 6 and product whose e extreme is ten time the product of the two intermediate terms

Answer 1

Answer:

15 , 6 , -3  & -12

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Step-by-step explanation:

let say terms are

a , a+d , a+2d , a+3d

Sum = 4a + 6d = 6

2a + 3d = 3

d = (3-2a)/3

a (a +3d) = 10 (a+d)(a+2d)

a^2 + 3ad = 10 ( a^2 + 3ad + 2d^2)

9 a^2 + 27ad + 20d^2 = 0

9 a^2 + 27a(3-2a)/3  + 20{ (3-2a)/3}^2 = 0

81 a^2 + 81a(3-2a) + 20 (3-2a)^2 = 0

81 a^2 +243a - 162a^2 + 20 (9+4a^2 -12a) = 0

-81a^2 + 243a +180 + 80a^2 - 240a = 0

-a^2 + 3a + 180 = 0

a^2 - 3a - 180 = 0

(a-15)(a+12) = 0

a = 15   then d = -9

a = -12 then d =  9

so four terms are

15 , 6 , -3  & -12

or

-12 , -3 , 6 & 15