Answer:
The three consecutive terms of the AP are;
-5, 2, and 9
Explanation:
The question is with regards to arithmetic progression, AP
Let, 'a₁', 'a₂', and 'a₃', represent the three consecutive terms of the AP
Let a₁ = x, a₂ = y, and a₃ = z
∴ y = x + d, z = x + 2·d
Where;
d = The common difference of the AP
The formula for the sum of 'n' consecutive terms of an AP, Sₙ, is given as follows;
Sₙ = n·(a₁ + aₙ)/2
∴ The sum of the three terms of the AP, S₃ = 3·(x + z)/2 = 6
∴ 3·(x + z)/2 = 3·(x + (x + 2·d))/2 = 3·(2·x + 2·d))/2 = 3·(x + d) = 6
∴ x + d = 6/3 = 2
x + d = 2 = y = The second term of the AP
∴ x = 2 - d
The product of the three terms of the AP is given as follows;
x × (x + d)×(x + 2·d) = -90
Substituting the value of (x + d) = 2 and the value of 'x', gives;
(2 - d) × 2 × (2 - d + 2·d) = -90
(2 - d) × 2 × (2 + d) = -90
(2 - d) × (2 + d) × 2 = -90
(2² - d²) = -90/2 = -45
2² - d² = -45
2² + 45 = d²
4 + 45 = 49 = d²
∴ d² = 49
d = √49 = 7
The first term, x = 2 - d = 2 - 7 = -5
x = -5
The second term, y = 2
The third term, z = x + 2·d = -5 + 2×7 = 9
z = 9
The three consecutive terms of the AP are -5, 2, and 9