The sum of three consecutive terms of an AP is 6 product is -90, find the terms

The three consecutive terms of the AP are;

-5, 2, and 9

Step-by-step 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

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