Question on Progression

Each term in an AP is added to the corresponding term from a GP to form a new progression S. The first 3 terms of this new progression S are -1,-2 &6. The common ratio of GP is equal to the first term of the AP. Prove that the first term of the AP is a root of the equation a^3 - a^2 - a +10=0. Verify that a=-2 is a root of this equation.

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AP has first term a, common difference d
First 3 terms: a, a+d, a+2d

GP has first term b, common ratio a
First 3 terms: b, b*a, b*a²

a + b = −1
a + d + ba = −2
a + 2d + ba² = 6

From 1st equation we get b = −a − 1

Subtract 3rd equation from 2 * 2nd equation:
2a + 2d + 2ba = −4
a + 2d + ba² = 6
-------------------------
a + 0 + b(2a−a²) = −10

Replace b with −a−1

a + (−a−1)(2a−a²) = −10
a − 2a² + a³ − 2a + a² = −10
a³ − a² − a = −10
a³ − a² − a + 10 = 0

When a = −2, we get:
(−2)³ − (−2)² − (−2) + 10 = −8 − 4 + 2 + 10 = −12 + 12 = 0

A.P. terms a, a + d, a + 2d

G.P. terms p, pr, pr² ---> p, ap, a²p......as a = r

=> a + p = -1.....(1)

=> a + d + ap = -2.....(2)

=> a + 2d + a²p = 6.....(3)

(1) into (2) for p gives:

a + d + a(-1 - a) = -2.....(4)

(1) into (3) for p gives:

a + 2d + a²(-1 - a) = 6....(5)

Eliminating d by 2(4) - 5 gives:

a + 2a(-1 - a) - a²(-1 - a) = -10

=> a - 2a - 2a² + a² + a³ = -10

=> a³ - a² - a + 10 = 0

Substituting a = -2 we have:

(-2)³ - (-2)² - (-2) + 10 = -8 - 4 + 2 + 10 = 0......hence, a = -2 is a solution.

A.P. terms ---> -2, 0, 2

G.P. terms ---> 1, -2, 4

Hope that helps.

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