A lvl H2 Maths: APGP

Question from http://www.sgforums.com/forums/2297/topics/348959

Each term of an AP is added to the corresponding term of a GP to form a 3rd sequence S, whose 1st 3 terms are -1, -2, 6. The common ratio of the GP is equal to the 1st term of the AP.

Prove that the 1st term of the AP is a root of the equation a³ - a² - a +10 = 0. Verify that a = -2 is a root of this equation. With this value of a, find the nth term of the sequence S and obtain an expression for the sum of the 1st n terms of S.

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Answer:

Let first term of AP = a
Thus Common ratio of GP = a
Let first term of GP = b

First 3 terms of GP = b, ba, ba²
First 3 terms of AP = a, a+d, a+2d

Hence,
b + a = -1 ---(1)
ba + a + d = -2 ---(2)
ba² + a + 2d = 6 --- (3)

From (1), b = -1 - a ---(4)
From (2), a + d = -2 - ba
2a + 2d = -4 - 2ba ---(5)

Sub (5) into (3)
ba² - a + (-4 - 2ba) = 6
ba² - a - 2ba = 10 ---(6)

Sub (4) into (6)
(-1 - a)a² - a - 2(-1 - a)a = 10
-a² - a³ - a + 2a + 2a² - 10 = 0
-a³ + a² + a - 10 = 0
a³ - a² - a + 10 = 0

Thus, the 1st term of the AP is a root of the equation a³ - a² - a +10 = 0 (shown)

When a = -2,
(-2)³ - (-2)² - (-2) +10 = -8 - 4 + 2 + 10 = 0
Thus, a = -2 is a root of this equation (verified)

Sub a = -2 into (1)
b = -1 -a = 1
Sub b = 1, a = -2 into (2)
d = -2 - a - ba = -2 + 2 + 2 = 2

So,
nth term of S = nth term of AP + nth term of GP
= (-2) + (n - 1)(2) + (1)(-2)^(n-1)
= 2n - 4 + (-2)^(n-1)

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