soumyanakarmi
Answered

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If a, a²+ 1 and a+6 are in an AS, find the possible values of a. ​

it's of optional math. If someone know how to solve this. please help me I have been trying this for like half hour

Sagot :

LammettHash

a, a² + 1 and a + 6 are all in arithmetic progression, in which there is a fixed difference d between consecutive terms. This mean we have

a² + 1 = a + d

a + 6 = a² + 1 + d

Eliminate d and solve for a :

(a² + 1) - (a + 6) = (a + d) - (a² + 1 + d)

a² - a - 5 = -a² + a - 1

2a² - 2a - 4 = 0

a² - a - 2 = 0

(a - 2) (a + 1) = 0

a - 2 = 0   or   a + 1 = 0

a = 2   or   a = -1

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