Asked by jen
The common ratio r of a geometric sequence satisfies the quadratic equation 2r squared - 3r -2 = 0. If the sum to infinity of the same sequence is 6, explain why, in this case, r can only take on one value, and hence state the common ratio, r. Also, show that the first term 'a' of this sequence is 9.
Thanks so much xx
Thanks so much xx
Answers
Answered by
drwls
The only solutions to 2r^2 - 3r -2 = 0 are
r = 2 and r = -1/2. You can prove that by factoring the equation.
The sequence a + ra + r^2 a + ...will not converge if r = 2. Therefore r must be -1/2.
Use the fact that the sum is 6 to figure out the first term, a
6 = a (1 -1/2 + 1/4 -1/8 + ...)
= a [1/(1 + (1/2))] = a*(2/3)
a = 9
r = 2 and r = -1/2. You can prove that by factoring the equation.
The sequence a + ra + r^2 a + ...will not converge if r = 2. Therefore r must be -1/2.
Use the fact that the sum is 6 to figure out the first term, a
6 = a (1 -1/2 + 1/4 -1/8 + ...)
= a [1/(1 + (1/2))] = a*(2/3)
a = 9
Answered by
Anonymous
167/3