Let a₁, a₂, a₃,… be an arithmetic sequence. If a₁ + a₃ + a₅ = -12 and a₁a₃a₅ = 80, find all possible values of a₁₀.
Let the common difference of the arithmetic sequence be d. Then,
a1+a3+a5=−12
a1+(a1+2d)+(a1+4d)=−12
3a1+6d=−12
a1+2d=−4
a1a3a5=80
8(a1+2d)(a1−2d)(a1+4d)=80
(a1−2d)(a1+2d)(a1+4d)=64
(−4)(−2d)(−4+d)=64
8d(d−1)=64
d(d−1)=8
d=2,4
If d=2, then
a1=−4−2d=−8
a10=a1+9d=−8+18=10
If d=4, then
a1=−4−4d=−16
a10=a1+9d=−16+36=20
Therefore, the possible values of a10 are 10,20.
By simple inspection:
a(1) = 2
a(3) = - 4
a(5) = - 10
[2 - 4 - 10] = - 12
[2 * - 4 * -10] = 80
[-4 - 2] = - 6 / 2 =- 3 - this is the common difference
So, your sequence looks like this:
2 , -1 , -4 , -7 , -10 , -13 , -16 , -19 , -22 , -25 , >>Number of terms = 10>>Total Sum == -115
So, the 10th term = -25 - with this solution. There may be others. Look for them!
