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Let a₁, a₂, a₃,… be an arithmetic sequence. If a₁ + a₃ + a₅ = -12 and a₁a₃a₅ = 80, find all possible values of a₁₀.

 May 13, 2023
edited by Guest  May 13, 2023
 #1
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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

a1​a3​a5​=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​.

 May 13, 2023
 #2
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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!

 May 13, 2023

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