Question

Et $a_1, a_2, a_3,\dots$ be an arithmetic sequence.

if $a_1 + a_3 + a_5 = -12$ and $a_1a_3a_5 = 80$, find all possible values of $a_{10}$.

Answer 1

Since it's an arithmetic sequence,

$a_2 = a_1 + d$

$a_3 = a_2 + d = a_1 + 2d$

$a_4 = a_3 + d = a_1 + 3d$

and so on, up to

$a_n = a_1 + (n-1)d$

Now, by substitution,

$a_1 + a_3 + a_5 = -12$

$a_1 + (a_1+2d) + (a_1+4d) = -12$

$3a_1 + 6d = -12$

$a_1 + 2d = -4$

$\implies a_1 = -4-2d, a_3 = -4, a_5 = -4+2d$

Then the product $a_1a_3a_5=80$ depends only on d, so that

$(-4-2d) \times (-4) \times (-4+2d) = 80$

Solve for d :

$(4+2d) (4-2d) = -20$

$16 - 4d^2 = -20$

$4d^2 = 36$

$d^2 = 9$

$d = \pm3$

If d = 3, then the first term in the sequence is $a_1 = -10$, and the tenth term would be $a_{10} = \boxed{17}$.

If d = -3, then the first term would instead by $a_1 = 2$, and the tenth term would be $a_{10} = \boxed{-25}$.