What is the 32nd term of the arithmetic sequence where a1 = 14 and a13 = -58?

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Answer 1 · 2015-07-17T22:46:00

The 32nd term of the sequence is $color(red)(-172)$ .

We know that $a_1 = 14$ and $a_13 = -58$ .

We can use the $n^"th"$ term rule for an arithmetic sequence:

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

$a_13 = a_1 + (13-1)d$

$-58 = 14 + 12d$

$-58 -14 = 12d$

$-72 = 12d$

$d = -72/12$

$d = -6$

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

$a_32 = 14 +(32-1)(-6) = 14 +31(-6) = 14-186$

$a_32 = -172$