The second and fifth terms of a geometric sequence are 10 and 1250, respectively. Is 31,250 a term of this sequence? If so, which term is it?

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Answer 1 · 2018-07-19T03:30:07

Yes, it's the seventh term

$a_{5} = a_{2} {(r)}^{5 - 2}$ $a_5= a_2(r)^(5-2)$

$1250 = 10 r^{3}$ $1250= 10r^3$

$125 = r^{3}$ $125= r^3$

$r = 5$ $r=5$

$a_{1} = a_{2} {(r)}^{1 - 2}$ $a_1= a_2(r)^(1-2)$

$a_{1} = \frac{10}{5}$ $a_1=10/5$

$a_{1} = 2$ $a_1= 2$

$31, 250 = 2 {(5)}^{n - 1}$ $31,250= 2(5)^(n-1)$

$15625 = {(5)}^{n - 1}$ $15625=(5)^(n-1)$

${log}_{5} (15625) + 1 = n$ $log_5(15625)+1=n$

$6 + 1 = 7$ $6+1=7$

$n = 7$ $n=7$