Question

In a geometric series t5 + t7 = 1500 and t11 + t13 = 187500. Find all possible values for the first term.

You should also note that tn = ar^n-1 in a geometric series. Help please.

Answer 1

what would t5 look like with the given tn equation?

Answer 2

\[ar^4+ar^6=1500\]\[a=\frac{1500}{r^4+r^6}\]
\[ar^{10}+ar^{12}=187500\]\[a=\frac{187500}{r^{10}+r^{12}}\]

\[\frac{1500}{r^{4}+r^{6}}=\frac{187500}{r^{10}+r^{12}}\]
\[\frac{r^{10}+r^{12}}{r^{4}+r^{6}}=\frac{187500}{1500}\]
\[\frac{r^{10}(1+r^2)}{r^{4}(1+r^{2})}=125\]
\[{r^{6}}=125;\ r=125^{1/6}\]

Answer 3

\[t5=ar^4=a(125)^{4/6}\]\[t7=ar^6=a(125)^{6/6}\]\[1500=a(125+125^{2/3})\]
\[a=\frac{1500}{125+125^{2/3}}\]

check it with the other set up as well

Answer 4

Ok so a = 10. But how do I find out all possible values of a?