Geometric sequence with second term $12$, sum $50$, and common ratio greater than $0.5$
The second term of a geometric sequence is $12$ and the sum to infinity of the corresponding series is $50$. Find the first term and the common ratio, which is greater than $0.5$.
Answer: $a = 20, r = 0.6$. I know that the first term is $12/r$. I tried plugging it into the formula for sum to infinity, but I'm not getting the correct answer.
$\endgroup$2 Answers
$\begingroup$Perhaps you should be more careful in calculating next time?
We have$$a=\frac{12}r$$$$\frac a{1-r}=50$$Thus$$\frac{12/r}{1-r}=50$$$$\frac{12}{1-r}=50r$$$$12=50r(1-r)=50r-50r^2$$$$50r^2-50r+12=0$$Solving, we get $r=\frac25$ or $r=\frac35$, and take the latter due to the condition $r>\frac12$. Thus the first term is $\frac{12}r=20$.
$\endgroup$ $\begingroup$let me use the usual notation for geometric series. Then we have $ar=12$ and as this is a convergent series (otherwise you wouldn't have a real number for the infinite sum) $$\dfrac{a}{1-r}=50.$$ Now solve these two equations for $a$ and $r.$
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