The problem I'm trying to solve is this:
The first three terms of a geometric sequence are also the first, eleventh and sixteenth terms of an arithmetic sequence. The terms of the geometric sequence are all different. The sum to infinity of the geometric sequence is 18. Find the common ration of the geometric sequence, and the common difference of the arithmetic sequence.
What I've done so far is to write the following equations, where u is the first term, r is the common ratio, and d is the difference:
$ur=u+10d$
$ur^2=u+15d$
$u/(1-r)=18$
But I don't know what to do from there. Any suggestions?
Best Answer
You will get
$$a_2=a_1q,a_3=a_1q^2$$ and $$\sum_{k=0}^\infty a_1 q^k=a_1\frac{1}{1-q}$$ and $$a_1=b_1,a_2=b _1+10d,a_3=b_1+15d$$ Can you proceed? Using your equation you will get $$a_1=b_1$$ $$a_q=b_1+10d$$ $$a_1q^2=b_1+15d$$ Since $$a_1=b_1$$ we obtain $$a_1q=a_1+10d$$ $$a_1q^2=a_1+15d$$
eliminating $q$ we get $$a_1\left(\frac{a_1+10d}{a_1}\right)^2=a_1+15d$$
From here we get the equation $$5a_1d+100d^2=0$$
so $$a_1=-20d$$ if $$d\ne 0$$
Solving this we get $$q=\frac{1}{2}$$