A Problem About Arithmetic And Geometric Sequences

Question

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?

Answer 1

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 $$q=\frac{1}{2}$$

Answer 2

Note that: $$b_1,b_2,b_3 \iff a_1,a_{11},a_{16} \Rightarrow \\ \begin{cases}b_2-b_1=10d \\ b_3-b_2=5d\end{cases} \Rightarrow \begin{cases}b_1(q-1)=10d \\ b_1(q^2-q)=5d\end{cases} \Rightarrow q=\frac12.\\ \frac{b_1}{1-q}=18 \Rightarrow b_1=9;\\ b_1(q-1)=10d \Rightarrow d=-\frac9{20}.\\$$

Answer 3

From the first two equations of$$\begin{cases}ur=u+10d, \\ur^2=u+15d, \\\dfrac u{1-r}=18 \end{cases}$$

you draw $$\frac{u(r^2-1)}{u(r-1)}=r+1=\frac{15d}{10d}$$ and $r=\dfrac12$. Using the third, $u=9$, and finally $d=-\dfrac9{20}$.

Answer 4

I will use $a_1$ for the initial term of both sequences, $r$ for the common ratio of the geometric sequence, and $d$ for the common difference. Since the second term of the geometric sequence is the eleventh term of the arithmetic sequence, $$a_1r = a_1 + 10d$$ Subtracting the first term of the sequence from each side gives \begin{align*} a_1r - a_1 & = a_1 + 10d - a_1\\ a_1(r - 1) & = 10d \tag{1} \end{align*} Since the third term of the geometric sequence is equal to the sixteenth term of the geometric sequence, $$a_1r^2 = a_1 + 15d$$ Since the second term of the geometric sequence is equal to the eleventh term of the arithmetic sequence, equality is preserved if we subtract the second term of the geometric sequence from the left-hand side and the eleventh term of the arithmetic sequence from the right-hand side, which yields \begin{align*} a_1r^2 - a_1r & = a_1 + 15d - (a_1 + 10d)\\ a_1(r^2 - r) & = 5d\\ a_1r(r - 1) & = 5d \tag{2} \end{align*} Since the terms of the geometric sequence are all different, $a_1 \neq 0$ and $r \neq 1$. Moreover, the terms of the arithmetic sequence must be distinct, so $d \neq 0$. Thus, if we divide equation 2 by equation 1, we obtain $$r = \frac{1}{2}$$ Since the sum of the geometric series is $18$ and $r \neq 1$, $$a_1 \frac{1}{1 - r} = 18$$ Solve for $a_1$, then use the equation $a_1(r - 1) = 10d$ to solve for $d$.

Answer 5

sup your logic is correct, the next step to solve the system of equations

$\begin{cases}ur=u+10d \space (1)\tag{1'}\label{1'} \\ ur^2=u+15d \space(2) \\ \frac{u}{1-r}=18 \space(3) \end{cases}$

And the system has solutions because it has 3 unknown variables ($u,r,d$) and the number of equations is also 3.

from (3) = > $u=18(1-r) (4)$

substruct (1) from (2)

$ur^2-ur=5d$ => $d = \frac{1}{5}ur(r-1) (6)$

put (6) to (1)

$ur=u+10(\frac{1}{5}ur(r-1))$ => $ur=u+2ur(r-1)$ => $ur=u+2ur^2-2ur (7)$

the both parts of (7) can be divided by $u$

$r=1+2r^2-2r (7')$ => $2r^2-3r+1=0 (7'')$

solve quadratic equation (7'') for $r$ and note that $r$ can not be 1.

$r_{1,2} = \frac{3\pm\sqrt{9-2*4*1}}{4} = \frac{3\pm1}{4} (8)$

$r = \frac{1}{2} (9)$

from (3) $u=9 (10)$

from (6) $d = \frac{9*0.5*(0.5-1)}{5} = -\frac{9}{20} (1)$

Questions, comments, remarks?