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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?

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  • $\begingroup$ Please type your questions rather than posting an image. Images cannot be searched. $\endgroup$ – N. F. Taussig Sep 10 '18 at 13:42
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    $\begingroup$ @N.F.Taussig Okay, I've done that. $\endgroup$ – user590211 Sep 10 '18 at 13:44
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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 $d$.

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

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

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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 $d=-\dfrac9{20}$.

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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?

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  • $\begingroup$ If you number equations with the \tag command, the equation number will be separated from the equation. See the Tags & References section of this MathJax tutorial. $\endgroup$ – N. F. Taussig Sep 10 '18 at 14:45
  • $\begingroup$ @N-F-Taussig Thank you for noticing this. $\endgroup$ – yW0K5o Sep 10 '18 at 15:01

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