Three numbers that are consecutive terms of geometric sequence summed up equal to $93$. $$b_k+b_{k+1}+b_{k+2}=93$$ Those same numbers are the first, the second and the seventh term of arithmetic sequence. $$a_1=b_k,a_2=b_{k+1},a_7=b_{k+2}$$ Can someone help me determine what those terms are?
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3 Answers
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Hint: We have $b_k = a$, $b_{k+1}=a+t$, $b_{k+2}=a+6t$ and so $$ 3a+7t = 93 $$ Since $b_k,b_{k+1},b_{k+2}$ are consecutive terms of a geometric sequence, we have $b_{k+1}^2 = b_{k}b_{k+2}$ and so $$ (a+t)^2 = a(a+6t) $$
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$\begingroup$ I give up! I really don't know how to continue. $\endgroup$– HanlonJun 2, 2018 at 12:19
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$\begingroup$ @Hanlon, the second equation gives $t=0$ or $t=4a$. $\endgroup$– lhfJun 2, 2018 at 12:26
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$\begingroup$ How? Can you explain how do you go from $b^2_{k+1}$ to $b_kb_{k+2}$? $\endgroup$– HanlonJun 2, 2018 at 12:33
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$\begingroup$ I understand that, but I still don't understand why is $b^2_{k+1}$ equal to $b_kb_{k+2}$. In other words, I don't understand what does that they are consecutive has to do with that inference. $\endgroup$– HanlonJun 2, 2018 at 12:42
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Alternatively, note that: $$\begin{align}\frac{\frac{\frac{a_1+a_7}{2}+a_2}{2}+a_1}{2}&=a_2 \Rightarrow\\ \frac{\frac{a_1+a_7+2a_2}{4}+a_1}{2}&=a_2 \Rightarrow \\ a_1+a_7+2a_2+4a_1&=8a_2 \Rightarrow \\ a_7-6a_2+5a_1&=0 \Rightarrow \\ b_kq^2-6b_kq+5b_k&=0 \Rightarrow \\ q^2-6q+5&=0 \Rightarrow \\ q_1=1, q_2=5.\end{align}$$ Hence: $$b_k(1+q+q^2)=93 \Rightarrow \\ b_k=\frac{93}{1+1+1^2}=\color{blue}{31}; b_{k+1}=31\cdot 1=\color{blue}{31}; b_{k+2}=31\cdot 1^2=\color{blue}{31};\\ b_k=\frac{93}{1+5+5^2}=\color{red}{3}; b_{k+1}=3\cdot 5=\color{red}{15}; b_{k+2}=3\cdot 5^2=\color{red}{75}.$$
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Let the three terms be $a,b, c$.
Hence we have
$$\begin{align}
a&=a&&=a\\
b&=ar&&=a+d\\
c&=ar^2&&=a+7d\\\\
b-a&=a(r-1)&&=d\tag{1}\\
c-a&=a(r^2-1)=a(r-1)(r+1)&&=6d\tag{2}\\
\text{Putting $(1)$ into $(2)$}:\qquad \\
&\qquad d(r+1)&&=6d\\
&\qquad d(r-5)&&=0\\
&\qquad d=0 \qquad \text{or}&& r=5
\end{align}$$
If $\boxed{r=5}$, then
$$ a(1+5+5^2)=93\\
\color{}{\qquad \lbrace a,b,c\rbrace}=\color{red}{\lbrace 3,15, 75 \rbrace}
$$
If $\boxed{d=0}$, then all three terms are equal, i.e. $$\lbrace a,b,c\rbrace=\color{red}{\lbrace31,31,31\rbrace}$$
answered Jun 2, 2018 at 13:19
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$\begingroup$ $\langle3,15,75\rangle$ is only one solution. The second one is $\langle31,31,31\rangle$. $\endgroup$– HanlonJun 2, 2018 at 13:34
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$\begingroup$ When you canceled $r-1$ after Dividing gives: you lost the solution $r=1$ $\endgroup$ Jun 2, 2018 at 13:36
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$\begingroup$ I have the solution here with me and both $\langle3,15,75\rangle$ and $\langle31,31,31\rangle$ are listed. $\endgroup$– HanlonJun 2, 2018 at 14:21
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$\begingroup$ Solution edited to include the obvious trivial case with all equal terms. $\endgroup$ Jun 3, 2018 at 5:15