Find three numbers that can be consecutive terms of geometric sequence and first, second and seventh term of arithmetic sequence and whose sum is $93$

Question

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?

Answer 1

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 $$ (a+t)^2 = a(a+6t) $$

Answer 2

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. $$\lbrace a,b,c\rbrace=\color{red}{\lbrace31,31,31\rbrace}$$

Answer 3

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: $$b_k(1+q+q^2)=93 \Rightarrow \\ b_k=\frac{93}{1+1+1^2}=\color{blue}{31}; b_{k+1}=31\cdot 1=\color{blue}{31}; b_{k+2}=31\cdot 1^2=\color{blue}{31};\\ b_k=\frac{93}{1+5+5^2}=\color{red}{3}; b_{k+1}=3\cdot 5=\color{red}{15}; b_{k+2}=3\cdot 5^2=\color{red}{75}.$$