Write the first ten terms of the arithmetic sequence given the first term and some other information

Question

The first term of the arithmetic sequence is 24. So, $$a_1=24$$ The first, fifth and the eleventh term are the consecutive terms of the geometric sequence. So, $$...,a_1,a_5,a_{11},...=...,b_1r^x,b_1r^{x+1},b_1r^{x+2},...$$

Write the first ten terms of the arithmetic sequence.

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The solution is

$$24,27,30,33,36,39,42,45,48,51$$ or $$24,24,24,24,24,24,24,24,24,24$$

Please explain to me how do we arrive at that solution.

Answer 1

Let $a$ be the arithmetic sequence step and let $b$ be the geometric sequence step. Also let $r_i$ be the arithmetic sequence's elements.

For the fifth element:

$r_5=r_1+4a$

$r_5=r_1 b$

For the eleventh element:

$r_{11}=r_1+10a$

$r_{11}=r_1 b^2$

Hence:

$r_{1}(b-1)=4a$

$r_{1}(b^2-1)=10a$

In case $b \neq 1$ then

$b=1.5$ and $a=3$

In case $b=1$ then

$b=1$ and $a=0$

Answer 2

Hint. $a_{5} =a_{1} +4d$ and $a_{11} =a_{1} +10d$ for some $d$. Also, $a_{5}= r a_{1}$ and $a_{11} =r^{2} a_{1}$ for some $r$. Try to come up with a $2 \times 2$ nonlinear system for $d$ and $r$.

Answer 3

Assume that the GP has a common ratio of $r$.

For the case where $\boxed{r\neq 1}$, the three terms are $24, 24r, 24r^2$.

As these are the first, fifth and eleventh terms of an AP, $$\frac {24r^2-24r}{24r-24}=\frac {11-5}{5-1}\\ r=\frac 32 \qquad (r\neq 1)$$ i.e. the GP is $24, 36, 54$.

It can then be easily worked out that the common difference of the AP as $3$.
Hence the AP is $\color{red}{24, 27, 30, \cdots }$.

For the trivial case where $\boxed{r=1}$, all terms of the GP and AP are equal to $24$.