# If an arithmetic sequence is added to a geometric sequence, term by term, is the resulting sequence arithmetic or geometric?

##### 1 Answer
Aug 10, 2015

Typically neither. The exceptions are when the arithmetic sequence is constant $0$ and/or the geometric sequence has common ratio $1$ (i.e. it is a constant sequence).

#### Explanation:

Let the arithmetic sequence be defined by ${a}_{n} = a + k n$ for $n = 0 , 1 , 2 , \ldots$

Let the geometric sequence be defined by ${b}_{n} = b \cdot {r}^{n}$ for $n = 0 , 1 , 2 , \ldots$

Then the result of adding these sequences has general term:

${c}_{n} = a + k n + b \cdot {r}^{n}$

Arithmetic sequence ?

${c}_{2} - {c}_{1} = \left(a + 2 k + b {r}^{2}\right) - \left(a + k + b r\right) = k + b \left({r}^{2} - r\right)$

$= k + b r \left(r - 1\right)$

${c}_{1} - {c}_{0} = \left(a + k + b r\right) - \left(a + b\right) = k + b \left(r - 1\right)$

If this is an arithmetic sequence, then these two expressions must be equal, so:

$\textcolor{red}{\cancel{\textcolor{b l a c k}{k}}} + b r \left(r - 1\right) = \textcolor{red}{\cancel{\textcolor{b l a c k}{k}}} + b \left(r - 1\right)$

Hence $b = 0$ or $r = 1$ (or both).

That means that the geometric sequence ${b}_{0} , {b}_{1} , {b}_{2} , \ldots$ must be constant.

Geometric sequence ?

If $b \ne 0$ and $\left\mid r \right\mid > 1$ then ${c}_{n + 1} / {c}_{n} \to r$ as $n \to \infty$.

If ${c}_{0} , {c}_{1} , {c}_{2} , \ldots$ is a geometric sequence then the ratio between successive terms must be constant, so we require ${c}_{1} / {c}_{0} = r$

$r = {c}_{1} / {c}_{0} = \frac{a + k + b r}{a + b}$

Multiply both ends by $\left(a + b\right)$ to find:

$a + k + \textcolor{red}{\cancel{\textcolor{b l a c k}{b r}}} = a r + \textcolor{red}{\cancel{\textcolor{b l a c k}{b r}}}$

Also

$r = {c}_{2} / {c}_{1} = \frac{a + 2 k + b {r}^{2}}{a + k + b r}$

Multiply both ends by $\left(a + k + b r\right)$ to get:

$a r + k r + \textcolor{red}{\cancel{\textcolor{b l a c k}{b {r}^{2}}}} = a + 2 k + \textcolor{red}{\cancel{\textcolor{b l a c k}{b {r}^{2}}}}$

Hence we have $a + k = a r$ and $a + 2 k = a {r}^{2}$

So:

${a}^{2} + 2 a k = a \left(a + 2 k\right) = {a}^{2} {r}^{2} = {\left(a r\right)}^{2} = {\left(a + k\right)}^{2}$

$= {a}^{2} + 2 a k + {k}^{2}$

So $k = 0$

Then $a r = a + k = a$, so $a = 0$ or $r = 1$

Well we assumed $\left\mid r \right\mid > 1$ so $r \ne 1$, so $a = 0$

If $a = 0$ and $k = 0$ then the arithmetic sequence is constant $0$.

Now consider the case $\left\mid r \right\mid \le 1$.

If the arithmetic sequence is non-constant, then ${c}_{n + 1} / {c}_{n} \to 1$ as $n \to \infty$. Hence we require the common ratio of the geometric sequence to be $1$. Then it's a constant sequence, therefore arithmetic and we require $b = 0$ or $r = 1$ (or both). As a result we also require the arithmetic sequence to be constant, contradicting our supposition.