What can we say if $a, x_1, x_2, x_3, x_4, x_5, b$ are in arithmetic progression?

What can we say if $a, x_1, x_2, x_3, x_4, x_5, b$ are in arithmetic progression?

Question

1. The seven numbers $a, x_1, x_2, x_3, x_4, x_5, b$ are in arithmetic progression. Express $x_2$ in terms of $a$ and $b$, and show that $x_1+x_3+x_5=\dfrac{3}{2}(a+b)$. Given also that the numbers $a, x_2, b$ are in geometric progression, and that $b\neq a$, express $b$ in terms of $a$.

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1. A geometric series has common ratio $r,$ where $|r|<1.$ The sum of the first $n$ terms of the series is $S_n$ and the sum to infinity is $S$. Express $r$ in terms of $S_n, S$ and $n,$ and prove that the sum of the first $2n$ terms of the series is $\dfrac{S_n(2S-S_n)}{S}$.

Suggestion

Can we work out $d$, the common difference in the arithmetic progression?

Solution

Let’s call the common difference of the arithmetic progression $d$. This means that $x_1=a+d,\ x_2=a+2d,\ \dots,\ b=a+6d.$ Thus $b-a=6d$, and $d=\dfrac{b-a}{6}$.

We can now substitute this value for $d$ into the expression for $x_2$, yielding $x_2=a+2\Bigl(\frac{b-a}{6}\Bigr)=\frac{b+2a}{3}.$

We’ll now rewrite

$x_1+x_3+x_5$

in terms of

$a$

and

$d$

, and then substitute

$d=\dfrac{b-a}{6}$

, giving

$$\begin{align*} x_1+x_3+x_5&=(a+d)+(a+3d)+(a+5d)\\ &= 3a+9d\\ &= 3a+9\Bigl(\frac{b-a}{6}\Bigr)\\ &= \frac{3a+3b}{2}\\ &= \frac{3}{2}(a+b), \end{align*}$$

as required.

For the final part, since $a, x_2, b$ are in geometric progression, they have a common ratio $r$, so $x_2=ar\quad\text{and}\quad b=ar^2.$

We wish to eliminate $r$. We have $x_2^2=a^2r^2$, which immediately implies that $x_2^2=ab$.

We already know that

$x_2=\dfrac{b+2a}{3}$

, so putting the two different expressions for

$x_2$

together, we have

$$x_2^2=\left(\frac{b+2a}{3}\right)^2=ab.$$

Expanding the brackets and multiplying both sides by

$9$

gives

$$\begin{align*} &b^2+4ab+4a^2=9ab\\ \implies\quad&b^2-5ab+4a^2=0\\ \implies\quad&(b-a)(b-4a)=0, \end{align*}$$

so either $b=a$ or $b=4a$. Since $b\neq a$, we must have $b=4a$.

1. A geometric series has common ratio $r$, where $|r|<1$. The sum of the first $n$ terms of the series is $S_n$ and the sum to infinity is $S$. Express $r$ in terms of $S_n,$ $S$ and $n,$ and prove that the sum of the first $2n$ terms of the series is $\dfrac{S_n(2S-S_n)}{S}$

Let

$a$

be the first term of this geometric series with common ratio

$r$

. Then

$$S_n=\frac{a(1-r^n)}{1-r}$$

and

$$S=\frac{a}{1-r}.$$

To find an expression for

$r$

in terms of

$S_n$

,

$S$

and

$n$

, we must first eliminate

$a$

. We can rearrange the second expression to get

$$a=S(1-r),$$

which we can substitute into the first expression to get

$$S_n=\frac{S(1-r)(1-r^n)}{1-r}=S(1-r^n).$$

We want an expression for

$r$

, so we rearrange to get

$$\begin{align*} 1-r^n =& \frac{S_n}{S}\\ \implies r^n =& 1-\frac{S_n}{S}\\ \implies r =& \left(1-\frac{S_n}{S}\right)^{\frac{1}{n}}. \end{align*}$$

To find the sum of the first

$2n$

terms, we substitute

$2n$

into the formula for the sum of the first

$n$

terms, giving

$$S_{2n}=\frac{a(1-r^{2n})}{1-r}.$$

We now rewrite the equation using our expressions for

$a$

and

$r^n$

, giving

$$\begin{align*} S_{2n}&=S(1-r^{2n})\\ &=S\bigl(1-(r^n)^2\bigr)\\ &=S\left(1-\left(1-\frac{S_n}{S}\right)^2\right)\\ &=S\left(1-\left(1-\frac{2S_n}{S}+\frac{S_n^2}{S^2}\right)\right)\\ &=S\left(\frac{2S_n}{S}-\frac{S_n^2}{S^2}\right)\\ &=2S_n-\frac{S_n^2}{S}\\ &= \frac{S_n(2S-S_n)}{S}. \end{align*}$$