Review question

# What can we say if $a, x_1, x_2, x_3, x_4, x_5, b$ are in arithmetic progression? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6405

## Solution

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 $a$.

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*}