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

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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 \(b=4a\).

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