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 \(b=4a\).

- 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 \(\dfrac{S_n(2S-S_n)}{S}\)