The coefficient of ${x^2}$ in the expansion of ${\left( {\frac{1}{x} + 5x} \right)^8}$ is equal to the coefficient of ${x^4}$ in the expansion of ${\left( {a + 5x} \right)^7},{\text{ }}a \in \mathbb{R}$. Find the value of $a$.

METHOD 1

$^8{C_r}{\left( {\frac{1}{x}} \right)^{8 - r}}{(5x)^r} = {}^8Cr{\left( 5 \right)^r}{x^{2r - 8}}$   (M1)

$r = 5$     (A1)

$^8{C_5} \times {5^5}{ = ^7}{C_4}{a^3} \times {5^4}$     M1A1

$56 \times 5 = 35{a^3}$

${a^3} = 8$     (A1)

$a = 2$     A1

METHOD 2

attempt to expand both binomials     M1

$175000{x^2}$     A1

$21875{a^3}{x^4}$     A1

$175000 = 21875{a^3}$     M1

${a^3} = 8$     (A1)

$a = 2$     A1

[6 marks]