Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P

Let’s consider a₁, a₂, … a_m be m numbers such that 1, a₁, a₂, … a_m, 31 is an A.P.

And here,

a = 1, b = 31, n = m + 2

So, 31 = 1 + (m + 2 – 1) (d)

30 = (m + 1) d

d = \( \dfrac{30}{(m + 1)}\) ……. (1)

Now,

a₁ = a + d

a₂ = a + 2d

a₃ = a + 3d …

Hence, a₇ = a + 7d

a_m–1 = a + (m – 1) d

According to the question, we have

\( \dfrac{a+7d}{a+(m-1)d}=\dfrac{5}{9}\)

\( \dfrac{1+7\dfrac{30}{(m+1)}}{1+(m-1)\dfrac{30}{m+1}}=\dfrac{5}{9}\)

\( \dfrac{m+1+210}{m+1+30m-30}=\dfrac{5}{9}\)

\( \dfrac{m+211}{31m-29}=\dfrac{5}{9}\)

9m+1899=155m-145

155m - 9m=1899 +145

146m=2044

m = 14

Therefore, the value of m is 14.