If a₁, a₂, a₃, ……., a_n is a G.P. having common ratio r and k is a natural number such that 3 < k < n, then r is equal to :

1. $\dfrac{a$k}{a{1}}

2. $\dfrac{a$1}{a2}

3. $\dfrac{a$k}{a{n-3}}

4. $\dfrac{a${k-1}}{a{k-2}}

We know that,

T_n = ar^{n - 1},

In the G.P.,

a₁, a₂, a₃, ……., a_n

a₁ is the first term and r is the common ratio.

$\Rightarrow \dfrac{a${k - 1}}{a{k - 2}} \\[1em] = \dfrac{a1r^{(k - 1) - 1}}{a1r^{(k - 2) - 1}} \\[1em] = \dfrac{a1r^{k - 2}}{a1r^{k - 3}} \\[1em] = r^{k - 2 - (k - 3)} \\[1em] = r^{k - 2 - k + 3} \\[1em] = r^{k - k + 3 - 2} \\[1em] = r.

Hence, option 4 is the correct option.