The 24th term of an AP is twice its 10th term. Show that its 72nd term is 4 times its 15th term.

Let a be the first term and d be the common difference of the AP. Then,

$a_{24}=2 a_{10}$              (Given)

$\Rightarrow a+23 d=2(a+9 d) \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow a+23 d=2 a+18 d$

$\Rightarrow 2 a-a=23 d-18 d$

$\Rightarrow a=5 d$           $\ldots \ldots(1)$

Now,

$\frac{a_{72}}{a_{15}}=\frac{a+71 d}{a+14 d}$

$\Rightarrow \frac{a_{72}}{a_{15}}=\frac{5 d+71 d}{5 d+14 d} \quad[$ From $(1)]$

$\Rightarrow \frac{a_{72}}{a_{15}}=\frac{76 d}{19 d}=4$

$\Rightarrow a_{72}=4 \times a_{15}$

Hence, the 72nd term of the AP is 4 times its 15th term.