Question

The ${24}^{th}$ term of an A.P. is twice its ${10}^{th}$ term. Show that its ${72}^{nd}$ term is $4$ times its ${15}^{th}$ term.

Answer 1

Let a$=$ first term of A.P.,

$d=$ common difference

$n^{th}$ term, $a_{24}=2a_{10}$

$\Rightarrow a+23d=2(a+9d)$

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

$\Rightarrow a=5d$

Now, $\frac{a_{72}}{a_{15}}=\frac{a+71d}{a+14d}=\frac{76d}{19d}=4$.