The $24^{t h}$ term of an A.P. is twice its $10^{t h}$ term. Show that its $72^{n d}$ term is $4$ times its $15^{t h}$ term.

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Solution

Let a $=$ first term of A.P.,
$d =$ common difference
$n^{t h}$ term, $a_{24} = 2 a_{10}$
$\Rightarrow a + 23 d = 2 (a + 9 d)$
$\Rightarrow a + 23 d = 2 a + 18 d$
$\Rightarrow a = 5 d$
Now, $\frac{a_{72}}{a_{15}} = \frac{a + 71 d}{a + 14 d} = \frac{76 d}{19 d} = 4$ .

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