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

a - first term
d - common difference

$T 24 = 2 (T 10)$
$a + 23 d = 2 a + 18 d$
$a = 5 d$

Here remember that this is relation is followed throughout the AP.
So, $T 72 = 4 (T 15)$
$a + 71 d = 4 a + 56 d$
$3 a = 15 d$
$a = 5 d$

Hence proved

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