Question

By using AP, how to solve?

Answer 1

its done ! ! ! ! ! ! ! ! ! ! ! ! ! ! !

Answer 2

$the \: 24th \: term \:is \: twice \: its \: 10th \: term.$
$a24 = {a} + (24 - 1)d \\ \: \: \: \: \: \: \: \: = \:a + 23d \\ \\ a10 = a + (10 - 1)d \\ \: \: \: \: \: \: \: \: = a \: + 9d$
Therefore,
$a + 23d = 2(a + 9d) \: = \: 2a + 18d \\ 23d - 18d = 2a - a = a \\ a = (23 - 18)d \: = \: 5d \\ a = 5d$
So,

To prove:
$72nd \: term \: is \: 4 \: times \: its \: 15th \: term$
$a72 = a + (72 - 1)d \\ \: \: \: \: \: \: \: \: = a + 71d \\ \: \: \: \: \: a = 5d. \: therefore \: \: a72 = \: 76d \: \: \\ \\ a15 = a + (15 - 1)d \\ \: \: \: \: \: \: \: \: = a + 14d \\ \: \: \: \: \: \: \: \: = 19d \: \: (as \: \: \: a = 5d)$

We know that 76 is 4 times 19.
Hence proved