Answer :

Let a be the first term and d be the common difference.

Given: a24 = 2(a10)


To prove: a72 = 4 × a15


Now, Consider a24 = 2a10


a + 23d = 2[a + 9d]


a + 23d = 2a + 18d


a = 5d …………………. (1)


Consider a72 = a + (72 - 1)d


a72 = 5d + 71d (from equation (1))


a72 = 76d ………………. (2)


Now, consider a15 = a + (15 - 1)d


a15 =5d + 14d (from equation (1))


a18 = 19d ………………….(3)


From equation (2) and (3), we get,


a72 = 4 × a15


Hence, proved.


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