If 100 times the 100^{{th}{~}} term of an A.P. with non-zero common difference equals the 50 times its 50^{{th}{~}} term, then the 150^{{th}{~}} term of this A.P. is: Sequence & Series (Mathematics)

\[\n\n\text{Given: } 100 \times T_{100} = 50 \times T_{50}\n\n\]\n \nLet

a

be the first term and

d \, (d \neq 0)

be the common difference of the AP.\n \n\[\n\nT_{50} = a + (50 - 1)d = a + 49d\n\n\]\n \n\[\n\nT_{100} = a + (100 - 1)d = a + 99d\n\n\]\n \n\[\n\nT_{150} = a + (150 - 1)d = a + 149d\n\n\]\n \nAccording to the given condition,\n \n\[\n\n100T_{100} = 50T_{50}\n\n\]\n \n\[\n\n100(a + 99d) = 50(a + 49d)\n\n\]\n \n\[\n\n2(a + 99d) = (a + 49d)\n\n\]\n \n\[\n\n2a + 198d = a + 49d\n\n\]\n \n\[\n\na + 149d = 0\n\n\]\n \n\[\n\n\therefore \ T_{150} = a + 149d = 0\n\n\]\n\n

Explanation

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