Question

the third term of an ap is 128 , seventh term is 24 find the tenth term

Answer 1

Step-by-step explanation:

Given:

$\because \: t_3 = 128 \\ \therefore \: a + (3 - 1)d = 128 \\ \implies \: a + 2d = 128.....(1) \\ \because \: t_7 = 24 \\ \therefore \: a + (7 - 1)d = 24 \\ \implies \: a + 6d = 24.....(2) \\ equation \: (2) - equation \: (1) \\ a + 6d - (a + 2d) = 24 - 128 \\ a + 6d - a - 2d= 24 - 128 \\ 4d= - 104 \\ d = \frac{ - 104}{4} \\ d = - 26 \\ \implies a + 2 \times( - 26) = 128 \\ \implies a - 52 = 128 \\ \implies a = 128 + 52\\ \implies a = 180 \\ \therefore \:t_{10} = a + 9d \\ \therefore \: t_{10} = 180 + 9( - 26) \\ \therefore \: t_{10} = 180 - 234 \\ \therefore \: t_{10} = - 54 \\ thus \: the \: tenth \: term \: of \: the \: ap \: is \: \\ - 54.$