Question

If the 3rd and loth terms of an ap
are 4 and -8 respectively, which term of
this AP is zero?​

Answer 1

Step-by-step explanation:

Pls Mark the answer as brainest

Answer 2

Answer:

hope it will helo

Step-by-step explanation:

$a3 = 4 \\ a9 = - 8 \\ a3 = 4 \\ a + (n - 1)d = 4 \\ a + (n - 1)d = 4 \\ a + 2d = 4 \: equation \: 1 \\ a9 = - 8$

$a9 = - 8 \\ a + (n - 1) = - 8 \\ a + (9 - 1)d = - 8 \\ a + 8d = - 8 \: \: equation \: 2 \\ substracting \: 1 \: from \: 2 \\ a + 8d = - 8 \\ - a + 2d = 4$

$a + 8d = - 8 \\ a + 2d = 4 \\ plus \: becomes \: minus \\ 6d = - 12 \\ d = \frac{ - 12}{6} \\ d = - 2 \\ putting \: d = 2 \: in \: equation \: 1$

$putting \: d = - 2 \: in \: equation \: \: 1 \\ a + 2d = 4 \\ a + 2 \times \: ( - 2) = 4 \\ a - 4 = 4 \\ a = 4 + 4 \\ 8$

$a \: n = 0 \\ a + (n - 1)d = 0 \\ 8 + (n - 1)( - 2) = 0 \\ (n - 1)( - 2) = 0 - 8 \\ (n - 1)( - 2) = - 8$

$(n - 1) = \frac{ - 8}{ - 2} = 4 \\ n = 4 + 1 \\ n = 5$