Question

What is a21 of the arithmetic sequence for which a7 = -19 and a10 =-28

Answer 1

$a_{n} = a_{1} + (n - 1)d \\ a_{7} = a_{1} + (7 - 1)d \\ a_{7} = a_{1} + 6d \\ a_{10} = a_{1} + 9d$

$- 19 = a_{1} + 6d \\ - 28 = a_{1} + 9d \\ \\ - 19 = a_{1} + 6d \\ 28 = - a_{1} - 9d \\\ - 3d = 9 \\ d = - 3$

$a_{1} = - 19 - 6d = - 19 - 6( - 3) = - 1$

$a_{21} = a_{1} + (21 - 1)d \\ a_{21} = - 1 + 20( - 3) = - 61$