Question

If the 4th term of an A.P is zero show that the 8th is double the 6th term.

Answer 1

$t_{8} = 2 \times t_{6}$

Step-by-step explanation:

$Let, \: \: first \: \: term = a \: \: \: and \: \: \: difference = d \\ Given, \: \: \: t_{4} = 0 \\ = > \: a + (4 - 1)d = 0 \\ = > \: a + 3d = 0 \\ = > \: \boxed{ a = - 3d} \\ Now, \: \: \: t_{8} = a + 7d \\ = > t_{8} = - 3d + 7d \\ = > \: \boxed{ t_{8} = 4d} \\ and \: \: \: t_{6} = a + 5d \\ = > \: t_{6} = - 3d + 5d \\ = > \: \boxed{t_{6} = 2d} \\ Now, \: \: \: \frac{t_{8}}{t_{6}} = \frac{4d}{2d} \\ \: \: = > \frac{t_{8}}{t_{6}} = 2 \\ so, \: \: \boxed{t_{8} = 2 \times t_{6}}$