Question

is 9 times the 9th term of an ap is equals to 13 times 13th term then the 22nd term of the ap is

Answer 1

[tex]9t_{9} = 13 t_{13} \\ t_{n} = a + (n-1)d \\ 9(a+8d) = 13(a+12d) \\ 9a+72d = 13a + 156d \\ 4a+84d = 0 \\ 4(a+21d) = 0\\ a+21d = 0\\ t_{22} = a+21d = 0[/tex]

Answer 2

The 22nd term of the AP will be 0.