If 19th term is equal to …

Let the first term of the A.P. be 'a'.
and the common difference be 'd'.
19^th term of the A.P., t₁₉ = a + (19 - 1)d = a + 18d
6^th term of the A.P., t₆ = a + (6 - 1)d = a + 5d
9^th term of the A.P., t₉ = a + (9 - 1)d = a + 8d
t₁₉ = 3t₆
{tex}\Rightarrow{/tex}a + 18d = 3(a + 5d)
{tex}\Rightarrow{/tex}a + 18d = 3a + 15d
{tex}\Rightarrow{/tex}18d - 15d = 3a - a
{tex}\Rightarrow{/tex}3d = 2a
{tex}\therefore \mathrm { a } = \frac { 3 \mathrm { d } } { 2 }{/tex}
t₉ = 19
{tex}\Rightarrow \frac { 3 \mathrm { d } } { 2 } + 8 \mathrm { d } = 19{/tex}
{tex}\Rightarrow \frac { 3 d + 16 d } { 2 } = 19{/tex}
{tex}\Rightarrow \frac { 19 \mathrm { d } } { 2 } = 19{/tex}
{tex}\Rightarrow{/tex} d = 2
{tex}\Rightarrow{/tex}a = 3
t₂ = 3 + (2 - 1)2 = 5
t₃ = 3 + (3 - 1)2 = 7
The series will be 3, 5, 7......