the 19th term of an A.P is equal to 3 times its 6th term. If the 9th term is 19, find the A.P

a₁₉=3(a₆)
a₉=19
we have to find a and d

a₁₉=3(a₆)
a+18d=3(a+5d)
a+18d=3a+15d
18d-15d=3a-a
​3d=2a
a=3d/2


substitute 'a' value in a₉=19
a+8d=19
3d/2+8d=19
taking LCM, we get,

3d+16d/2=19

3d+16d=19x2
19d=38
d=2

put d=2 in 'a' value

a=3d/2
  =3x2/2
  =6/2
  a =3


so the AP will be

a,a+d,a+2d,........
3,3+2,3+2(2), .........
3,5,7,........