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,........
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,........