The 19th term of an ap is equal to 3 times its6th term.if its 9th is 19 .find the ap
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let in anAP first term =a, Common difference =d
9th term =19
a+8d=19----(1)
19th term =3*(6th term)
a+18d=3(a+5d)
a+18d=3a+15d
0=3a+15d-a-18d
2a-3d=0---(2)
subtract (2) from 2*(1)
-19d=-38
d=2
substitute d=2 in (1)
a=3
therefore
a=3,d=2
required AP is
3,5,7,9,....
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