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15find the 15th term of the Arithmetic progression 3, 6, 9.... using suitableformula

Answer»

For AP 3, 6, 9...a1 = 3, d = 6 - 3 = 3

nth term of AP an = a1 + (n-1)d

Therefore, 15th term of APa15 = a1 + 14d = 3 + 14*3 = 3 + 42 = 45

a=3,d=a2-a1=6-3=3n=15an=a+(n-1)dan=3+(15-1)3an=3+(14)3an=3+42an=45


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