A) \[f(n)=an+b;\,n\in N\]
B) \[f(n)=k{{r}^{n}};\,n\in N\]
C) \[f(n)=(an+b)\,k{{r}^{n}};\,n\in N\]
D) \[f(n)=\frac{1}{a\left( n+\frac{b}{n} \right)};\,n\in N\]
Correct Answer: A
Solution :
Sequence \[f(n)=an+b;\ n\in N\] is an A.P. Putting \[n=1,\ 2,\ 3,\ 4,\ ..........,\] we get the sequence \[(a+b),\ (2a+b),\ (3a+b),.........\] which is an A.P. Where first term \[(A)=(a+b)\] and common difference \[d=a\]. Aliter: As we have mentioned in theory part that \[{{n}^{th}}\] term of an A.P. is of the form\[an+b,\ \ n\in N\].
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