Question

# Is $$-150$$ a term of the $$A.P$$. $$11,8,5,2,.....$$?

Solution

## Ans.given A.P is$$11,8,5,2,.....$$first term of this A.P is  $$a_{1}=11$$second term of this A.P is  $$a_{2}=8$$ common difference of an A.P is given by$$(i.e., d=a_{n+1}-a_{n})$$putting n=1 in above equation$$d=a_{2}-a_{1}=8-11=-3$$hence common difference for this A.P is $$d=-3$$the nth term of an A.P is given by.$$a_{n}=a_{1}+(n-1)d$$$$\implies a_{n}=11+(n-1)(-3)$$$$\implies a_{n}=11-3n+3$$$$\implies a_{n}=-14-3n...eq(1)$$we have to find is  $$a_{n}=-150$$ is a term of this A.Pput $$a_{n}=-150$$ in eq(1)$$\implies -150=-14-3\times n$$$$\implies 3\times n=-150+14$$$$\implies n=\frac{-136}{3}$$$$\implies n=45.33$$but we know that n can't be an fractional number, it must be an integer(i.e., we can't have 45.33th term in our A.P)$$hence\ -150 \ is \ not \ an \ term \ of \ this \ A.P$$Maths

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