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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.P
put $$a_{n}=-150$$ 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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