Question
Check whether $–150$ is a term of the $AP: 11, 8, 5, 2, …..$
✓
Answer
The given list of numbers is $11, 8, 5, 2,.....$
$a_2- a_1= 8 - 11 = -3$
$a_3 - a_2 = 5 - 8 = -3$
$a_4 - a_3 = 2 - 5 = - 3$
i.e. $a_{k+1} - a_k$_ is the same every time.
So, the given list of numbers forms an $AP$ with first term $a = 11$ and the common difference $d = -3.$
Let $-150$ be the nth term of the given $AP$
Then, $a_n = -150$
$ \Rightarrow a + (n - 1) d = -150$
$ \Rightarrow 11+ (n - 1)(-3) = -150$
$ \Rightarrow (-3) (n - 1) = -150 - 11$
$ \Rightarrow (-3) (n - 1) = -161$
$ \Rightarrow 3(n - 1) = 161$
$ \Rightarrow n - 1 = \frac{{161}}{3}$
$ \Rightarrow n = \frac{{161}}{3} + 1$
$ \Rightarrow n = \frac{{164}}{3}$
But n should be a positive integer. So, -150 is not a term of $11, 8, 5, 2,....$
$a_2- a_1= 8 - 11 = -3$
$a_3 - a_2 = 5 - 8 = -3$
$a_4 - a_3 = 2 - 5 = - 3$
i.e. $a_{k+1} - a_k$_ is the same every time.
So, the given list of numbers forms an $AP$ with first term $a = 11$ and the common difference $d = -3.$
Let $-150$ be the nth term of the given $AP$
Then, $a_n = -150$
$ \Rightarrow a + (n - 1) d = -150$
$ \Rightarrow 11+ (n - 1)(-3) = -150$
$ \Rightarrow (-3) (n - 1) = -150 - 11$
$ \Rightarrow (-3) (n - 1) = -161$
$ \Rightarrow 3(n - 1) = 161$
$ \Rightarrow n - 1 = \frac{{161}}{3}$
$ \Rightarrow n = \frac{{161}}{3} + 1$
$ \Rightarrow n = \frac{{164}}{3}$
But n should be a positive integer. So, -150 is not a term of $11, 8, 5, 2,....$
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