Solve this

To Find: $-47$ is a term of the AP or not.

Given: The series is $5,2,-1,-4,-7, \ldots .$

$a_{1}=5, a_{2}=2$, and $d=2-5=-3($ Let suppose an $=-47)$

NOTE: n is a natural number.

(Where $a=a_{1}$ is first term, $a_{2}$ is second term, $a_{n}$ is nth term and $d$ is common difference of given AP)

Formula Used: $a_{n}=a+(n-1) d$

$a_{n}=-47=a+(n-1) d$

$-47=5+(n-1)(-3)$

$-47-5=(n-1)(-3)$ [subtract 5 on both sides]

$52=(n-1)(3)[$ Divide both side by '-' $]$

$17.33=(n-1)$ [Divide both side by 3 ]

$18.33=\mathrm{n}[$ add 1 on both sides $]$

As n is not come out to be a natural number, So –47 is not the term of this AP.