Question

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

Answer 1

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.

$a_n=a_1+(n-1)d$

$⟹a_n=11+(n-1)(-3)$

$⟹a_n=11-3n+3$

$⟹a_n=-14-3n...eq\left(1\right)$

we have to find is $a_n=-150$ is a term of this A.P

put $a_n=-150$ in eq(1)

$⟹-150=-14-3\times n$

$⟹3\times n=-150+14$

$⟹n=\frac{-136}{3}$

$⟹n=45.33$

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)

$hence-150isnotantermofthisA.P$