Question

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

• A. No
• B. Yes
• C. Data Insufficient
• D. Can't say

Answer 1

Answer: (A)

No

Clearly the given sequence is an AP with first term $a=11$ and the common difference $d=8-11=-3$.

Let $-150$ be the $n^{th}$ term of given AP. Then,
$a_n=-150\Rightarrow t_n=a+(n-1)d=-150$
$\Rightarrow 11+(n-1)\times -3=-150$
$\Rightarrow (n-1)\times -3=-161$
$\Rightarrow n-1=\frac{-161}{-3}$
$\Rightarrow n=\frac{161}{3}+1=\frac{164}{3}$
Clearly $n$ is not an natural number, Hence, $-150$ is not a term of given AP.