Question

Which term of the $A.P$ $121,117,113,..$ is its first negative term

• A. $32nd$
• B. $30th$
• C. $40th$
• D. $42nd$

Answer 1

The correct option is A $32nd$

The given sequence is an AP in which first term $a=121$ and common difference $d=-4$. We have to find value of n for which $t_n$ is first negative number. Then,

$t_n<0$

$\Rightarrow a+(n-1)d<0$

$\Rightarrow 121+(n-1)\times -4<0$

$\Rightarrow 125-4n<0$

$\Rightarrow 4n>125\Rightarrow n>31\frac{1}{4}$

Since, $32$ is the natural number just greater than $31\frac{1}{4}$. So,

$n=32$.

Thus, ${32}^{nd}$ term of the given sequence is the first negative term.