Question

For the following A.P., find the first value of $n$ and $t_n$ for which $t_n$ is negative.
$122,116,110,...$
(Note : find smallest n such that $t_n<0$)

• A. $n=22,-4$
• B. $n=20,-3$
• C. $n=24,-10$
• D. $n=25,-25$

Answer 1

The correct option is A $n=22,-4$

The given sequence is an AP in which first term $a=122$ and common difference $d=-6$. We have to find value of n for which $t_n$ is first negative number. Then,
$t_n<0$
$\Rightarrow a+(n-1)d<0$
$\Rightarrow 122+(n-1)\times -6<0$
$\Rightarrow 128-6n<0$
$\Rightarrow 6n>128\Rightarrow n>21\frac{2}{6}$
Since, 22 is the natural number just greater than $21\frac{2}{6}$. So, $n=22$

Thus, $t_{22}$ term of the given sequence is first negative term and given by

$t_{22}$ = a+(n-1)d = 122+(22-1)(-6) = -4