Question
Find:
Which term in the $A.P. 121, 117, 113, ..... $ is its first negative term?
Which term in the $A.P. 121, 117, 113, ..... $ is its first negative term?
✓
Answer
In the given problem, we are given an A.P. and the value of one of its term. We need to find which term it is (n).
So here we will find the value of n using the formula, $a_n = a + (n - 1)d.$
Given,
$A.P. 121, 117, 113, .....$
Here,
First term $= 121$
Difference $= 117 - 121 = -4$
We have to find which term of A.P. is its first negative term is then,
$a_n < 0$
We know, $n^{th}$ term of A.P.
$a + (n - 1)d < 0$
$\Rightarrow 121 + (n - 1) -4 < 0$
$\Rightarrow 121 + (-4n + 4) < 0$
$\Rightarrow 121 - 4n + 4 < 0$
$\Rightarrow 125 - 4n < 0$
$\Rightarrow 4n > 125$
$\Rightarrow\ \text{n}>31\frac{1}{4}$
$\because\ \text{n}>31\frac{1}{4}$
⇒ n = 32
Hence, $32^{th}$ term of the given A.P. for getting first negative term.
So here we will find the value of n using the formula, $a_n = a + (n - 1)d.$
Given,
$A.P. 121, 117, 113, .....$
Here,
First term $= 121$
Difference $= 117 - 121 = -4$
We have to find which term of A.P. is its first negative term is then,
$a_n < 0$
We know, $n^{th}$ term of A.P.
$a + (n - 1)d < 0$
$\Rightarrow 121 + (n - 1) -4 < 0$
$\Rightarrow 121 + (-4n + 4) < 0$
$\Rightarrow 121 - 4n + 4 < 0$
$\Rightarrow 125 - 4n < 0$
$\Rightarrow 4n > 125$
$\Rightarrow\ \text{n}>31\frac{1}{4}$
$\because\ \text{n}>31\frac{1}{4}$
⇒ n = 32
Hence, $32^{th}$ term of the given A.P. for getting first negative term.
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