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Question: Find \({1^{st}}\) negative term of \[231, 228, 225, \ldots \]...

Find 1st{1^{st}} negative term of 231,228,225,231, 228, 225, \ldots

Explanation

Solution

Here, we will verify if the given sequence is an arithmetic progression or not by subtracting each term and its preceding term. If the common difference turns out to be the same, thus, it is an AP. Then, we will use the general formula of an AP and find the terms which is 0. This term comes just before the first negative term. Hence, next term will give us the required 1st{1^{st}} negative term of 231,228,225,231,228,225, \ldots

Formula Used:
General term of an AP is: a+(n1)da + \left( {n - 1} \right)d, where aa is the first term, dd is the common difference and nn is the number of the term.

Complete step by step solution:
Given sequence is: 231,228,225,231,228,225, \ldots
First, we will find whether this given sequence is an Arithmetic Progression or not.
We will subtract each term and its preceding term, therefore, we get
228231=3228 - 231 = - 3
225228=3225 - 228 = - 3
We can see that the difference between each term and its preceding term is same, thus this sequence forms an Arithmetic Progression (AP)
The first term of this AP, a=231a = 231
The common difference, d=3d = - 3
Now, in order to find the 1st{1^{st}} negative term, first, we will find the term which gives us zero.
Hence, let the nth{n^{th}} term of this AP be 0.
Hence, since, the nth{n^{th}} term of this AP is 0. Thus, equating it with the general term a+(n1)da + \left( {n - 1} \right)d, we get,
a+(n1)d=0a + \left( {n - 1} \right)d = 0
Now, substituting the known values of first term and common difference in the above equation, we get
231+(n1)(3)=0231 + \left( {n - 1} \right)\left( { - 3} \right) = 0
Multiplying the terms, we get
2313n+3=0\Rightarrow 231 - 3n + 3 = 0
Adding the like terms, we get
3n=234\Rightarrow 3n = 234
Dividing both sides by 3, we get
n=78\Rightarrow n = 78
Therefore, 78th{78^{th}} term of this AP is 0.
Thus, since, the terms of this AP are decreasing, thus, the 79th{79^{th}} term of this AP will definitely be a negative number.
Using the general form, 79th{79^{th}} term of an AP can be written as a+(791)da + \left( {79 - 1} \right)d.
Now, substituting the given values, we get,
79th{79^{th}}term=231+78(3) = 231 + 78\left( { - 3} \right)
Multiplying the terms, we get
79th{79^{th}}term =231234=3 = 231 - 234 = - 3

Therefore, the required 1st{1^{st}} negative term of 231,228,225,231, 228, 225, \ldots is 3 - 3
Hence, 3 - 3 is the required answer.

Note:
An Arithmetic Progression is a sequence of numbers such that the difference between any term and its preceding term is constant. This difference is known as the common difference of an Arithmetic Progression (AP). A real-life example of AP is when we add a fixed amount in our money bank every week. Similarly, when we ride a taxi, we pay an amount for the initial kilometer and pay a fixed amount for all the further kilometers, this also turns out to be an AP.